BG Fernandes Department of Electrical Engineering II T - Power ...

EE 660 

Application of Power Electronics 

in 

Power Systems 

B. G. Fernandes 

Department of Electrical Engineering 

I. I. T Bombay 

bgf@ee.iitb.ac.in 

Application of Power Electronics in Power Systems 


1/454

• Introduction 

• Load Compensation 

Course Outline 

• Shunt Compensation 

• Series Compensation 

• HVDC Transmission 

Theory 

Equipment 

Theory 

Equipment 



2/454

Books for Reference 

• T. J. E. Miller “Reactive power control in Electrical 

system,” John Wiley & Sons, New York, 1982. 

• K. R. Padiyar “FACTS CONTROLLERS in Power 

Transmission & Distribution,” New Age International 

(P) Ltd.,” 2007. 

• K. R. Padiyar “HVDC POWER TRANSMISSION 

SYSTEMS Technology and System Interactions,” New 

Age International (P) Ltd.,” 1990. 

• Hingorani N. G “Understanding FACTS Concepts & 

Technology of FACTS Systems,” IEEE PRESS, 2000. 



3/454

Introduction 

“Power Electronics has grown as a major & 

extremely important discipline in Electrical 

Engg.” 

• What are major applications of Power 

Electronics ? 

• Major role in Power Transmission & 

Distribution 



4/454

• Consumption of Electricity are Demanding 

Customers 

• Loss of Power for single cycle can make 

computer screen go blank 

• Can interrupt sensitive Electronic equipment 



5/454

• Consumption of Electricity is also 

• Transmission lines are being operated close 

to their limits 

• Power is being transmitted through long 

overhead transmission lines & they are 

interconnected 



6/454

• Thermal limit (depends on ambient 

conditions) 

• Voltage limit 

P 

THERMAL LIMIT 

• Stability limit 

Voltage and Stability 

Constraints 

SIL 

Distance 



7/454

Type of conductors 

• Thermal limit No. of Conductors 

Ambient conditions 

• Voltage limitations 

• For typical 400 kV line Z c = 300 Ω 

SIL = 540 MW 

• For cable SIL is large 



8/454

• Voltage profile along the line is flat 

if P = SIL 

• If V S = V R = 1, V ↓ as we move towards 

the midpoint, if Ps > SIL 

P < SIL 

P = SIL 

V S 

P > SIL 

V R 



9/454

• Line absorbs reactive power 

• V ↑ if P S < SIL 

• Voltage swell, line generates ‘Q’ 

P, Q 

P, Q 

V s 

i s i R 

Transmission Line 

V R 



10/454

• To control V R & ↑ power transfer capacity 

of the line, ‘Q’ generation is required at the 

receiving end 



11/454

Q 

V 

2 

= ↓ As V R ↓ 

X C 

‘Q’ requirement ↑ as V R ↓ 

• Other limitations 

• ‘L’ required during over voltage 

• Separate ‘L’ & ‘C ’ are required 



12/454

• High ‘V’ & high KVar source 

• 3-ph inverter can supply 

± 

Q 

• Requires only ΔP 



13/454

O/P V => PWM 

• 2- level inverter 

• Harmonic spectrum depends on switching 

frequency (F S ) 

• PWM 

Constant F S 

Variable F S => Not suitable 



14/454

• What sort of PWM technique to use ? 

• With low switching frequency how to 

improve the harmonic spectrum 

• Do we need to change the power circuit 

configuration ? 



15/454

Ρ = 

V 

S 

V 

X 

R 

Sinδ 

• To have sufficient stability margin max. 

length of line = 450 km 

• Provide shunt reactive power 

compensation, there by P↑ & maintain 

V profile. 

• Use a mid point compensator 



16/454

V = V = V = 

m S R 

V 

It can be shown, for loss- less line 

2V 

2 ⎛ δ ⎞ 

P = Sin⎜ 

⎟ = 2 

X ⎝ 2 ⎠ 

P Uncompensated 



17/454

• “If shunt compensation is applied at 

sufficient close interval, it may be possible to 

transmit power up to thermal limit of line” 

• P transmitted over long lines is limited by 

series reactance ‘X’ 



18/454

Provide 

• Series capacitive compensation to cancel a 

portion of series ‘X’ 

δ 



19/454

V = V = 1pu 

S R 

P 

2 

V 

= 1 

( − K ) 

X 

Sinδ 

K = Degree of compensation = X 

X C 

• C is not permanently connected in series 

• During fault condition, X eff should be 

increased 

• May require ‘L’ also 



20/454

• Is it possible to change the phase angle 

difference between two ends of the line 

and there by control the power flow 

• “Phase angle regulator” ? 

• Inject a voltage in series with the line & 

proportional to the current flow (voltage 

should lag the I ) 



21/454

δ 



22/454

• Injecting V in series with line and with 

any phase angle with respect to V S 

δ 

• Both magnitude & phase angle of I has 

changed 

• Both P & Q flow has changed 



23/454

• Consider an AC network 

• Power flow in Line-1 & 2 depends on circuit 

conditions 

• Lower X line may be over loaded 

• Not possible to set the amount of power that 

should flow through a particular line! 



24/454

• Definite amount of power that should flow 

through HVDC line can be set 

• If power transfer over long distances 

• Two near by areas having different 

frequencies ( Back to Back connection) 



25/454

Review 

• Power flow control through AC lines is not 

“FLEXIBLE” 

• Depending upon the loading, there could be 

voltage swell or sag as we go towards the 

mid point 

R+jX 

V 1 

V 2 



26/454

• To control the power flow & to maintain 

voltage profile, provide 

• Shunt compensation 

{ 

Passive elements with 

P.E switches or 

• Series compensation Inverter 

• At Tr. voltage levels PWM with high 

switching frequency may not be possible 

• Modify the existing power circuit 

• Can we regulate the power flow by converting 

AC-DC-AC => HVDC Transmission ? 



27/454

Introduction ( contd…) 

Load compensation 

• Loads are unbalanced 

• P.F is lagging 

No compensation 

of harmonics 

• Source should supply only active power & 

see a balanced load 



28/454

• Most of the loads are Non-linear 

• Harmonics are generated 

• Voltage at P.C.C is non sinusoidal 

• P.F is lagging 

• Circuit to filter the harmonics (on-line) + 

compensate the loads 



29/454

P.C.C 

→ 

Point of common coupling 



30/454

Current drawn by the load fed from P.E. equipment 

flows through system impedance. 

Voltage at P.C.C is non-sinusoidal 

(We had assumed that 'V' is sinusoidal). 



31/454

2 3 ⎡ 1 1 

⎤ 

i 

a= I0 

sinωt- sin5ωt+ sin7ωt-............. 

π ⎢ 

5 7 

⎥ 

⎣ 

⎦ 

= 6N ± 1 , Harmonics 

⇒ Line Commutated converter → causes notches 

in the source voltage waveform. 

→ Source current has harmonics. 



32/454

Effect of harmonics: 

A. In the Rotating machine → Increases heating. 

→ They produce noise. 

→ Torque pulsations. 

B. In Transformers → Cu losses ↑ . 

→ Audible noise & heating. 

C. In Cables → Additional heating. 

D. P.F correction capacitors. 

→ 

Thermal voltage stress. 

E. Electronic Equipments → Affects control system. 

→ Maloperation of relays. 



33/454

• Load compensation + Active filter 

• Depending upon the voltage & power level, 

circuit configuration & control should 

change 



34/454

Conclusions 

• Load compensator + Active filter to 

compensate non-linear loads 

• Power flow in AC network is determined by 

circuit conditions 

• Power transfer capability can be increased 

through shunt & series compensation 

• HVDC can be used for bulk power 

transmission & to inter connect the systems of 

different frequencies 



35/454


• In ideal power system 

• V & F should be constant 

• V should be sinusoidal 

• P.F = 1 

• The above should be independent of size & 

characteristics of load 

• No interference between different loads 



36/454

Notation of quality of supply 

• How nearly constant are V & F at the 

supply point ? 

• How near to unity is the P.F ? 

• In 3-ph system, degree to which V & I are 

balanced 



37/454

• What are the characteristics of power system 

& loads which can deteriorate the quality of 

supply ? 

• How to compensate ? 

Objectives of load compensation 

• Power factor correction 

• Improvement in voltage regulation 

• Load balancing 



38/454

Ideal compensator 

• Correct the power factor to unity 

• Reduce the voltage regulation to an 

acceptable value 

• Balance the load current => not expected to 

compensate harmonics in V & I, also will 

not generate harmonics 

• Should consume zero avg. power 

• Response time = 0 



39/454

Load requires P.F correction 

• Large no. of uncompensated industrial loads, 

P.F is less than 0.8 ( they are non linear also) 

• Arc furnace, induction furnace, steel rolling 

mills, large motor loads 

• ‘S’ rating of the compensator (P=0) 

P L 

= 

Q 

L 

= 

S 

L 

sinΦ 

L 

= 

S 

L 

2 

1− 

cos 

Φ 

L 

Ф L 

S L 

Q L 



40/454

Voltage regulation 

• Which is the most important parameter of the 

load & supply system affects regulation ? 

E 

I S 

R S +jX S 

V 

I L 

S l = P L +jQ L 

Y L = G L +jB L 



41/454

V reg 

= 

E 

− 

V 

V 

= 

E −V 

V 

No compensator I L = I S 

E 

ΔV = Z S 

I L 

V 

ΔV 

I S X S 

ΔV X 

I S R S 

ΔV R 

* 

L 

VI = P + 

L 

jQ 

L 

I L = I S 



42/454

I 

L 

= 

P 

L 

− 

V 

jQ 

L 

ΔV 

= 

( R + jX ) 

S 

S 

P 

L 

− jQ 

V 

L 

= 

R 

S 

P 

L 

+ Q 

V 

L 

X 

S 

+ 

j 

X 

S 

P 

L 

− 

V 

R 

S 

Q 

L 

= ΔV 

+ jΔV R X 

• Change depends on both active & reactive 

power of the load 



43/454

Adding a compensator in parallel with load 

E 

So that E = V 

I S 

Replace Q L by 

R S +jX S 

V 

E 

Q = Q + 

2 

S 

Such that 

= 

L 

Q 

C 

( V + Δ ) 2 

+ ( Δ ) 2 

V R 

V X 

I L 

I C 

= 

⎧ RS 

PL 

+ Q 

⎨V 

+ 

⎩ V 

S 

X 

S 

⎫ 

⎬ 

⎭ 

2 

+ 

⎧ 

⎨ 

⎩ 

X 

S 

P 

L 

− 

V 

R 

S 

Q 

S 

⎫ 

⎬ 

⎭ 

2 

− 

−( 

A) 



44/454

Vary Q S => ΔV rotates till 

E = 

V 

Solve (A) with 

E = V 

E 

jI S X S 

I C 

ΔV 

• There is always a 

solution for Q C for any 

value of P 

I S 

V 

I S R S 

I L 



45/454

• If the compensation is used to make 

P.F unity then 

ΔV 

= 

R 

P 

S L 

+ 

V 

jX 

S 

P 

L 

= 

( ) 

P 

R jX V 

S 

+ 

S 

L 

• Independent of Q L 

• Not under the control of compensator 

• Passive reactive compensator can not 

maintain constant V & unity P.F at the same 

time 



46/454

• Approximate relationship for voltage regulation 

Short circuit at the load bus 

S = P + jQ = EI = 

SC 

SC 

Z = R + 

SC S 

* 

SC 

Z SC 

Z = 

SC 

jX 

S , 

* 

SC 

E 

Z 

2 

* 

SC 

I SC → S.C Current 

R 

X 

S 

S 

2 

E 

= Z 

SC 

cos Φ 

SC 

= cos Φ 

S 

SC 

2 

E 

= Z 

S 

sin Φ 

SC 

= sin 

S 

SC 

Φ 

SC 

SC 



47/454

• Change in V influenced by ΔV R 

• Neglect ΔV X 

RS 

PL 

+ 

ΔVR 

= 

V 

ΔV 

V 

R 

= 

P 

L 

Assume 

Z 

Sc 

E 

V 

Q 

L 

cosΦ 

≈ 1 

X 

SC 

S 

+ QLZ 

V 



SC 

sinΦ 

Z 

SC 

= 

2 L SC L 

V 

1 

= PL 

cosΦ 

SC 

+ QL 

sin Φ 

S 

SC 

SC 

{ P cosΦ 

+ Q sin Φ } 

SC 

{ } 

SC 

V 

E 

ΔV R 

ΔV X 

48/454

• If short circuit resistance of source=0 

=> CosФ SC = 0 

ΔV = 

V 

Q 

S 

L 

SC 

E −V 

V 

= 

Q 

S 

L 

SC 

⎡ 

E V ⎢1 

+ 

⎣ 

V 

Q 

= L 

S 

SC 

⎡ 

⎢1 

+ 

⎣ 

Q 

= L 

E 

S 

SC 

≈ 

⎡ 

E ⎢1 

− 

⎣ 

Q 

S 

L 

SC 

⎤ 

⎥ 

⎦ 

⎤ 

⎥ 

⎦ 

⎤ 

⎥ 

⎦ 

−1 

Slope = -E/S SC 

V 

ΔV 

Q L 



49/454

Load balancing 

• Assume all loads are fully compensated for 

reactive VA 

V 

V 

V 

ab 

bc 

ca 

I 

I 

I 

a 

b 

c 

= 

= 

= 

V 

V 

V 

L 

L 

L 

ca 

∠ 

0, 

∠ − 

∠120 

bc 

120 

ab 

bc 

, 

= 

ab 

− 

ca V ca 

V ab 

= 

= 

I 

I 

I 

− 

− 

I 

I 

I 

V bc 



50/454

I 

a 

= 

V 

R 

ab 

Vca 

− 

jX 

VL∠0 V∠120 

VL∠0 

V∠30 

= − = − 

R jX R X 

= 

= 

V 

V 

L 

L 

⎧ 1 

⎨ 

⎩ R 

⎧ 1 

⎨ 

⎩ R 

− 

− 

1 

X 

3 

2X 

( cos 30 + j sin 30) 

− 

2 

j 

X 

⎫ 

⎬ 

⎭ 

⎫ 

⎬ 

⎭ 



51/454

I 

c 

I 

b 

= 

= 

Vbc 

− jX 

Vca 

jX 

− 

V 

− 

R 

Vbc 

− jX 

ab 

= 

= 

= 

= 

V 

L 

V L 

V L 

V 

VL∠30 

V∠ − 30 

= − 

X R 

VL 

= j − − − (3) 

X 

L 

∠ −120 VL∠0 

− 

− jX R 

∠30 

V − 

X R 

⎧ 3 

⎨ 

⎩ 2 X 



− 

∠120 

jX 

1 

R 

− 

− 

V 

L 

2 

j 

X 

⎫ 

⎬ 

⎭ 

− − − 

∠ −120 

− jX 

(2) 

52/454

I b 

= I c 

∠120 

⎛ 

⎜ 

⎝ 

3 

2X 

− 

1 

R 

⎞ 

⎟ 

⎠ 

− 

2 

j 

X 

= 

j 

X 

⎛ 

⎜ 

− 

⎝ 

1 

2 

+ 

j 

3 

2 

⎞ 

⎟ 

⎠ 

⎛ 

⎜ 

⎝ 

3 1 ⎞ 3 

− ⎟ = − 

2X R 

⎠ 2X 

X = 3R 



53/454

Review 

• Using passive reactive element, it is possible to 

achieve ΔV = 0 

• ΔV X has negligible effect on ΔV 

• Determined by ΔV R (≠ i S R S ) 

E 

E 

V 

ΔV 

I S X S 

ΔV X 

V 

ΔV R 

ΔV X 

I L = I S 

I S R S 

ΔV R 



54/454

Contd.. 

• Using passive reactive element it is not 

possible to have ΔV=0 & P.F =1 

• Load balancing 

• All three line currents are balanced if 

X = 3R 



55/454

Load balancing (Contd..) 

I 

a 

= V 

L 

⎧ 

⎨ 

⎩ 

1 

2 

R 

− 

j 

2 

1 ⎫ 

⎬ 

3R 

⎭ 

3R 

∠ − 30 

⎧ 1 1 ⎫ VL 

Ib 

= VL 

⎨− 

− j ⎬ = ∠210 

⎩ 2R 

2 3R 

⎭ 3R 

1 ⎫ 

⎨ 

⎧ V 

= 0 + 

L 

Ic 

VL 

j ⎬ = ∠90 

⎩ 3R 

⎭ 3R 

• Rule: For the load connected between line a-b, 

capacitor should be connected between b-c, and 

Inductor should be connected between c-a 

= 

V 

L 



56/454

Comments 

• Branch currents of Δ are unbalanced 

• Reactive power is balanced within Δ 

• Reactive power generated by C connected 

between line b & c = Q is absorbed by L 

connected between c & a 

• If the load is 

ab 

L 

ab 

L 

Y = G + 

jB 

ab 

L 

• Compensating susceptance 

B 

ab 

C 

= 

−B 

ab 

L 



57/454

• Each branch of Δ will have 3-parallel 

compensating susceptances 

B 

ab 

C 

= −B 

ab 

L 

+ 

⎛ 

⎜ 

⎝ 

G 

ca 

L 

− G 

3 

bc 

L 

⎞ 

⎟ 

⎠ 

B 

bc 

C 

= −B 

bc 

L 

+ 

⎛ 

⎜ 

⎝ 

G 

ab 

L 

− G 

3 

ca 

L 

⎞ 

⎟ 

⎠ 

B 

ca 

C 

= −B 

ca 

L 

+ 

⎛ 

⎜ 

⎝ 

G 

bc 

L 

− G 

3 

ab 

L 

⎞ 

⎟ 

⎠ 



58/454



59/454

Observations 

• Any linear unbalanced 3-Ф load can be 

transformed into a equal 3-Ф balanced load 

• Net real power is the same 

• Corresponding elements are purely reactive 

X = 

R 

3 

Corresponding to power 

consumed by the load 

As the power varies, X also should change 



60/454

• May not be possible 

• Most of the loads are non-linear => 

Harmonics + lagging P.F 

P.F ≠ cos 

I 

V 



61/454

P 

= 

VC1V 

X 

S 

sinδ 

If δ = 0 

If 

V 

C1 

> V S 

I C1 

V S V C1 

jωLI C1 

• I C1 is leading V S 



62/454

• Can be shown that if 

V < 

C1 

V 

S 

• I c1 is lagging 

Q 

= 

V 

S 

I 

C1 

⇒ 

V 

S 

⎛ 

⎜ 

⎝ 

V 

S 

−V 

ωL 

C1 

⎞ 

⎟ 

⎠ 

V α 

C 1 

mV dc 



63/454

I C1 

• Non ideal case 

V S 

δ 

• Var generated α m 

α V dc 

V C1 

jωLI C1 

I C1 R 

V C1 

δ 

V S 

I C1 

jωLI C1 



64/454

• M => Magnitude of sine wave (not very popular) 

• Magnitude of space vector 

• T1 & T2 are to be determined 

T 

T 

1 

2 

sin(60 −θ 

) 

= m 

sin 60 

sinθ 

= TC 

m 

sin 60 

. T 

c 

Intelligent controller 

is required 



65/454

• Vary V dc 

• Var supplied α V dc 

• Var generated is 

controlled by varying 

V C1 & i C1 

• O/P voltage of inverter 

• Indirect current controller Synchronous link 

converter Var compensator (SLCVC) or 

STATCOM 



66/454

Review 

• Linear lagging load can be balanced using 

passive elements 

• Difficult to realize in 

real life 

bc 

Y L 

ca 

Y L 

• Use V.S.I to supply ‘Q’ 

bc 

B C 

ab 

B C 

ca 

B C 

ab 

Y L 



67/454

Contd.. 

• Similar to over-excited 

Syn. motor on No-load 

• Draws only small ‘P’ 

• ‘δ’ is very small 

δ 

E 

V 

• In V.S.I δ = 

V C1 

V S 

• ‘V C1 ’ is synthesized using PWM 



68/454

Contd.. 

• If space vector PWM is 

used at the Z.C instant of 

supply voltage, V S* should 

lag by angle ‘δ’ 

• In sinusoidal PWM 

technique, fundamental 

component of V C1 is in 

phase with modulating 

wave 



69/454

Harmonic elimination Techniques 

Undesirable harmonics can be eliminated 

and fundamental can be controlled by creating 

notches at pre-determined angles 

• At the Z.C of supply voltage, modulating 

wave should lag by ‘δ’ 

⇒ 

1 

4 

If 'n' switchings / cycle 



70/454

⇒ (n-1) harmonics are 

eliminated & magnitude 

of fundamental can be 

controlled 

⇒ 4 switchings /(1/4) cycle 

(α 1 , α 2 , α 3 , α 4 ) 

α 1 < α 2 < α 3 < α 4 < π/2 



71/454

• 3 significant harmonics = 0 

• Fundamental can be controlled 

• Square wave has quarter wave odd symmetry 

• Coefficient of the fundamental & harmonic 

components are given by 

b 

n 

m 

4 ⎧ 

= ⎨1 

+ 2∑ 

nπ 

⎩ k = 1 

( ) 

k 

−1 

cos( nα 

) ⎬ ⎫ 

⎭ 

k 



72/454

• Assume that there are 5 switchings / (1/4) cycle 

• 4 harmonics can be made zero 

• In 3 phase, 3 wire system, triple harmonics 

can be ignored 

• So harmonics to be eliminated are 5 th , 7 th , 

11 th and 13 th 

4 

b1 = {1 − 2cosα1 

+ 2cosα 

2 

− 2cosα3 

π 

+ 2cosα 4 

− 2cosα5} 



73/454

4 

b 

5 

= {1-2cos5 α1+2cos5α2-2cos5 α3+2cos5α4 

5π 

-2cos5 α5 

} = 0 

4 

b 

7 

= {1-2cos7 α1+2cos7α2-2cos7α3 

7π 

+2cos7α4-2cos7 α5 

} = 0 

4 

b 

11 

= {1-2cos11 α1+2cos11 α2........................ 

11π 

-2cos11 α5 

} = 0 

4 

b 

13 

= {1-2cos13 α1+2cos13 α2........................ 

13π 

-2cos13 α5 

} = 0 



74/454

• Non-linear transcendental equations 

• Solve numerically 

• Choose required value for b 1 

⇒ Fundamental component 

α 1 = 10.514, α 2 = 23.228, α 3 = 29.289, 

α 4 = 46.421, α 5 = 50.157 

b 1 = 0.986 p.u. 

• Immediate dominant harmonic ‘V’ gets 

amplified 



75/454

• Var supplied α V dc 

• Var generated is 

controlled by varying 

V C1 or i C1 

• O/P voltage of inverter 

• Indirect current controller Synchronous link 

converter Var compensator (SLCVC) or 

STATCOM 



76/454

How to calculate Ref. Var ? 

i 

= 

I 

m 

( ωt 

− Φ) & V V cosωt 

cos = 

m 

= I 

P 

cos ωt 

+ 

Multiply by cosωt 

I 

q 

sin 

ωt 

∴i 

= 

= 

I 

P 

I 

2 

P 

ω 

cos 2 t + 

I 

q 

sin 

ω 

t. 

cos 

( ) 

q 

1− 

cos 2ωt 

+ sin 2 t 

I 

2 

ω 

ωt 



77/454

• Use a low pass filter ⇒ I P /2 ≈ average 

• Remaining ⇒ Reactive power 

• Limitations: Response time is poor 

⇒ min. one cycle 



78/454

Controlled current SLCVC 

• Compensator current is actually sensed & 

controlled to follow the reference 

• Source should supply 

active component of load 

current + compensate 

inverter loss 



79/454

• Reactive component of load current (i qL ) 

should come from inverter 

i C 

= i PC + i qL 

i qL ⇒ obtained from Var calculator 

i PC ⇒ Accounts for loss 

• If there is a mismatch in power supply and 

consumed ⇒ V dC will change 



80/454

Control strategy -I 



81/454

• To ↑i C close S 4 & S 3 , To ↓i C open S 4 & S 3 

• Response is fast 

• Switching frequency 

varies 

• Var calculator is 

required 



82/454

Review 

• In harmonic elimination technique, if there are 

‘n’ switchings / (¼) cycle, (n-1) harmonics can be 

eliminated & fundamental can be controlled 

⇒ If ‘F’ of pre-dominant harmonic is > 2kHz 

at 50Hz, up to 40 th harmonic should be absent 

⇒ 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37 

⇒ 12 harmonics should be eliminated 



83/454

Contd.. 

• 13 switchings / (¼ ) cycle 

• 13 non linear transdential equations to be 

solved 

• H. S. Patel & R. G. Hoft “Generalized 

technique of harmonic elimination and voltage 

control in thyristor inverters,” Part-1 harmonic 

elimination., IEEE Trans. Ind. Applicat., vol. 

IA-9, pp 310-317, May 1973. 



84/454

Contd.. 

Controlled current SLCVC 

• Compensator current 

i C = i PC + i qL ⇒ 

sinusoidal if load is 

linear 

• If i qL has the 

information about the 

non-linear, ⇒ i C is non 

- sinusoidal 



85/454

Control strategy -II 

• Sense source current i S 

⇒ Compare with sinusoidal reference current i S 

* 



86/454

• i S * is in phase with v S 

• i S is also in phase with v S 

• V dC is held constant 

• All the active power is supplied by the source 

• Rest (‘Q’ + Harmonic I) supplied by inverter 

• i S = i L + i C 

⇒ To ↑i S , ↑ i C 

⇒ To ↓ i S , ↓i C 

} 

Using inverter switchings 



87/454

How often i S* is changed ? 

• Once in every cycle 

• If active power demand of the load has changed 

in between +ve Zero crossings 

• Power is supplied by inverter 

⇒ V dC will ↓ 

• V dC > V m ⇒ peak of V S 

⇒ Large size ‘C’ is required 



88/454

• If Inverter i S * is changed in between the cycle 

• Source ‘I’ will have a DC component 

• Smaller size ‘C’ may be sufficient 



89/454

• Current control is suitable for low power 

• For high power loads switching ‘F’ ↓ 

• Inverter ⇒ Voltage control 

• Harmonic spectrum is inferior 

• Load current has harmonics 

• In addition inverter with voltage control 

also generates harmonics 



90/454

• Use two compensators & connect them in 

parallel 

• Var generator ⇒ High power inverter 

• High V & high I 

• Harmonic filter ⇒ Low power inverter 

• Switching frequency is high 

• Since low power, use current controlled 

PWM technique 



91/454

Active filter +Var compensator for high power 



92/454

• Main compensator ⇒ Voltage control mode 

• Aux. compensator ⇒ controlled current mode 

• Generate i ref ⇒ ref. I of suitable magnitude & 

in phase with source V 

• Force i S = i Cm + i Cx + i L to follow the reference 

within a hysterisis band 

• Error decides the switching instant of aux. 

compensator devices 



93/454

• To ↑ i S , ↑ i Cx ⇒ close S 4 & S 3 

• To ↓ i S , ↓ i Cx ⇒ open S 4 & S 3 

• Now i ref = i L(p) + i Cm(p) 

Where i L(p) = Real component of load I 

i Cm(p) = Real component of the main 

compensator current 



94/454

i 

Cm1 

= 

V 

S 

−VCm 

1∠ −δ 

Z∠θ 

= 

( V − mV cosδ 

) 

S 

dC 

+ 

Z∠θ 

jKV 

dC 

sinδ 

I 

Cm1 

= 

⎛ 

⎜ 

⎝ 

I 

2 

Cm1 

p( 

real ) 

2 

+ 

I 

2 

Cm1 

p( 

q) 

2 

⎞ 

⎟ 

⎠ 



95/454

Control block diagram 



96/454

• Var calculator determines V 

* 

dc (‘m’ is constant) 

V dc * - V dc ⇒ determines δ 

• µC ⇒ determines i ref using I p , δ, V dC & V S 

• Compare i S & i ref to generate switching 

signals for aux. inverter 



97/454

• For low power 

Review 

Var generator + Active filter 



98/454

Contd.. 

• For high power 

application 

Use high power inverter 

for Var generation 

To compensate harmonics 

use active filter 

• Used Var calculator to 

determine ‘Q’ required by 

the load 

• Linear load is assumed 



99/454

3-Phase to 2-phase conversion 

[v] = [z] [i] 

[v'] = [z'] [i'] 

[v] = [A] [v'] 

[i] = [A] [i'] 

[v] = [z] [i] 



100/454

[A] [v'] = [z] [A] [i'] 

[v'] = [A] -1 [z] [A] [i] 

Z' 

⇒ Inverse should exist 

p = i 1 v 1 + i 2 v 2 + i 3 v 3 = [i] t [v] 

p' = i 1 'v 1 ' + i 2 'v 2 '+ i 3 'v 3 ' 

= [i'] t [v'] 



101/454

p = p' 

[i t ][v] = { [A] [i'] } t [A] [v'] 

= [i'] t [A] t 

[A] [v'] 

[U] ⇒ Unit matrix 

[A] t = [A -1 ] or [A] = [A] t 

-1 



102/454

Vector representation of instantaneous 

3-phase quantities 

• 3-current vectors ⇒ one vector ⇒ space vector 

i S = K[i a + i b e j2π/3 + i c e -j2π/3 ] 

Has 2-components ⇒ (α, β) 

i α = K d [i a -(1/2) i b –(1/2) i c ] 

i β = K q [0 + √3/2 i b - √3/2 i c ] 

i 0 = K 0 [i a + i b + i c ] 

i β 

i b 

i a 

i C 

i α 



103/454



104/454 

⎥ 

⎥ 

⎥ 

⎦ 

⎤ 

⎢ 

⎢ 

⎢ 

⎣ 

⎡ 

⎥ 

⎥ 

⎥ 

⎦ 

⎤ 

⎢ 

⎢ 

⎢ 

⎣ 

⎡ 

− 

− 

− 

= 

⎥ 

⎥ 

⎥ 

⎦ 

⎤ 

⎢ 

⎢ 

⎢ 

⎣ 

⎡ 

c 

b 

a 

q 

q 

d 

d 

d 

i 

i 

i 

K 

K 

K 

K 

K 

K 

K 

K 

i 

i 

i 

0 

0 

0 

0 

2) 

3 

( 

2) 

3 

( 

0 

2) 

1 

( 

2) 

1 

( 

β 

α 

[C] 

[ ] 

⎥ 

⎥ 

⎥ 

⎦ 

⎤ 

⎢ 

⎢ 

⎢ 

⎣ 

⎡ 

− 

− 

− 

= 

− 

0 

0 

0 

1 

3 

1 

3 

1 

3 

1 

3 

1 

3 

1 

3 

1 

3 

1 

0 

3 

2 

K 

K 

K 

K 

K 

K 

K 

K 

C 

q 

d 

q 

d 

d

[ C] 

t 

= 

⎡ 

⎢ 

⎢− 

⎢ 

⎣− 

(1 

(1 

K 

d 

2) K 

2) K 

d 

d 

( 

( − 

3 

3 

0 

2) K 

q 

2) K 

q 

K 

K 

K 

0 

0 

0 

⎤ 

⎥ 

⎥ 

⎥ 

⎦ 

If K d = K q = 2/3 & K 0 =√2/3 

[C] -1 = 3/2 [C] t 



105/454

⎡i 

⎢ 

⎢ 

i 

⎢⎣ 

i 

α 

β 

0 

⎤ 

⎥ 

⎥ 

⎥⎦ 

= 

2 

3 

⎡ 

⎢ 

⎢ 

⎢⎣ 

1 

1 

0 

2 

−1 

1 

−1 

2 ⎤⎡i 

− 3 2 

⎥⎢ 

⎥⎢ 

i 

1 2 ⎥⎦ 

⎢⎣ 

i 

Similarly 3-ph AC voltages ⇒ two phase voltages 

3 

2 

2 

2 

a 

b 

c 

⎤ 

⎥ 

⎥ 

⎥⎦ 

⎡e 

⎢ 

⎢ 

e 

⎢⎣ 

e 

α 

β 

0 

⎤ 

⎥ 

⎥ 

⎥⎦ 

= 

2 

3 

⎡ 

⎢ 

⎢ 

⎢⎣ 

1 

1 

0 

2 

−1 

3 

1 

2 

2 

2 

−1 

2 ⎤⎡v 

− 3 2 

⎥⎢ 

⎥⎢ 

v 

1 2 ⎥⎦ 

⎢⎣ 

v 

a 

b 

c 

⎤ 

⎥ 

⎥ 

⎥⎦ 



106/454

⎡v 

⎢ 

⎢ 

v 

⎢⎣ 

v 

a 

b 

c 

⎤ 

⎥ 

⎥ 

⎥⎦ 

= 

⎡ 1 

⎢ 

⎢−1 

⎢ 

⎣ 

−1 

2 

2 

− 

0 

3 2 

3 2 

1 

1 

1 

2⎤⎡e 

⎥ 

2 

⎢ 

⎥⎢ 

e 

2⎥⎢ 

⎦⎣ 

0 

α 

β 

⎤ 

⎥ 

⎥ 

⎥⎦ 

p = v a i a + v b i b + v c i c 

p = e α i α +{ (-1/2 e α +√3/2 e β ) (-1/2 i α + √3/2 i β ) } 

+ { (-1/2 e α -√3/2e β ) (-1/2i α -√3/2i β ) } 

( ) 

p = 3/2 (e α i α +e β i β ) = 3 2 e . i + e . i 

α 

α 

β 

β 



107/454

Instantaneous reactive power compensation 

Instantaneous real power 

p = v a i a + v b i b + v c i c 

Definition of instantaneous reactive current: 

That part of the three phase current can be 

eliminated at any instant without affecting ‘P’ 



108/454

e 

e 

i 

i 

α 

β 

α 

β 

= 

= 

V 

V 

S 

S 

cosψ 

sinψ 

= i cos ϕ + ψ 

S 

( ) 

= i sin ϕ + ψ 

S 

( ) 

i β i S 

e β 

V S 

φ 

ψ 

i α e α 

3 

p = V i ψ ϕ+ ψ + ψ ϕ+ 

ψ 

2 S S 

{ cos .cos( ) sin .sin ( )} 

= 3 V { cos( )} 

3 

S 

iS ψ −ϕ− ψ = VSiS 

cosϕ 

2 2 



109/454

• Can be concluded that 3/2 i S sinφ component of 

current i S can be eliminated without effecting ‘P’ 

Reactive power 

q = 32V i sinϕ 

S 

S 

= 32V 

i sinϕ+ ψ −ψ 

S 

S 

( ) 

{ ( ϕ ψ) ψ ( ϕ ψ) 

ψ} 

= 32V 

i sin + cos − cos + sin 

S 

S 

{ V ψ i ( ϕ ψ) V ψ i ( ϕ ψ) 

} 

= 32 cos . sin + − sin . cos + 

S S S S 

{ ei ei} { e i e i} 

α β β α α β β α 

= 32 − = 32 × + × 



110/454



111/454 

⎥ 

⎦ 

⎤ 

⎢ 

⎣ 

⎡ 

⎥ 

⎥ 

⎦ 

⎤ 

⎢ 

⎢ 

⎣ 

⎡ 

⎥ 

⎦ 

⎤ 

⎢ 

⎣ 

⎡ 

− 

= 

⎥ 

⎦ 

⎤ 

⎢ 

⎣ 

⎡ 

⎥ 

⎦ 

⎤ 

⎢ 

⎣ 

⎡ 

⎥ 

⎦ 

⎤ 

⎢ 

⎣ 

⎡ 

− 

= 

⎥ 

⎦ 

⎤ 

⎢ 

⎣ 

⎡ 

− 

q 

p 

e 

e 

e 

e 

i 

i 

i 

i 

e 

e 

e 

e 

q 

p 

1 

3 

2 

2 

3 

α 

β 

β 

α 

β 

α 

β 

α 

α 

β 

β 

α 

In matrix form 

⎥ 

⎦ 

⎤ 

⎢ 

⎣ 

⎡ 

⎥ 

⎦ 

⎤ 

⎢ 

⎣ 

⎡ 

− 

+ 

= 

q 

p 

e 

e 

e 

e 

e 

e 

α 

β 

β 

α 

β 

α 

2 

2 

1 

* 

3 

2

⎡i 

⎢ 

⎣ 

i 

α 

β 

C 

⎤ 

⎥ 

⎦ 

= 

e 

1 

+ e 

⎡e 

⎢ 

⎣e 

− 

e 

e 

⎤⎡ 

⎥⎢ 

⎦⎣− 

2 

0 

C α β 

3 

. 

2 

α 

2 

β 

β 

α 

⎤ 

q 

⎥ 

⎦ 



112/454

e . q 

i 

* 

= 

β 

α C 

3 2 + 

( 

2 2 

) e e 

α 

− e . 

i 

* 

= 

α 

β C 

3 2 + 

α 

q 

β 

( 

2 2 

) e e 

β 

Where 

q 

= 2 

3 

[ ] e i − e i 

α 

β 

β 

α 



113/454



114/454

• Frequency of e α , i α , e β & i β 

frequency 

is same as supply 

• ‘p’ & ‘q’ are calculated based on instantaneous 

values 

• Assume supply voltages & currents are nonsinusoidal 

and have few common harmonic 

components 



115/454

• Avg. power due to these common harmonic 

components is finite 

• We can not eliminate these frequency 

components from source i ! 

• Source ‘i’ is non-sinusoidal 



116/454

Review 

Instantaneous real power 

P = v a i a + v b i b + v c i c 

i β i S 

e β 

V S 

φ 

ψ 

i α 

e α 

P 

= 

3 

2 

V 

( e . i + e i ) 

S 

I 

S 

cosϕ 

= 3 2 

α α β 

. 

Instantaneous reactive current: 

That part of the three phase current can be 

eliminated at any instant without affecting ‘P’ 

β 



117/454

Contd.. 

q= 

32V i sinϕ 

= 3 2 

S 

S 

{ e × i + e × i } 

α 

β 

β 

α 

• If ‘v’ is sinusoidal, i L is non-sinusoidal 

⇒ 

If q=0, then i S will be sinusoidal and in phase 

with V s ( since average of the product of 

fundamental ‘ω’ & higher ‘ω’ term = 0) 

p 

n 

= 

vsinωt 

∞ ∑ 

n= 

2 

Avg. of p n = 0 

i 

n 

sin 

nωt 



118/454

Contd.. 



119/454

Contd.. 

• If ‘v’ is non-sinusoidal & i L is also non-sinusoidal 

⇒ 

i S will have component corresponding to 

common frequency term of voltage & current 

• H. Akagi, Y. Kanzawa, and A. Nabae 

“Instantaneous Reactive Power Compensators 

Comprising Switching Devices without Energy 

Storage Components,” Part-1 harmonic elimination., 

IEEE Trans. Ind. Applicat., vol. IA-20, No. 3,pp 625- 

630, May 1984. 



120/454

Change of reference frame 

q S 

d 

S 

dt 

ω 

S 

q r 

ω S 

d r 

⎡d 

⎢ 

⎣q 

⎡d 

⎢ 

⎣q 

s 

s 

r 

r 

⎤ 

⎥ 

⎦ 

⎤ 

⎥ 

⎦ 

= 

= 

⎡cosθS 

⎢ 

⎣sinθS 

⎡ cosθS 

⎢ 

⎣− 

sinθS 

θ = 

⎥ ⎦ 

⎤ 

− sinθS 

⎤⎡d 

cosθ 

⎥⎢ 

S ⎦⎣q 

sinθS 

⎤⎡d 

cosθ 

⎥⎢ 

S ⎦⎣q 

r 

r 

s 

s 

⎤ 

⎥ 

⎦ 

θ S 

d S 



121/454

3 - phase 

(St. Frame) 

50 Hz 

⇒ 

2 - phase 

(St. Frame) 

50 Hz 

⇒ 

2 - phase 

(rotating. 

Frame at ω S ) 

D. C 

⎡dr 

⎢ 

⎢ 

qr 

⎢⎣ 

0 

⎤ 

⎥ 

⎥ 

⎥⎦ 

= 

2 

3 

⎡ cosθ 

S 

⎢ 

⎢ 

− sinθ 

S 

⎢⎣ 

1 2 

cos 

− sin 

( θ − 2π 

3) cos( θ + 2π 

3) 

s 

( θ − 2π 

3) − sin( θ + 2π 

3) 

s 

1 2 

1 2 

s 

s 

⎤⎡a⎤ 

⎥⎢ 

⎥ 

⎥⎢ 

b 

⎥ 

⎥⎦ 

⎢⎣ 

c⎥⎦ 



122/454

• Let us assume that v S is along d r - axis in the 

syn. Rotating frame & i S is making an angle φ 

3 

2 

V I S S 

cosϕ 

q r 

q S 

and 

P = θ S d r 

= 3 2V S I 

q 3 

= 

r q 

2 V S I 

r 

r d r 

φ 

i S 

v S 

ω S 

d S 



123/454

• Transform all the variables to Syn. rotating 

frame (rotating at ω S ) 

• Fundamental component of v & i will become dc 

• Other components will pulsates 

• Use a filter to eliminate these pulsating 

component 

• (Could have used a filter to eliminate harmonics 

from input signal) 



124/454

• AC filtering ⇒ phase shift 

• V S is filtered component 

• i q is made zero 

q r 

q s 

i S 

d r 

i q 

φ 

ψ 

i d 

V S 

d s 



125/454



126/454

• Information about system frequency is 

required 

• Frequency varies over a narrow range 

• Should be insensitive to harmonics or multiple 

zero crossings 



127/454

Harmonic Oscillator 

. 

⎡ ⎤ 

⎢ 

x 

⎥ 

⎢ . ⎥ 

⎣y⎦ 

= 

⎡ 0 ω⎤⎡x⎤ 

⎢ ⎥⎢ 

⎥ 

⎣-ω 

0⎦⎣y⎦ 

• Has Eigen values at S = 

• If x(0) = 0 and y(0) =1 

± jω 

x( t) 

= sinωt 

y( t) 

= cosωt 

x& 

= ω y 

y& 

= 

− 

ω 

x 



128/454

* 

x 

∫ 

ω 

xt () 

x 

y 

n 

+ 1 

− 

Δ t 

n +1 

− 

Δ t 

y 

x 

n 

n 

= 

= 

− 

ω y 

xω 

* 

y 

−ω 

∫ 

y() 

t 

x 

n 

= x + ωyΔt 

+ 1 n 

y 

n 

= y − ωxΔt 

+ 1 n 



129/454

How to generate 3-phase sinusoids? 

x sinωt 

a 

= y = cosωt 

v = Cosωt = y 

1 3 

vb 

= Cos( ωt− 120) =− y+ 

x 

2 2 

1 3 

vc 

= Cos( ωt− 240) =− y− 

x 

2 2 

• Let e a , e b and e c are the 3φ instantaneous system 

voltages 



130/454

e 

α 

= 

e 

a 

− 

1 

2 

e 

b 

− 

1 

2 

e 

c 

= 

3 

2 

e 

a 

e 

β 

= 

3 

2 

3 

e b 

− e c 

2 

e = e + je s α β 

• Space vector representation of v a , v b and v c 

v 

s 

= 

v 

a 

+ 

v 

b 

e 

j 

2 π 

2 

− j 

π 

3 

+ v e 

3 

c 



131/454

2π 

2π 

2π 

2π 

= cosωt 

+ cos( ωt 

−120)(cos 

+ jsin 

) + cos( ωt 

−240)(cos 

− jsin 

) 

3 3 

3 3 

1 3 1 3 1 3 1 3 

= cosω 

t + ( − cosω 

t + sinωt 

)( − + j ) + ( − cosω 

t − sinωt 

)( − − j ) 

2 2 2 2 2 2 2 2 

3 3 

= cosωt + j sinωt 

= vα 

+ 

2 2 

jv 

• Projection of e s on d r and 

v s 

q r 

• ( is aligned along d r ) 

β 



132/454

e 

d 

e d 

= 

= 

= 

e 

e s 

e 

s 

α 

cos( θ −ωt) 

{ cos θ cosωt 

+ sinθsinωt} 

cos ωt 

+ 

e 

β 

sinωt 

e 

q 

= 

= 

e 

e s 

s 

sin( θ −ωt) 

{ sinθ 

cosωt 

−cosθ 

sinωt 

} 

e q 

= 

e 

β 

cos 

ωt 

− 

e 

α 

sin 

ωt 



133/454

Objective 

• To make the phase and frequency of v a , v b ,v c and 

e a , e b ,e c same 

• v s and e s are in phase 

• e q =0 

v a 

= 

y 

v 1 3 

= − y + x 

b 

2 2 v 1 3 

= − y − x 

c 

2 2 



134/454

Review 

• In synchronous rotating frame (speed of the 

frame = ω s ), supply frequency terms will 

become DC 

• If input ‘v’ are unbalanced 

→ 

+ve sequence terms DC 

-ve sequence terms → oscillate at 2 

ω s 



135/454

• Other higher frequency terms in the synchronous 

reference frame can be filtered out 

• They can also be filtered out in the input side 

• Phase shift is introduced – not an issue 

• Active filter control 

Contd.. 



136/454

To change MI using harmonic elimination PWM 

technique 

10.9091, 23.2907, 29.8505, 46.3408, 50.6781 

} 

10.7120, 23.2678, 29.5761, 46.3867, 50.4260 5, 7, 11, 13 are eliminated and 

10.5138, 23.2278, 29.2896, 46.4210, 50.1567 Magnitude of fundamental is 

different 

• Frequency information is required. 

• C. Schauder and H. Mehta, “Vector analysis and 

control advanced static Var compensators” IEE 

proc, vol.140, pp. 299-306, 1993 



137/454

Through Hardware 

• Digitize the sine wave and store in EPROM 

(1024 part) 

• Address the EPROM using 10 bit counter 

( 2 10 =1024 ) 

• Use a PLL as a multiplier 



138/454



139/454

Software approach 

Harmonic oscillator 

. 

⎡ ⎤ 

⎢ 

x 

⎥ ⎡ 0 ω⎤⎡x⎤ 

= 

⎢ . ⎥ ⎢ ⎥⎢ 

⎥ 

⎣-ω 

0⎦⎣y⎦ 

⎣y⎦ 

x( t) 

= 

x 

y 

n 

sinωt 

− x 

Δ t 

− y 

n 

Δt 

+ 1 

n +1 

ω → Instantaneous frequency 

• Input to harmonic oscillator is ω 

n 

y( t) 

= cosωt 

= ω y 

= − xω 



140/454

• 3φ sinusoids which are in phase with supply 

fundamental component of the supply voltage 

are required 

• Input voltage may have harmonics 

• e a , e b ,e → c input system voltages may have 

harmonics + may be unbalanced 

e = e + je s α β 



141/454

• Let v a 

, v b 

,v c 

are the 

3φ pure sinusoids 

e S 

• e s 

should be in phase with v s 

ω S t 

v S 

v a 

= 

y 

v 1 3 

= − y + x 

b 

2 2 v 1 3 

= − y − x 

c 

2 2 



142/454

• This voltage waveform can be used as 

reference current waveform in hystersis 

current control PWM technique 

• Source current follows this reference ‘i’ 

• Source current is in phase with fundamental 

component of input voltage 



143/454

One cycle control of 3φ Var compensator 

and Active filter 

• No zero crossing detection 

} 

No reference wave 

• No PLL 

generation 

Basic Analysis : 

• Switching frequency is much higher than supply 

frequency 

• Let x(t) be an input to a switch operating at 

variable ON and OFF times 



144/454

1 1 

• = = Switching frequency 

T T T 

ON 

+ 

OFF s 

• Produces switched output with average 

y 

( t 

) 

= 

1 

T 

s 

T ON 

∫ 

0 

x 

( t 

) dt 

= x(t) D(t) 

D= duty cycle 



145/454

• Duty ratio has to be generated as control input 

based on some reference signal V ref (t) 

• If the duty ratio is controlled so that 

• Average output 

T 

ON T S 

∫ x( 

t) 

dt = 

0 

∫ 

0 

y( 

t) 

= 

V 

1 

T 

s 

ref 

T s 

∫ 

0 

( t) 

dt 

V 

ref 

( t) 

dt 



146/454

• Assume that over one cycle V ref (t) is roughly 

constant 

y(t)=V ref (t) 

• Works for constant switching frequency 

• V ref could be a variable feedback signal 

• Can be implemented using a simple integrator 

with reset 



147/454

• Generate reset pulse at required frequency 

• At the start of every cycle switch is turned ON 

by the reset pulse 

• Integrate the input 

• When the output of the integrator just exceeds 

V ref turn OFF the switch 

• Start the cycle again after T s when integrator 

resets 



148/454

Rule to be followed 

• A term in the control equation which is being 

multiplied with duty cycle of the switch has to 

be passed through a reset integrator and 

compared with the appropriate reference 



149/454

1φ AC-DC Active filter + Var generator 

Assumption: 

• In one switching cycle input is constant 

• V dc 

is constant and ripple free 



150/454

S 4 , S 3 ON for DT S : 

di 

L = V s 

+ V DC 

dt 

S 1 , S 2 ON for (1-D)T S : 

L 

di 

dt 

= 

V s 

− V DC 



151/454

• Assume i(t) is continuous and i(0) = i(T s ) 

• Average ‘V’ across L = 0 

( V + V ) DT = ( V −V 

)(1 − D) 

T 

s 

DC 

S 

DC 

S 

s 

V 

DC 

Vs 

= 

1−2D 



152/454

Aim 

• i s and V s should be in phase 

Vs= i s R e (R e = Emulated resistance) …..(a) 

(1-2D)V dc = i s R e 

i s = (1-2D)V dc /R e ……(b) 

• In each switching cycle if the duty ratio D is 

controlled in such a way that equation (b) is 

satisfied , equation (a) also gets satisfied 

• Control requirement is (1-2D)V m = i s 

Where V m = V dc /R e 



153/454

Review 

• One cycle control 

} 

→ 

→ 

No PLL 

No ZCD 

Rule to be followed: 

• A term in the control equation which is being 

multiplied with duty cycle of the switch has to be 

passed through a reset integrator and compared 

with the appropriate reference 



154/454

Contd.. 

• Generate reset pulse at required frequency 

• At the start of every cycle switch is turned ON 

by the reset pulse 

• Integrate the input 

• When the output of the integrator just exceeds 

V ref turn OFF the switch 



155/454

• Start the cycle again after T s when integrator 

resets 

• K. M. Smedley & C. Qiao, “Unified constantfrequency 

integration control of active power 

filters –steady –state and dynamics” IEEE 

Transaction on power electronics, vol. 16, No. 

3, May 2001 



156/454

1φ AC-DC 

Control technique 

(1 − 2D ) V = i m s 

V 

R 

c 

V 

m 

= → Emulated resistance 

e 



157/454

DT 

s 

1 

Vm 

− ∫Vmdt 

= 

Ti 

0 

Vm 

V 

m 

− DTs 

= i 

T 

i 

s 

i 

s 

T i 

= Integrator time constant 

F s = 1/T S 

= Switching frequency 

• V m remains constant in one cycle 

• If 

1 

T i 

T s 

2 

(1 − 2D ) V m 

= i 

= s 



158/454

Alternate Approach DC-DC Converter 

( V ) 

i 

avg 

= 

−V DT 

c 

+ V 

T 

c 

(1 − 

D) 

T 

= V c 

( 1− 

2D) 



159/454

• ‘L’ is small 

V + iω 

L = 

V 

i 

Buck Converter 

c 

V s 

= V 

s 

= V ( 1−2D) 

• ‘V o ’ to be maintained 

constant 

c 

• Compare with reference 

and vary D or depending 

upon V s change ‘D’ 



160/454

• Information regarding V s should be known 

• Assume that V s and i s are in phase (required) 

• Instead of varying ‘D’ as function of V s 

• Vary ‘D’ as a function of i s 

• If V s and i s are not in phase chosen values of 

‘D’ may not give the desired V o 

• If ‘V o ’ is regulated, our assumption that V s 

and i s are in phase is valid 



161/454



162/454



163/454

• DC link voltage has to be regulated 

• Generate fixed frequency clock 

• At the rising edge reset the integrator and turn 

ON the switches S4 and S3 

• i s ↑ 

• As t ↑ X ↓ When i s = X ; R = 1 

• Turn OFF the S 4 , S 3 and Turn ON S 1 , S 2 



164/454

Inverter topology for high power application 

• For high power applications 

• Conventional 3φ Inverter with ‘V’ control 

• Switching ‘F’ is low 

• ‘F’ of predominant harmonic is low 

• 

• 



165/454

• 2 converters 

→ Var Compensator 

→ 

Low power inverter for 

active filtering 

• There are only two levels 

Instead 



166/454

• Number of pulse should be high for superior 

harmonic spectrum 

• Instead modify the Inverter structure 

• More than two levels 

• Multi-level inverter 



167/454

Diode clamp multilevel inverters 

3 Level Inverter: 

• Consider only 

one leg 

• Any time two switches are ON = (n-1) 



168/454

Switches ON 

V AX 

S1, S2 V dc 

S2, S3 

S3, S4 

V dc 

0 

2 

• Number of capacitors required = 2 =(n-1) 

• Number of switches required = 4/phase = 2(n-1) 



169/454



170/454



171/454

• Voltage across each capacitor = V dc /2 = V dc /(n-1) 

• Number of diodes = 2 ? 

4 level Inverter 

• Number of switches ON = 3 = (n-1) 

• Number of switches/leg = 6 = 2(n-1) 

• Number of capacitors = 3 = (n-1) 

• Voltage across each capacitor = V dc /3 = V dc /(n-1) 



172/454

Review 

• In one cycle control ‘i S ’ is compared with 

(1-2D)V m 

• V m is passed through 

reset integrator & 

compared with V m -R S i S 

⇒ R S is sensing resistor 

• No reference current waveform generation 



173/454

Contd.. 

• For high power ⇒ Use multi-level inverter 

• For 3-level ⇒ V AX = V dC , ½ V dC , 0 



174/454

Contd.. 

• At any time 2-devices (n-1) devices are ON 

• No. of Switches = 2(n-1) 

• ‘V’ across each ‘C’ = V dC / 2 = V dC /(n-1) 

• ‘V’ rating of switch = V dC /2 = V dC /(n-1) 

• ‘V’ rating of diode = V dC /2 

• No. of diodes = 2 = (m-1)*(m-2) 



175/454

References 

• Bum-Seok Suh and Dong-Seok Hyun “A New N- 

Level High Voltage Inversion System,” IEEE Trans. 

Ind. Electron., vol. 44, No. 1,pp 107-115, Feb 1997. 

• Nam S. Choi, Jung G. Cho and Gyu H. Cho “A 

General Circuit Topology of Multilevel Inverter,” in 

Proc. IEEE Power electron specialist conf. Rec., pp 96- 

103, 1991. 



176/454

4-level inverter 

• Number of switches ON = 3 = (n-1) 

• Number of switches/leg = 6 = 2(n-1) 



177/454

• Number of capacitors = 3 = (n-1) 

• Voltage across each capacitor = V dc /3 = 

V dc /(n-1) 

S1, S2, S3 ON: ⇒ V AX = V dc 

• ‘V’ rating of each 

device = V dc /3 



178/454

S2, S3, S4 ON : 

⇒ V AX = 2V dc /3 

S3, S4, S5 ON : 

⇒ V AX = V dc /3 



179/454

S4, S5, S6 ON: 

⇒ V AX = 0 

Observations: 

• Duty cycle of switch is not the same 

• Lower switches are ON for longer time 

• Switch utilization is poor 



180/454

• ‘V’ rating of D B = 2V dc /3 

• ‘V’ rating of D A = V dc /3 



181/454

• ‘V’ rating of diodes is not the same 

• Number of diodes = (n-1) (n-2) = 6 



182/454

Voltage space vectors for 3 level inverter 

Large voltage vectors 

CBA 

NNP→ NPP → NPN → PPN → PNN → PNP → NNP 

• Similar to conventional 2-level inverter 

• 6 active vectors and 2 zero vectors 



183/454



184/454 

⎥ 

⎥ 

⎥ 

⎦ 

⎤ 

⎢ 

⎢ 

⎢ 

⎣ 

⎡ 

⎥ 

⎥ 

⎥ 

⎦ 

⎤ 

⎢ 

⎢ 

⎢ 

⎣ 

⎡ 

− 

− 

− 

− 

− 

− 

= 

⎥ 

⎥ 

⎥ 

⎦ 

⎤ 

⎢ 

⎢ 

⎢ 

⎣ 

⎡ 

co 

bo 

ao 

cn 

bn 

an 

V 

V 

V 

V 

V 

V 

2 

1 

1 

1 

2 

1 

1 

1 

2 

3 

1 

⎥ 

⎥ 

⎥ 

⎦ 

⎤ 

⎢ 

⎢ 

⎢ 

⎣ 

⎡ 

⎥ 

⎦ 

⎤ 

⎢ 

⎣ 

⎡ 

− 

− 

− 

= 

⎥ 

⎦ 

⎤ 

⎢ 

⎣ 

⎡ 

cn 

bn 

an 

qs 

ds 

V 

V 

V 

V 

V 

2 

3 

2 

3 

0 

2 

1 

2 

1 

1

( NNP ) ⇒ ( 001 ) ⇒ 

( PPN ) ⇒ ( 110 ) ⇒ 

( NPN ) ⇒ ( 010 ) ⇒ 

( PNP ) ⇒ ( 100 ) ⇒ 

V dC 

∠0 

V dC 

∠π 

V dC 

∠2π / 

V dC 

∠ −π / 

3 

3 

( NPP ) ⇒ ( 011 ) ⇒ 

/ V dC 

∠π 

3 

( PNN ) ⇒ ( 100 ) ⇒ 

V dC 

∠ − 2π / 

3 



185/454

Small voltage vectors 

C B A 

O P P 

P O P 

P P O 

C B A 

O O P 

O P O 

P O O 



186/454

C B A 

O O N 

O N O 

N O O 

C B A 

O N N 

N O N 

N N O 



187/454

C B A ⇒ O P P 

⇒ V AO = V BO = V dC /2, V CO = 0 

⇒ V an = V dC /6, V bn = V dC /6, V cn = - V dC /3 

VdC 

1 VdC 

1 VdC 

V 

d 

= − − = 

6 2 6 2 3 

V 

4 

dC 



188/454

V 

q 

∴V 

S 

= 

= 

3 

2 

V 

2 

dC 

⎡V 

⎢ 

⎣ 6 

dC 

∠π / 

V 

+ 

3 

3 

dC 

⎤ 

⎥ 

⎦ 

= 

3 

4 

V 

dC 

OPP 

∴POO 

⇒ 

V 

S 

= 

V 

2 

dC 

∠4π 

/ 

3 

POO 



189/454

NOO : 

V AO = V BO = 0, V CO = -V dC /2 

V an = V dC /6, V bn = V dC /6, V cn = -V dC /3 

VdC 

V 

ds 

= 

2 6 

V 

4 

3 

dC 

3 ⎡VdC 

VdC 

⎤ 3 

= , Vqs 

= + = VdC 

V 

V 

dC 

dC 

∴VS 

= ∠π / 3 ⇒ ONN ⇒ VS 

= ∠4π 

/ 3 

2 

2 

2 

⎢ 

⎣ 

6 

3 

⎥ 

⎦ 

4 



190/454

OOP : 

V AO = V dC /2, V BO = V CO = 0 

V an = V dC /3, V bn = V cn = -V dC /6 

VdC 

V 

ds 

= , V qs 

= 0 

2 

PPO 

OON 

OOP 

NNO 

∴ V 

S 

= 

VdC 

2 ∠0 

⇒ 

PPO 

⇒ 

V 

S 

= 

V 

2 

dC 

∠π 



191/454

NNO : 

V AO = 0, V BO = V CO = -VdC / 2 

V an = 1/3[0 +V dc /2 + V dc /2] = V dC /3, 

V bn = V cn = 1/3[-2V dC / 2 + V dC / 2] = - V dC /6 

VdC 

V 

ds 

= , V qs 

= 0 

2 

∴ V 

S 

= 

VdC 

2 ∠0 

⇒ OON 

⇒ V 

S 

= 

V 

2 

dC 

∠π 



192/454

OPO : 

V AO = V CO = 0, V BO = V dC /2 

V an = V cn = -V dC /6, V bn = V dC /3 

V 

ds 

V 

4 

, 

3 

2 

⎡ 

⎢ 

⎣ 

V 

3 

V 

6 

3 

4 

dC 

dC dC 

= − Vqs 

= 6. + = VdC 

∴ V VdC 

S 

= ∠2π 

/ 3 

VdC 

⇒ POP ⇒ V = ∠5π 

/ 

2 

2 

3 

S 

⎤ 

⎥ 

⎦ 



193/454

NON : 

V AO = V CO = -V dC /2, V BO = 0 

V an = V cn = -V dC /6, V bn = V dC /3 

V 

ds 

= 

V 

− 

4 

dC 

, 

3 

V qs 

= V dC 

4 

OPO 

NON 

VdC 

VdC 

∴VS 

= ∠2π 

/ 3 ⇒ ONO ⇒ VS 

= ∠5π 

/ 3 

2 

2 

POP 

ONO 



194/454

Medium voltage vectors 

ONP : 

V AO = V dC /2, V BO = -V dC /2 , V CO = 0 

V an = V dC /2, V bn = -V dC /2 , V cn = 0 

3 

3 

V 

ds 

= V dC 

, V qs 

= − VdC 

4 

4 

3 

∴V S 

= V 

2 

dC 

∠ −π / 

6 



195/454

NOP : 

V AO = V dC /2, V BO = 0 , V CO = -V dC /2 

V an = V dC /2, V bn = 0 , V cn = -1/2 V dC 

3 

3 

V 

ds 

= V dC 

, V qs 

= VdC 

4 

4 

3 

∴V S 

= V 

2 

dC 

∠π / 

6 



196/454

NPO : 

V AO = 0, V BO = V dC /2 , V CO = -V dC /2 

V an = 0, V bn = V dC /2 , V cn = -V dC /2 

Vds 

= 0, 

3 

V qs 

= V 

2 

dC 

3 

∴V S 

= V 

2 

dC 

∠π / 

2 



197/454

PNO : 

V AO = 0, V BO = -V dC /2 , V CO = V dC /2 

V an = 0, V bn = -V dC /2 , V cn = V dC /2 

Vds 

= 0, 

3 

∴V S 

= V 

2 

3 

V qs 

= − V 

2 

dC 

∠3π 

/ 

2 

dC 



198/454



199/454

Review 

3-Level Inverter 

• No. of large voltage vectors = 6 

⇒ V S = V dC 

• No. of small voltage vectors = 6 

⇒ V S = 1/2V dC 

⇒ 12 possible combinations 

+ ve or –ve bus 

& 

mid point 



200/454

Contd.. 

• No. of medium voltage vectors = 6 

⇒ + ve, - ve & mid-point bus 

⇒ 

V = 3 2 

S 

V dC 



201/454



202/454

Voltage control 

• Space vector PWM 

⇒ Depending upon the position of space 

vector, switch the corresponding switch 

NPP 

OPP 

NOO 

NOP 

PPP 

NNN 

OOO 

OOP 

NNO 

NNP 



203/454

Voltage unbalance between DC-Line 

capacitance 

• Each leg ⇒ 3 possibilities 

• There are 27 switching instances are possible 



204/454

• Unbalances has no effect on load 

• Load is connected across the DC bus 

• Somewhat effective 

in reducing voltage 

unbalance 



205/454

• C 1 supplies the power 

• C 2 does not supply the power 

• ‘V’ across C 2 ↑ 

• For remaining 2 configuration, V across C 1 ↑ 



206/454

• Passive elements 


• Inverter 

⇒ Current control 

⇒ Voltage control 

⇒ Main compensator 

⇒ Aux. compensator 

• Instantaneous reactive power theory 

• One cycle controlled inverter 

• Multi level inverter 



207/454

Transmission line voltage support 

• Provide mid-point compensation 

⇒ Shunt 

⇒ Series 

⇒ Combination of shunt & series 

⇒ Combination of series & series 

P < SIL 

P = SIL 

V S 

P > SIL 

V R 



208/454

Shunt Compensation : 

• Inject current in to the system 

• If injected ‘I’ is in phase quadrature with the ‘V’ 

• Only reactive power transfer 

• Else, it has to handle real ‘P’ as well 

Series Compensation : 

• Inject voltage in series with the line 



209/454

• If ‘V’ is in quadrature with line ‘I’, only reactive 

power transfer 

Combination of series & Shunt Compensation : 

• Inject ‘I’ with the shunt part & 

• Inject ‘V’ with the series part 

• When combined there can be real power 

exchange between the series & shunt controllers 



210/454

Mid point voltage regulator 

• Two machine model 

Ρ 

= 

V S 

V 

X 

R 

Sinδ 

⇒ If V s = V r = V 

P 

max = 

V 

X 

2 



211/454

• Connect a compensator at the mid point & 

V m = V s = V r = V 

• Whether active power transfer is require ? 

• System is loss-less 



212/454



213/454

• Let V sm & V mr are fictitious voltages in phase 

with I sm & I mr respectively 

Vsm = Vmr 

= 

V. Cos δ 

( / 4) 

( δ / 4) 

2V 

. Sin 

I 

sm 

= I 

mr 

= 

= 

X 2 

4V 

X 

Sin 

( δ / 4) 

P 

P 

= V 

. I 

r 

= 

sm sm 

= 

4V 

X 

2 

Sin 

2V 

2 

= Sin δ 

X 

( δ / 4) . Cos( δ / 4) 

( / 2) 



214/454

• Reactive power supplied by the 

compensator 

= V I = VI 

m 

c 

c 

= 

2. 

V. 

I Sin δ 

sm 

( / 4) 

= 

2 

8V 

2 

X 

Sin 

( δ / 4) 

= 

4V 

2 δ 

X 

( 1− 

Cos( / 2) 

) 



215/454



216/454

• Shunt compensator can increase ‘P’ 

• ‘Q’ demand also ↑ 

• Can have multiple compensators located at 

the equal distances 

• Theoretically ‘P’ would double for each 

doubling of the segments 



217/454

• ↑ the no. of segments results in flat 

‘V’ profile 

• Expensive 



218/454

Review 

Mid-point shunt compensation 

⇒ If V s = V r = V 

P = 

2V 

2 Sin δ 

X 

( / 2) 

Q 

= 

4V 

2 δ 

X 

( 1− 

Cos( / 2) 

) 

⇒ ‘I’ is injected into the line 

(in quadrature with ‘v’) 



219/454

Contd.. 

• For each doubling of the segments, 

transmittable ‘P’ also doubles 

• ‘V’ profile is almost flat 

• Large no. of shunt compensators ⇒ expensive 



220/454

Summary 

• Compensator must remain in synchronism 

with the ac system under all operating conditions 

including major disturbances 

• Must regulate the bus voltage 

• For the inter connecting two systems, best 

location is in middle 

• For radial feed to a load, best location is 

at the load end 



221/454

Methods of controlling Var generation 

• Mechanically switched capacitor and/or 

inductor ⇒ course control 

⇒ in-rush current 

• Continuously variable Var generation or 

absorption ⇒ originally over excited syn. motor 

• Modern Var generators → use power 

semiconductor devices/equipment + energy 

storing elements 



222/454

Variable impedance type S.V.C 

1. Thyristor controlled reactor (TCR): 

• T 1 & T 2 is triggered in the + ve 

& - ve half cycles respectively 

α ⇒ Can be measured w. r. t 

zero crossing or peak of ‘V’ 



223/454

• ‘i’ flows from α to β 

di 

L = VmSinωt 

dt 

Vm ∴i ω 

ωL 

() t = ( Cosα 

− Cos t) 

i(t) =0 at ωt = β 

Cos α = Cosβ 

∴β 

= 2π 

−α 

⇒ β = extinction angle 



224/454

• ‘i’ is continuous when α = π/2 

• ‘i’ is sinusoidal 

• No control ⇒ ’L’ is fixed & it is minimum 

• As α↑, all odd harmonics are introduced 



225/454

• As α↑, L ↑ 

∴ 

I LF 

α V ⎛ 2 1 ⎞ 

( ) = ⎜1− 

α − sin α ⎟ 

ωL 

⎝ π π 

2 

⎠ 

⇒ 

B L 

α 1 ⎛ 2 1 ⎞ 

( ) = ⎜1− 


ωL 

⎝ π π 

2 

⎠ 

• V L(MAX) ⇒ Voltage limit 

• I L(MAX) ⇒ current limit 

• B L(MAX) ⇒ Max. admittance of TCR 



226/454

2. Thyristor switched capacitor (TSC): 

• Small ‘L’ is required to 

limit the surge current 

• Thyristors are switched 

when v c = v 

• ‘V’ rating of the switch ? 



227/454



228/454

3. Fixed Capacitor, Thyristor controlled 

Reactor (FC-TCR): 

In TCR 

• ‘i L ’ is varied by varying ‘α’ 

• i L = i L(max) when α = π/2 

• In FC-TCR, for any value of 

i L , net effect of C ↓ 

• ‘C’ also provides a low impedance path for 

harmonics generated by TCR 



229/454

• ‘Q C ’is constant 

• Net Q = Q C when Q L = 0 (α = π) 

• To ↓ net Q, ↓ α 

• Net Q = 0, when Q C = Q L 

• If α is ↓ further, net Q is inductive 



230/454

• At α = π/2, Q L = Q L(max) 

• Operating V-I region of FC-TCR 



231/454

STATCOM 

• VSI can supply ± Q 

• Also known as static 

synchronous condenser 

• Similar to syn. motor 

I 

V − E 

X 

V − E 

= Q = . V 

X 

Q ⇒ reactive power received by the source 



232/454

Control 

• ‘Q’ is controlled by M.I & δ ⇒ accounts for losses 

• Assumed that inverter is capable of injecting ‘Q’ 

demand of the line 



233/454

• If ‘Q’ demand >Var rating of inverter 

• It may fail due to over load 

• Have a inner ‘I’ loop 



234/454

Operating V-I region 



235/454

Review 

T.C.R 

• If α = π/2 ⇒ i = i max 

• As α↑, L eff ↑ 

• Harmonics 



236/454

Contd.. 

T.S.C 

• Thyristors are triggered 

when v c = v 

F.C.T.C.R 

• T.S.C – T.C.R scheme 

is also possible 



237/454

Contd.. 

• Above schemes are variable impedance types 

STATCOM 

• Variable source type 

I 

= 

V 

− 

X 

E 

V − E 

Q = . V 

X 



238/454

Advantages 

• Since voltage profile is maintained 

(in radial system) 

⇒ Voltage instability is prevented 

⇒ Improves transient stability 

⇒ Damping of power oscillations 

⇒ Able to maintain ‘V’ profile 



239/454

Series compensation 

• Reciprocal of shunt compensation 

• Shunt compensator : Controlled reactive 

‘I’ source connected in parallel with the 

Tr. Line to control ‘V’ 

• Series compensator : Controlled reactive 

‘V’ source connected in series with the 

Tr. Line to control ‘I’ 



240/454

Series compensation 

• Injects voltage in series with the line 

• Could be variable ‘Z’ (such as ‘C’ or ‘L’) 

• Voltage source 

• Effective in controlling the power flow 



241/454

Concept of series capacitive compensation 

⇒ To decrease reactance of the line 

P 

= 

V . V 

X 

S R 

. 

Sinδ 

X 

= 

( − ) 

X L 

X C 



242/454

X 

eff 

= 

( X − X ) 

L C 

= ( 1− K ) X 

L 

K = 

X C 

X L 

⇒ 0 < K < 1 

⇒ Degree of series 

compensation 



243/454

• If V S = V R =V 

I 

= 

V. Sinδ 

2 2V 

. Sinδ 

2 

= 

( 1− 

K ) X ( ) 

L 

2 1− 

K X 

L 

P 

= 

V 

m 

I 

= 

2V 

. Sinδ 

2 

( VCosδ 

2 ). 

( 1− 

K ) X 

L 

2 

V . Sinδ 

= 1 

( − K ) X 

L 

2 2 

2 4V 

. Sin 2 

Q 

C 

= I X 

C 

= 

. X 

2 

X 

( 1− 

K ) 

2 

δ 2V 

.( 1− 

Cosδ 

) 

C 

= 

2 

( 1− 

K ) . X 

L 

2 

L 

. K 



244/454

Q 

Q 

se 

sh 

= 

tan 

⎛ 

⎜ 

⎝ 

2 δ 

max 

2 

⎞ 

⎟ 

⎠ 

δ max ⇒ maximum angular difference 

between the two ends of the line 



245/454

• If δ max ⇒ 30 - 40 o 

• Q se = 7- 13% of Q SL 

• Cost of series capacitor ? 

• Location of series capacitor is not very 

critical 



246/454

Approaches to controllable series compensation 

Variable Z type : 

1. GTO controlled series capacitor (GCSC) 

Objective : Vary V C 

• GTO is closed when v c = 0 

• Open when ‘i’ charges ‘C’ 

• Duality between TCR & GCSC 



247/454

v 

• GTO is turned ON when v c = 0 

for α < ωt < α+γ 

c 

ωt 

1 

∫ ωC 

() t = i() t . d( ωt) 

∴i( t) = I. 

Cos t 

α 

ω 

= 

I 

ωC 

( Sinωt 

− Sinα 

) 

• v c is maximum when 

ωt = π/2 & v c = 0 

when ωt = π-α 



248/454

• Amplitude of the fundamental 

V 

π 

4 2 

c1 = ∫ π 

0 

= 

π 

4 2 

= ∫ 

v 

c 

() t . Sinωt. 

d( ωt) 

I 

C 

.( Sinωt 

− Sinα 

) Sinωt. 

d( ωt) 

π ω 

0 

IX c 

⎡ 2α Sin2α 

⎤ 

⎢ 

1− 

− 

⎣ π π ⎥ 

⎦ 



249/454

Controlling modes 

(a). Voltage compensation mode: 

• GCSC ⇒ Should maintain rated compensation 

voltage when I min 

⇒ V comp = V rated = I min X c 

⇒ As I↑, ↑ αSo that 

V comp is maintained constant 



250/454

(b). Impedance compensation mode: 

V 

c(max) 

I 

max 

= 

X 

c 

Protection issues: 

• Required to have higher short time rating 

• During S.C, ‘I’ could be much higher than I rated 

• I fault > I GTO(rating) 



251/454

• If it flows through ‘C’, V c ↑ 

• ‘V’ across GTO ↑ 

• Use MOV 

Limitations: 

• Harmonics are generated 



252/454

Review 

GTO controlled series capacitor (GCSC) 

• ‘α’ is measured w.r.t peak 

of ‘i’ 

1 

⎛ 

⎞ 

X ( α ) = ⎜1− 

α − Sin α ⎟ 

ωC 

⎝ π π 

2 

C 

⎠ 

α ⇒ extinction angle 

2 

1 

• ‘V C ’ has harmonics 



253/454

TCR 

Contd.. 

GCSC 

• Switch is series with ‘L’ 

• Supplied from a ‘V’ 

source 

• ‘α’ (turn-ON delay) is 

measured w.r.t peak of ‘v’ 

• Switch is parallel with ‘C’ 

• Supplied from a ‘i’ 

source 

• ‘α’ (turn-OFF delay) is 

measured w.r.t peak of ‘i’ 



254/454

Contd.. 

• Control ‘i’ in ‘L’ . 

Parallel with the source 

representing variable 

admittance to the source 

• Control ‘v’ across ‘C’ 

developed by ‘i’ source 

representing variable 

reactance to the source 

V 

ωL 

⎡ 

⎤ 

⎡ 

⎤ 

I LF 

( α ) = 1− 

− 

⎥ ( α ) = 1− 

− 

⎦ 

⎥ ⎦ 

⎢ 

⎣ 

2α 

π 

Sin2α 

π 

V CF 

I 

ωC 

⎢ 

⎣ 

2α 

π 

Sin2α 

π 

⇒ 

α 1 ⎛ 2 1 ⎞ 

( ) = ⎜1− 


ωL 

⎝ π π 

2 

⎠ 

1 2α 

Sin2α 

α = 

⎢ 

1− 

− 

ωC 

⎣ π π 

B ⎡ 

⎤ 

L ( ) 

⎥ ⎦ 

⇒ 

X C 



255/454

Thyristor switched series capacitor (TSSC) 

• Capacitors are disconnected by turning ON 

the thyristors 

• They turn OFF naturally (at Z.C of I ) 



256/454

Voltage compensating mode : 

• Reactance of ‘C’ bank is chosen so as to 

produce average rated V comp = n X C I min 

(‘n’ is the no. of banks) 

• As I ↑ above I min , ↓ n 

• By-pass ‘C’ 



257/454

Impedance compensating mode : 

• TSSC should maintain maximum rated 

compensating reactance at any line current 

up to Rated current (I max ) 

• Maximum series compensation 

nX 

C 

= 

V 

C(max) 

I 

max 

at rated ‘I’ 



258/454

• In FCTCR continuously varying capacitive 

compensation is achieved by varying ‘α’ 

of TCR 



259/454

Thyristor controlled series capacitor (TCSC) 

• If ‘V’ is the applied voltage across the TCR 

• Fundamental component of ‘I’ for ‘α’ 

(measured w.r.t peak of voltage) is 



260/454

V 2 1 

I1 ⎜ Sin 

X ⎝ π π 

⎛ 

⎞ 

( α ) = 1− 

α − 2α 

⎟ 

⎠ 

X 

L 

L 

π 

⎜ 

⎝ π − 2α 

− Sin2α 

⎛ 

⎞ 

( α ) = X 

⎟ 

⎠ 

L 

X < X ( α ) < ∞ 

L L 

⇒ Combined ‘Z’ of TCR & fixed ‘C’ 

X 

TCSC 

= 

⎛ 

⎜ 

⎝ 

− X 

X 

L 

( ) 

( ) ⎟ ⎞ 

C. 

X 

L 

α 

α − X 

C ⎠ 



261/454

• When α = π/2, X L (α) = ∞ 

& X TCSC = -X C 

• When X L (α) = X C X TCSC ⇒ undefined 

• When X L (α) < X C X TCSC ⇒ Inductive 

At X L (α) = X L ⇒ 

X 

TCSC 

= 

⎛ 

⎜ 

⎝ 

X 

X 

L 

C 

. X 

− X 

L 

C 

⎟ ⎞ 

⎠ 



262/454



263/454

• Continuously varying series capacitor by 

‘α’ control 

ωL 

< X α 

L 

( ) < ∞ 

• When 

X 

L 

( α ) < ∞, 

X 

TCSC 

= 

X 

C 

=1 

ωC 

• At X L (α) = X C ⇒ parallel resonance, 

X TCSC 

⇒ 

∞ 

∴ω =1 

LC 

As L(α) > L 

( ) 

⇒ ω > ω α 

o 



264/454

• If X L (α) < X C , There are two operating zones 

α 

C(lim) 

≤ 

α 

≤ 

π 

2 

⇒ Capacitive, ‘i’ leads V C 

0 ≤α≤ α L(lim) ⇒ X TCSC is inductive 

• Not exactly similar to TCR 

connected in parallel With 

‘V’ source 

• Input ‘V’ is sinusoidal 



265/454

• In TCSC, the ‘V’ is voltage across ‘C’ 

• Switch is open ⇒ TCR is O.C, ‘i’ flows through ‘C’ 

• Turn-on TCR at ‘α’ 

(w.r.t peak of ‘v’) 

⇒ ‘i’ is +ve & ‘v c ’is -ve 



266/454

• ‘V C ’ gets distorted 

• In phasor form ‘i’ leads V C 

in capacitor zone 



267/454

• In inductive zone, ‘i’ lags V C 

• TCR current is high 

X 

TCSC 

= 

⎛ 

⎜ 

⎝ 

− jX 

j( 

X 

C 

TCR 

. jX 

− X 

TCR 

C 

) 

⎞ 

⎟ 

⎠ 

= 

⎛ 

⎜ 

⎝ 

− jX 

C 

( 1− 

X 

C 

X 

TCR 

⎞ 

⎟ 

) ⎠ 

i 

TCR 

= 

− 

j( 

X 

TCR 

jX 

− 

C 

X 

C 

) 

. I 

= 

I 

( 1− 

X X ) 

TCR 

C 



268/454

• If X TCR = 1.5X C ⇒ Capacitive 

X 

X 

TCSC 

C 

= 

⎛ 

⎜ 

⎝ 

(1 − 

X 

C 

1 

X 

TCR 

) 

⎞ 

⎟ 

⎠ 

= 

• If X TCR = 0.75X C ⇒ Inductive 3 

1 

1−1 1.5 

= 

I TCR 

I 

= 

1 

1−1.5 

= −2 

X = 0. 75X 

TCR 

C 



269/454

X 

X 

TCSC 

C 

= 

1 

1−1 

0.75 

= −3 

I TCR 

I 

= 

1− 

1 

0.75 

= 

4 

• For same magnitude of X TCSC , I TCR in ‘C’ 

zone = (1/2)I TCR in ‘L’ zone 



270/454

Modes of operation 

By pass mode : 

• ‘i L ’ is continuous & sinusoidal 

• Each thyristor conducts for 180 o 

• X TCSC ⇒ inductive 

• Most of the line ‘I’ flow through ‘L’ not ‘C’ 

• Used to protect ‘C’ against over voltage 



271/454

Thyristor blocked mode : 

• No ‘i’ through ‘L’ 

• Fixed ‘C’ ⇒ Avoided 

Vernier control 

• Thyristors are gated and they conducts 

for part of cycle 

• X TCSC ↑ as conduction angle ↑ from zero 

to α C(lim) 



272/454

Static Synchronous Series Compensation 

• Function of series capacitor ⇒ produces an 

appropriate voltage of fundamental ‘F’ in 

quadrature with Tr. Line ‘I’ 

P 

= 

V V 

( X − X ) 

L 

S 

R 

C 

Sinδ 



273/454

• Instead: Use VSI to inject a voltage in 

quadrature with ‘i’ 

V = ± j. 

q 

V q 

( γ ) 



274/454

• Voltage across ‘L’ ⇒ V ( ) 

L 

= 2VSin 

δ 2 + Vq 

I 

= 

2VSin 

δ 

( 2) 

X 

+ V 

q 

( ) 

( δ 2) . 2VSin( δ ) 

P = VCos 

2 + 

V q 

= 

V 

X 

2 

Sinδ 

+ 

V. 

V 

X 

q 

Cos 

( δ 2) 

• If V q > I.X, power flow will reverse 



275/454

T.C.S.C : 

Review 

• Used for vernier control of ‘C’. 

GCSC also provides this feature 

• Cost of GTO > that of thyristor 

• Effective capacitive 

compensation increases 

as α↓from π/2 to α C(lim) 



276/454

Contd.. 

• For both region X L < X C (inductive & capacitive) 

• In inductive zone, I TCR > I Line and are in phase 

• In capacitive zone, I Line is out of phase with I TCR 

• ‘V’ across ‘C’ gets distorted 



277/454

Contd.. 

Static Synchronous Series Compensation: 

• Instead of passive elements 

use VSI 

P 

= 

V 

X 

2 

Sinδ 

+ 

V. 

V 

X 

q 

Cos 

( δ 2) 

• Reverse power flow is possible 



278/454

Control range: 

• Voltage compensation mode : SSSC can 

maintain the rated capacitive or inductive 

compensating ‘V’ for ‘I’ till I q(max) 

• Ideal condition (‘I’ line 

can not be zero) 

• ΔP is required for SSSC 



279/454

impedance compensation mode : 

• Maintain rated X C or X L 

up to rated I 

Exchange of Active power by SSSC: 

• Can exchange active as well as reactive power 

• Some active source should be connected to 

DC side 



280/454

• Compensation for both reactive and resistive 

compensation of series line impedance to keep 

X/R ratio high (3-10 is desirable) 

• With series compensation 

effective ( X X ) R ratio ↓ 

L − 

C 



281/454

• X/R ratio in case1 > X/R ratio in case2 

• Reactive component of 

I q 

( 

1 

2 ) 

= I. Sin δ + ϕ 

↑ 

( 

1 

δ 2 + ) 

• Real component of I = Ia 

= I. Cos ϕ 

transmitted to the receiving end decreases 

corresponding to R=0 



282/454

• If V S = V R =V 

Per phase power received by the receiving end 

P 

( 90 +δ −ϕ) 

= V. I. 

Cos 2 

( ϕ − 2) 

= V. I. 

Sin δ 

2VSinδ 

/ 2 

= V. . Sin δ 

Z 

( ϕ − 2) 

= 

2 V 

2 . Sinδ 

/ 2 

Z 

δ 

{ Cosδ 

/ 2. Sinϕ 

− Cosϕ. 

Sin / 2} 



283/454

= 

2 2 

V 

2 

Z 

{ Sinϕ. 

Sinδ 

/ 2. Cosδ 

/ 2 − Cosϕ. 

Sin δ / 2} 

2 

V 

= 1 

Z 

{ Sinϕ. 

Sinδ 

− Cosϕ. 

( − Cosδ 

)} 

2 

V ⎧ X R 

= ⎨ . Sinδ 

− . δ 

Z ⎩ Z Z 

( 1− 

Cos ) ⎬ ⎫ 

⎭ 

2 

V 

= 1 

2 2 

R + X 

{ X. 

Sinδ − R. 

( − Cosδ )} 



284/454

⇒ Reactive VA associated with the receiving end 

Q 

( 90 +δ / −ϕ) 

= VI.Sin 2 

2V 

2 Sinδ 

/ 2 

= Cos 2 − 

Z 

( δ / ϕ) 

2 

= V 

R + X 

1 

2 2 

{ R. 

Sinδ + X ( − Cosδ )} 



285/454

• Maximum transmittable active power ↓ 



286/454

Voltage & phase angle regulators 

Voltage regulator: 

• Injection of appropriate in phase 

component in series with ac system 

• Similar to transformer tap changer 



287/454

Phase angle controller : 

• Inject ‘V’ at an angle ±90 o 

relative to the system ‘V’ 

• Resultant angular change approx. proportional 

to injected ‘V’. Magnitude of ‘V’ is constant 



288/454

Power flow control : 

• Optimal loading of transmission line in 

practical system can not always be achieved 

at the prevailing angle 

Occur when ? 

• Power between two buses is transmitted 

over parallel lines of different length, use 

phase angle regulator (PAR) 



289/454

PAR : A sinusoidal synchronous ac voltage 

source with controllable amplitude and 

phase angle 

V + 

Seff 

and 

V = 

V 

= VS 

Vr 

S Seff 



290/454

• Basic idea is to keep the transmittable 

power at the desirable level 

independent of prevailing ‘δ’ 

also 

V r 

V S 

> 90 o 

⇒ angle to be controlled 

is (δ-σ ) 

2 

V 

P = Sin 

X 

( δ −σ ) 



291/454

• Multi functional FACTS controller : 

based on back-back VSI with a common 

DC-link 

• One converter in series (SSSC) and other 

is in shunt (SVC) ⇒ unified power flow 

controller (UPFC) 

• Both converters are connected in series but 

in two different lines (Inter line Power Flow 

Controller-IPFC) 



292/454

UPFC : 

• Able to control simultaneously or 

selectively all the parameters affecting the 

power flow in Tr. line 



293/454

• Converter-1 supplies active power 

required by converter-2 



294/454

• Independently control the reactive power 

flow at the point of connection 

UPFC can fulfill 

• Reactive power control 

• Series compensation 

• Phase angle regulator 



295/454



296/454

Case1 : 

Control capabilities 

ρ = 0, 

Voltage regulator 

V pq 

= ± ΔV 

• Similar to tap changing transformer with 

large no. of steps 

Reactance compensator : Series reactive 

compensator 

V pq = V q at 90 o with I 

⇒ Similar to SSSC 



297/454

Phase angle regulator : 

V pq 

= V σ 

⇒ at any angular relationship w.r.t V S 

so that desired phase shift is achieved 

Multi functional feature : 

V 

pq 

= ΔV 

+ Vq 

+ V σ 



298/454

U.P.F.C : 

Review 

• Two VSI connected back to back with 

common DC-link 

• One connected in series with line and other is 

connected across the line 

• DC-link ‘V’ is maintained 

constant by converter-1 



299/454

Contd.. 

• Active power required by the 

system is drawn by converter-1 

Can function as 

• Voltage regulator ⇒ V+ΔV 

• SSSC ⇒ injects ‘V’ in quadrature with ‘I’ 

• Phase angle regulator ⇒ injects ‘ΔV’ in 

quadrature with ‘V’ 



300/454

Using UPFC 

• Active power flow and 

• Reactive power flow can be set 

• In SSSC : Quadrature injected ‘V’ 

results in increase in power flow 

⇒ Magnitude of injected ‘V’ determines ‘P’ 

⇒ Circuit conditions determines ‘Q’ 



301/454

• Main function : Control the flow of ‘P’ & ‘Q’ 

by injecting a voltage in series with the 

Tr. line 

• Both magnitude & phase angle are varied 

• Control of ‘P’ & ‘Q’ allows power flow in 

prescribed routes 

⇒ 2 port representation 



302/454

⇒ A common DC-link voltage is regulated 

Re( ) 

* * 

V I 1 V I 2 − P 0 

u1 + 

u2 

loss 

= 

• In addition to maintain real power balance, 

shunt branch can independently exchange 

reactive power with the system 



303/454

• Transmitted active power and reactive power 

supplied by receiving end 

P 

r 

− 

jQ 

r 

= V 

r 

⎛V 

. ⎜ 

⎝ 

S 

+ V 

pq 

jX 

−V 

r 

⎞ 

⎟ 

⎠ 

* 

V = Ve 

S 

jδ 2 

− 

V r 

= Ve 

jδ 2 

V 

pq 

= 

V 

pq 

e 

j 

( δ 2+ρ 

) 



304/454

= Ve 

⎧V 

⎨ 

⎩ 

( Cosδ 2 − jSinδ 

2 − Cosδ 

2 − jSinδ 

) Vpq 

− j( δ 2+ 

ρ ) 

− jδ 

2 2 

− 

jX 

− 

jX 

e 

⎫ 

⎬ 

⎭ 

= Ve 

⎧ 

⎨ 

⎩ 

VSin 

X 

Vpq 

− 

jX 

− jδ 2 2 δ 2 

− j 2 

e 

( δ + ρ ) 

⎫ 

⎬ 

⎭ 

= 

V V V 

Sinδ 

2 

. 

X 

jX 

2 2 

− j( δ + ρ ) 

( Cosδ 

2 − jSinδ 

2) − 

pq e 

= 2 2 

V 

V. 

V 

2 

jSin 

X 

jX 

pq 

( Sinδ 2. Cosδ 

2 − jSin δ 2) − Cos( δ + ρ ) − ( δ + ρ ) 

( ) 



305/454

P 

r 

− 

jQ 

r 

= 

V 

X 

2 

V. 

Vpq 

Sinδ − Sin + 

X 

( δ ρ ) 

− 

⎧ 

j⎨ 

⎩ 

V 

X 

2 

2 2 

Sin 

V. 

V 

δ 2 − 

X 

pq 

Cos 

( δ + ρ ) 

⎫ 

⎬ 

⎭ 

∴P 

r 

= 

V 

X 

2 

V. 

Vpq 

Sinδ − Sin + 

X 

( δ ρ ) 

∴Q 

r 

= 

2 

2 2 

V 

X 

Sin 

V. 

Vpq 

δ 2 − Cos + 

X 

( δ ρ ) 



306/454

• ‘ρ’ can vary from 0 to 2π 

• ‘P’ & ‘Q’ are controllable from 

P 

V. 

V 

pq 

( δ ) − to P( δ ) 

X 

+ 

V. 

V 

X 

pq 

⇒ Transmitted real power 

2 

V V. 

V 

= Sinδ 

± 

X X 

( ) 

pq max 



307/454

Control strategy: 

• There are 3 degrees of freedom 

• Magnitude and angle of series V 

• Shunt reactive current 

⇒ Both are VSI 

⇒ Series injected ‘V’ can be instantaneously 

changed 



308/454

⇒ Shunt current is controlled indirectly by 

varying output of shunt converter 

Series injected ‘V’ control : 

• Injected ‘V’ can be split into two components 

1. In phase with line ‘I’ 

2. In quadrature with line ‘I’ 

• ‘P’ can be controlled by varying series reactance 

of the line 



309/454

• Reactive ‘V’ injection ⇒ similar to series 

connection of reactance except that injected ‘V’ 

is independent of Tr. Line ‘I’ 

Shunt current control : 

• Shunt current can be split into real & reactive 

components 

• Magnitude of real component ⇒ DC link ‘V’ 

• Magnitude of reactive component ⇒ Bus ‘V’ 

magnitude regulator 



310/454

FACTS installments in India 

• TSC+TCR (400 kV) at Kanpur ⇒ ±240 MVar 

• TCR (400 kV) at Itarsi ⇒ ±50 MVar 

• TCSC (400 kV ) at ⇒ Raipur - Rourkela 

(Double ckt.) 

⇒ Gorakhpur - Mazaffarpur 

⇒ Kanpur - Ballabhgarh 



311/454

Kanpur – Ballabhgarh 400 kV line: 

Fixed capacitor 

TCSC 

Rated V L-L 420 kV 420 kV 

Nominal Var 151.60 MVar 79.87 MVar 

Rated continuous 

‘V’ across ‘C’ 

42.2 kV 16.6 kV 

TCR/ph 

- 

4.4 mH 



312/454

HVDC 

• Long distance transmission ( Competing 

technology : AC with FACTS) 

• Cable transmission (> 40 Km) ⇒ HVDC 

• Asynchronous link ⇒ HVDC 

• HVDC lines are cheaper than AC lines 

• Terminal equipment costs are higher 



313/454

In India : 

• Long distance HVDC 

• Rihand – Dadri : 1500 MW, ±500 kV 

• Chandrapur – Padghe : 1500MW, ±500 kV 

• Talcher – Kolar : 2000MW, ±500 kV 

• Barsur– Lower Sileru : 200MW, 200 kV 



314/454

Back to Back : 

• Chandrapur – Ramagundam : 1000 MW 

(Asynchronous link) 

• Jeypore – Gajuwaka : 500 MW 

(Asynchronous link) 

• Vindhyachal 

: 500 MW 

• Sasaram : 500 MW 



315/454

• ‘P’ through DC link can be regulated. 

• Power control through firing angle control 



316/454

• ‘P’ through link can not be regulated 



317/454

• P 1 + P 2 can be regulated 

• If alternator-1 generates 1000 MW & 

load 1100 MW 



318/454

• If alternator-2 generates 1000 MW & 

load 900 MW 

• P 1 +P 2 has to be -100 MW 

(frequency of alternator-1 &2 are same) 

• P 1 + P 2 can be set 



319/454

Types of HVDC system 

Two terminal : with DC transmission line 

One rectifier terminal + one inverter terminal 

Back to Back : 

• Two terminals with no DC line ⇒ used for 

asynchronous link 

Multi terminal : with DC line and several rectifier 

and/or inverter terminals connected to more than 

two nodes of AC network 



320/454



321/454

Types of links : 

• Mono-polar 

• Bi-polar 

Mono-polar HVDC link : 



322/454

• One conductor (generally –ve) 

• Return path ⇒ ground ⇒ Resistance should 

be low 

• Instead metallic return 



323/454

Bi-polar HVDC link : 

• Has two conductors 

+ve 

-ve 



324/454

• Each terminal has two converters of equal 

rating ‘V’ connected in series on the DC side 

• Junction is grounded 

• ‘I’ in two phases are equal 

• No ground ‘I’ 

• Two poles can operate independently 

• If one is faulty, then other can operate with 

ground as the return 



325/454

Review 

HVDC 

• Asynchronous link 

• Back to back 



326/454

Components of HVDC transmission 

Bi-polar 

HVDC 



327/454

Converter : 

• Perform AC – DC conversion 

DC – AC conversion 

• 12 pulse converter 

Transformer with tap changer 



328/454

Smoothing Reactor : Large value of ‘L’ in 

Series with each pole 

Purpose : 

• ↓ harmonic voltage & current in DC line 

• Prevents ‘I’ from being discontinuous on 

light load 

• Limit the ‘I’ during S. C in the DC line 



329/454

Harmonic filter : 

• Converter generates 

harmonic currents 

• Because of source ‘L’, ‘V’ gets distorted 

• Affects the other loads & interference 

with communication network 



330/454

Reactive power support : 

• Both converter & inverter absorb 

reactive power 

• As α↑, ‘Q’ requirement ↑ 

• ‘Q’ source is a must 

• If bus is strong, shunt capacitor can be used 

• ‘C’ associated AC filter also supply ‘Q’ 



331/454

Basic module of converter : 

• 3-ph full bridge 

V an 

= V∠0 

= V∠ −120 

V cn 

= V∠ − 240 

V ab 

= 3V∠ 

π 6, 

= 3V∠− 

π 2, 

V bn 

V bc = 3V∠ 

− 210 

V ca 



332/454

• If α 1 is trigger angle for bridge-1 

• If α 2 is trigger angle 

for bridge-2 

⇒ Neglect i dc r dc & 

Assuming ideal devices 

α = π −α 2 1 



333/454

⇒α = 30 o (w.r.t natural 

commutation) 

or 

⇒ corresponding to Z.C of 

phase-A α = 60 o 



334/454

⇒T 1 is turned off at ωt= 30+ (30+120) = 180 o 

When T 3 is triggered, ‘V’ across T 1 = V ab 



335/454

V a 

= Sin180 = 0, 

2 

V b 

2 

= Sin60 = 

3 

∴V 

ab 

= − 

3 



336/454

At ωt = 210 o 

V a 

= Sin210 = −1 

= Sin90 = 1 

V b 

3 

2 

V ab 

= −1.5 

At ωt = 240 o 

V a 

− 3 2, V = 2 

= 

b 

3 

∴V 

ab 

= − 



337/454

At ωt = 270 o 

V a 

−1 , V = 1 

= 

b 

2 

∴V 

ab 

= −1.5 

At ωt = 300 o - 

V − 3 2, V = 0 V = − 3 2 

a 

= 

b 

∴ ab 

At ωt = 300 o + , T 5 is triggered, ‘V’ across T 1 is V ac 

V − 3 2, V = 3 2 V = − 3 

a 

= 

c 

∴ ac 



338/454

At ωt = 330 o 

V a 

−1 2, V = 1 

= 

c 

∴V 

ac 

= −1.5 

At ωt = 360 o 

V 0 , V = 3 2 V = − 3 2 

a 

= 

c 

∴ ac 

At ωt = 30 o 

V 1 2, V = 1 2 V = 0 

a 

= 

c 

∴ ac 



339/454

At ωt = 60 o - 

V 3 2, V = 0 V = 3 2 

a 

= 

c 

∴ ac 

⇒ T 1 is reverse biased for 210 o 

What happen when α = 150 o 

T 1 is turned off at ωt = 30+150+120 = 300 o 

(w.r.t +ve Z.C of Ph- A) 



340/454

At ωt = 300 o 

V − 3 2, V = 0 V = − 3 2 

a 

= 

b 

∴ ab 

At ωt = 330 o 

V −1 2, V = −1 

2 V = 0 

a 

= 

b 

∴ ab 

V 

a 

At ωt = 360 o 

0, 

V = − 2 

= 

b 

3 

∴V ab 

= 3 2 = + ve 



341/454

⇒ T 2 must attain forward voltage blocking 

capability within 30 o 



342/454

Vdc = 2.34V 

ph. 

Cosα 

=1.35V LL 

.Cosα 

For α = 30 o 



343/454

Review 

HVDC 

• Two six pulse converters 

connected in series 

α 

= π − 

2 

α 1 



344/454

Contd.. 

• As α 1 ↑ (AC-DC converter), ‘Q’ requirement 

also ↑ 

• As α 2 ↑, duration for which 

the devices is reverse biased↓ 

• When α = 150 o , duration for which the devices 

is reverse biased = 30 o 



345/454

Harmonic component in converter i/p : 

• No even harmonics, only odd harmonics 

π 3 

2 

2I Ln 

= ∫ I0Cosnθ. 

dθ 

π 

−π 

3 

6 

I L 1 

= . I 0 

, I 

L 3 

= 0 

2 ⎛ nπ 

⎞ π 

I Ln 

= . I0⎜2Sin 

⎟ 

2nπ 

⎝ 3 ⎠ 

I 

I 

L1 

L1 

I 

L5 

= − , I 

L7 

= − 

7 

5 



346/454

Phase relationship between phase V & I 1 

Neglect losses 

V 

⎛ Vm 

⎞ 

= 3⎜ 

⎟ 

⎝ 2 ⎠ 

dcI0 . I 

L1 

Cosϕ 

⎛ Vm 

⎞ 6 3 3 

3. 

⎜ ⎟. 

I0 Cosϕ 

= VmCosα. 

I 

⎝ 2 ⎠ π 

π 

Cosϕ 

= Cosα 

∴ϕ 

= α 

0 



347/454

α + 60 

6 

V0 = ∫Vab. 

dωt 

2π 

α 

α + 60 

6 

= ∫ 

2π 

α 

3V 

m 

Sin 

( 

o 

ωt 

+ 60 ). 

d t 

ω 

= 

3 3 

Vm Cosα 

= V 

π 

dco 

Cosα 

= 

3 3 

π 

2V rms 

Cosα 

= 2 .34V 

Cosα = 1. V Cosα 

rms 

35 

LL 



348/454

As α↑: 

• V dc ↓ 

• Displacement angle ↑ & P.F ↓ 

• Q ↑ 

Effect of source L : 

• T 1 , T 2 when conducting 

T 3 is triggered 



349/454

i + = 

i I 

1 3 0 

di 1 

di 

= − 

3 

dt dt 

V 

ba 

= 2L 

c 

di 

dt 

3 V Sinωt 

= 2L 

m 

3 

c 

di 

dt 

3 

∴i 

3 

= − 

3V 

m 

Cosωt 

+ 

2ωL 

c 

K 



350/454

Boundary conditions : 

At ωt = α, i = I i = −I 

, i 0 

, = 

1 0 2 0 3 

= α+μ, 

i 0 i = I 

1 

= , i2 

= −I0, 

3 

0 

∴i 

3 

= 

3V 

2ωL 

m 

c 

( Cosα 

− Cosωt) 

At ωt = α+μ, i 

3 

= I0 

∴ I 

0 

= 

3V 

2ωL 

m 

c 

( Cosα 

− Cos( α + μ) 

) 



351/454

V 

V 

pn 

pn 

= V 

= V 

an 

bn 

− 

− 

L 

L 

di 

dt 

di 

dt 

1 

3 

2V 

pn 

= V 

an 

+ V 

bn 

− 

⎛ 

L⎜ 

⎝ 

di 

dt 

1 

+ 

di 

dt 

3 

⎞ 

⎟ 

⎠ 

∴V 

pn 

= 

V 

an 

+ V 

2 

bn 

= 

V 

− 

2 

cn 



352/454

∴V 

0 

= 

V pn 

−V mn 

V 

= − 

2 

cn 

−V 

cn 

= −1.5V 

cn 

Reduction in V 0 = (ΔV 0 ) : 

ΔV 

α + μ 

6 

0 

= 

2π 

∫ 

α 

( V + 1.5V 

). 

d t 

bc 

cn 

ω 

α + μ 

6 

= 

2π 

∫ 

α 

3V 

m 

Sin 

o 

( ωt 

+ 60 ) + 1.5V 

Sin( ωt 

−π 

2 ). 

d t 

m 

ω 



353/454

V m 

3 3 

= Cos 

2π 

( Cosα 

− ( α + μ) 

) 

= 

V 

2 

dco 

I 

0 

2ωL 

3V 

c 

m 

= 

3ωL c 

π 

I 

0 

∴V 

0 

= V 

dc0 

Cosα − 

3ωL 

π 

c 

I 

0 



354/454

Representation of inverter mode of 

operation in presence of μ 

−V 

d 

= 

V 

dco 

cosα 

− 

R 

c 

I 

d 



355/454

V = −V 

cosα 

+ 

d 

dco 

R 

c 

I 

d 

= V cos( π −α) 

+ R 

dco 

= V cos β + R 

dco 

c 

I 

d 

c 

I 

d 

α → 

delay angle 

β 

→ Angle of advance 



356/454

Converter 

α ⇒ delay angle 

μ ⇒ overlap angle 

Inverter 

β = π-α ⇒ advance angle 

μ ⇒ overlap angle 

γ = β-μ ⇒ extinction angle 

γ = π-(α+μ) 



357/454

ΔV 

o 

= 

V 

2 

dco 

[cosα − cos( α + μ)] 

V 

d 

= 

V 

dco 

− 

ΔV 

o 

V 

= dco 

2 

[cosα + cos( α + μ)] 

Also 

V 

d 

V 

2 

= dco 

[cosα + cos( α + μ)] 

= [cos( π − β ) 

+ 

cos( π −γ 

)] 



358/454

V dco 

= [cos + cos γ ] − − − − − 

2 

− ( A) 

I 

d 

= 

3V 

2ωL 

m 

c 

β 

)] 

[cosα 

− cos( α + μ 

= 

3V 

2ωL 

m 

c 

[cos( π − β ) 

− cos( π −γ 

)] 

3V 

m 

= [cosγ 

− cos β ] − − − − − − − ( B) 

2ωL 

c 



359/454

Eq. A+B ⇒ 

∴2cos 

2V 

V 

d 

γ = + 

dco 

I 

d 

2ωL 

3V 

c 

m 

V 

d 

= V 

dco 

cosγ − 

3 3 

π 

V 

m 

ωL 

c 

3V 

m 

= V 

dco 

dco 

cosγ 

− 

3ωL 

π 

= V cosγ 

− R 

c 

c 

I 

I 

d 

d 



360/454

12-pulse converter 

• Series connection of two 6-pulse converters 

3-Φ voltages supplied to one 

bridge is displaced by 30 o 

from those applied to 2 nd bridge 



361/454

• DC voltage is doubled 

• Harmonic spectrum has improved 

12n ± 1 on AC side 

12n on DC side 



362/454



363/454

Relation between Ac and DC quantity : 

With multi phase bridge 

If ‘Β’ no. of bridges in series 

∴V =1.35. B. 

T. 

do 

V L 

⇒ No load 

Corresponding voltage drop : 

Output 

3 

V = Vd 

= VdoCosα − Id 

. B. 

X 

π 

3 

I d 

X C 

π 

C 

bridge 



364/454

V 

d 

⎛ 3 

= VdoCosα 

− Id 

. B. 

⎜ X 

C 

⎝ π 

⎞ 

⎟ 

⎠ 

⎛ 3 

= VdoCosγ 

− I 

d 

. B. 

⎜ X 

C 

⎝ π 

⎞ 

⎟ 

⎠ 



365/454

Summary of technical data of Padghe 

• Nominal line voltage ⇒ 400 kV 

• Maximum line voltage ⇒ 

430 kV 

• Minimum line voltage ⇒ 380 kV 

• Total ‘Q’ at both stations ⇒ 800 MVar 

⇒ 4*200 MVar 

• 12 th harmonic filter ⇒ 2*120 MVar 



366/454

• 24/36 harmonic filter ⇒ 2*80 MVar 

Power : 

• Nominal Power ⇒ 2*750 MW 

• Minimum (single pole) ⇒ 2*75 MW 

• 2 hours overload ⇒ 

2*825 MW 

• 5 Sec. overload ⇒ 2*1000 MW 



367/454

Direct voltage : 

• Nominal line voltage ⇒ 

500 kV 

• Maximum line voltage ⇒ 512 kV 

• Minimum line voltage ⇒ 488 kV 



368/454

Direct current : 

• Nominal I ⇒ 1500 A 

• Maximum I at nominal load ⇒ 1542 A 

• Max. I at 2 hour over load ⇒ 1695 A 

• Max. I at 5 sec. over load ⇒ 2140 A 

‣ Nominal line resistance = 7.5 Ω 



369/454

Rectifier firing angle : 

• Minimum ‘α’ ⇒ 5 o 

• Mini. ‘α’ during normal operation ⇒ 12.5 o 

• Max. ‘α’ during normal operation ⇒ 17.5 o 

Inverter firing angle : 

• Minimum ‘γ’ ⇒ 16 o 

• Max. ‘γ’ during normal operation ⇒ 18 o 



370/454

Basic control : 

• DC voltage or I (or power) can be controlled 

by controlling the internal voltage (V dcor Cosα) 

and V dcoi Cosγ 

⇒ Gate control or using tap changing of 

converter transformer 

⇒ Gate control is fast 



371/454

⇒ Tap changing : Slow ( 5-6 sec/step) 

⇒ Gate control is used for initial rapid 

control action 

⇒ Followed by tap changing to restore the 

converter quantities ( ‘α’ of rectifier & ‘γ’ 

for inverter) to their normal ranges 



372/454

Basis for selection of control : 

Following considerations influences the selection 

of control characteristics 

• Prevention of large fluctuations of DC 

current due to variation in AC system 

• Maintaining DC voltage near rated value 

• Maintaining power factor at the sending & 

receiving end that are as high as possible 

• Prevention of commutation failure in inverter 



373/454

• Rectifier control ⇒ To prevent large 

fluctuations in DC current 

I 

d 

= 

V 

dcor 

Cosα 

−V 

R + R − 

cr 

L 

Cosγ 

R 

dcoi 

ci 



374/454

• Denominator is very small 

• A small change in V dcor or V dcoi cause a large 

change in I d 

• 25% change either in V dcor or V dcoi changes 

‘i d ’ by 100% 

• If ‘α’& ‘γ’ are kept constant, I dc can vary 

over a wide range for small change in i/p 

AC voltage at either end 



375/454

• Not acceptable 

• Rapid converter control prevents fluctuation 

of I dc 

• For a given power transmitted V dc profile 

along the line should be close to rated values 

• It minimizes I d & therefore line loss 



376/454

• P.F should be as high as possible 

• Minimize losses and current rating of 

equipment in the AC system 

• Reduce the voltage drop at the AC terminal 

as load ↑ 

• ↓ the cost of reactive power supply to line 



377/454

• So keep the rated power of the converter 

as high as possible for a given ‘V’ & ‘I’ rating 

of transformer 

• P.F depends on ‘α’& ‘γ’ 

α min = 5 o (a +ve ‘V’ should appear across 

the device) 

• Normally operate at 15 – 20 o , so that 

V dcor can be ↑ to control DC power flow 



378/454

• γ ⇒ necessary to maintain a certain minimum 

extinction angle to avoid commutation failure 

• Device should attain forward 

voltage blocking capability 

γ = 

β − μ 

= 15 o at 50 Hz 

μ ⇒ depends on I d & i/p ‘V’ 



379/454

Control of HVDC system 



380/454

I 

d 

= 

V 

dcor 

Cosα 

−V 

R + R + 

cr 

L 

Cosγ 

R 

dcoi 

ci 

Power at rectifier terminal, P dr = V dc .I d 

Power at inverter terminal = V di .I d 

= P dr -i d2 R L 



381/454

Control characteristics 

Ideal characteristics : 

• Voltage regulation & 

current regulation 

Kept distinct & are 

assigned to separate 

terminals 

• Under normal operation : 

⇒ Rectifier maintains current control (CC) & 

⇒ Inverter operates constant extinction angle 

(CEA) 



382/454

• Maintains adequate commutation margin 

• V dc ⇒ measured at the rectifier terminals 

• Inverter characteristics includes I d .R L drop 

V 

d 

= V Cosγ 

+ 

dcoi 

( R ) L 

− R ci 

I d 



383/454

• Rectifier characteristics can be shifted 

horizontally by adjusting reference current 

or current command or current order 

• If measured current < current command, 

controller ↓ α 

• Inverter characteristics can be raised or 

lowered by means of transformer taps 



384/454

• As taps are changed, CEA regulator quickly 

restores desired γ 

• I d changes 

• Current regulator of rectifier changes ‘α’ 

and control ‘i’ 

• Tap changer of rectifier acts to bring ‘α’in 

the desired range (10-20 o ) 



385/454

Review 

Rectifier firing angle : 

• Minimum ‘α’ ⇒ 5 o 

• Mini. ‘α’ during normal operation ⇒ 12.5 o 

• Max. ‘α’ during normal operation ⇒ 17.5 o 

Inverter firing angle : 

• Minimum ‘γ’ ⇒ 16 o 

• Max. ‘γ’ during normal operation ⇒ 18 o 



386/454

Basic control : 

• DC voltage or I (or power) can be controlled 

by controlling the internal voltage (V dco Cosα) 

and V dco Cosγ 

⇒ Gate control or using tap changing of 

converter transformer 

⇒ Gate control is fast 



387/454

⇒ Tap changing : Slow ( 5-6 sec/step) 

⇒ Gate control is used for initial rapid 

control action 

⇒ Followed by tap changing to restore the 

converter quantities ( ‘α’ of rectifier & ‘γ’ 

for inverter) to their normal ranges 



388/454

Basis for selection of control : 

Following considerations influences the selection 

of control characteristics 

(a). Prevention of large fluctuations of DC 

current due to variation in AC system 

R ≈ 10 Ω and L =250 mH ⇒ Back to back 

L =1H ⇒ for long line 

τ =20 m.sec ⇒ roughly 



389/454

• Simulation study taking line L, R & C in 

addition L filter is required 

(b). Maintaining DC voltage near rated value 

(c). Maintaining power factor at the sending & 

receiving end that are as high as possible 

(d). Prevention of commutation failure in inverter 



390/454

• Rectifier control ⇒ To prevent large 

fluctuations in DC current 

I 

d 

= 

V 

dcor 

Cosα 

−V 

R + R − 

cr 

L 

Cosγ 

R 

dcoi 

ci 



391/454

• γ ⇒ necessary to maintain a certain minimum 

extinction angle to avoid commutation failure 

• Device should attain forward 

voltage blocking capability 

γ = 

β − μ 

= 15 o at 50 Hz 

μ ⇒ depends on I d & i/p ‘V’ 



392/454

Control of HVDC system 



393/454

Control characteristics 

Ideal characteristics : 

• Voltage regulation & 

current regulation 

Kept distinct & are 

assigned to separate 

terminals 

• Under normal operation : 

⇒ Rectifier maintains current control (CC) & 

⇒ Inverter operates constant extinction angle 

(CEA) 



394/454

• Quantities forming the co-ordinates are 

measured at some common point in the DC line 

• Converter terminal can be one such possibility 

V 

d 

= V Cosγ 

+ 

dcoi 

( R ) L 

− R ci 

I d 

• Has a small –ve slope 



395/454

• Maintains adequate commutation margin 

• Inverter characteristics includes I d .R L drop 



396/454

• Rectifier characteristics can be shifted 

horizontally by adjusting reference current 

or current command or current order 

• If measured current < current command, 

controller ↓ α 

• Inverter characteristics can be raised or 

lowered by means of transformer taps 



397/454

• As taps are changed, CEA regulator quickly 

restores desired γ 

• I d changes 

• Current regulator of rectifier changes ‘α’ 

and control ‘i’ 

• Tap changer of rectifier acts to bring ‘α’in 

the desired range (10-20 o ) 



398/454

• Constant current characteristics could be a 

line parallel to y-axis 

• If proportional controller ⇒ slope could be -ve 

• Generally current control is 

given to both the converters 

• Ref. current for rectifier > Ref. current for 

inverter 



399/454

• I ref(conv) –I ref(inv) = I margin = +ve 

• Assume that power flows in the line to be ↑ 

• α conv ⇒ takes the value of α min 

• Incase I d approaches I ref(conv) , then 

⇒ rectifier is working under constant ignition control 

⇒ Inverter is working under constant extinction control 



400/454

• After some time, tap changer changes the tap 

⇒ ‘α’ of the converter ↑ to attain its normal 

operating value (12- 17 o ) 



401/454

Actual characteristics : 

• Rectifier maintains constant ‘I’ by changing ‘α’ 

• ‘α’ can not be < α min 

• Once α min is reached, no further ↑‘V’ is possible 

• Rectifier will operate constant ignition angle 

(CIA) 



402/454

• Therefore rectifier characteristics has two 

segments (AB & FA) 

• Constant current 

characteristics may not be 

truly vertical 

⇒ Depends on the current 

regulator 

• With proportional control 

C.C characteristics has – ve slope 



403/454

∴V 

dco 

Cosα 

= 

K 

[ I − I ] 

order 

d 

= V + R 

d 

cr 

I 

d 

I ord ⇒ current order 

V = KI − + 

d 

order 

( K R ) cr 

I d 

ΔV 

d 

= − 

( K + R ) 

cr 

ΔI 

d 

∴ 

ΔV 

ΔI 

d 

d 

= − 

( K + R ) 

cr 

⇒ (with PI it is vertical) 



404/454

• At normal voltage , characteristics is defined 

by FAB 

• At reduced ‘V’, it 

shifts down ⇒ F 1 A 1 B 1 

• CEA characteristics of the inverter intersect 

at ‘E’ for normal ‘V’ condition 



405/454

• At reduced ‘V’, it does not intersect F 1 A 1 B 

• A big reduction in rectifier ‘V’ would cause 

I d & ‘P’ ↓ 

⇒ System could shut down 



406/454

• In order to avoid the problem, inverter is 

provided with current control 

• Inverter I ord < rectifier I ord 

I ord(R) –I ord(I) ≈ 0.1 I rated 



407/454

• Under normal condition 

• Rectifier ⇒ C. C 

• Inverter ⇒ CEA 

• When i/p ‘V’ ↓ ⇒ rectifier ‘V’↓ 

⇒ Operating point E 1 

• Changes from one mode to another is known 

as mode shift 



408/454

• When inverter is on current control 

V 

V 

d 

doi 

= 

R 

L 

I 

d 

− 

Cosγ 

= V 

R 

d 

ci 

I 

− 

d 

R 

+ V 

L 

I 

d 

doi 

+ 

Cosγ 

With proportional controller 

R 

ci 

I 

d 

− 

( I ) 

ord 

− I 

d 

= Vd 

− RLI 

d 

RcrId 

K + 



409/454

Δ 

( V Cosγ 

) = −KΔ( I − I ), 

doi 

ref 

d 

K >1 

= ΔV 

d 

− ΔI 

d 

( R − R ) 

L 

ci 

d 

( R L 

− R ci 

K ) I d 

ΔV = Δ + 

ΔV 

ΔI 

d 

d 

= 

( R L 

− R ci 

+ K ) I d 

⇒ Slope is +ve 

⇒ ↑ V dor to ↑ i d 

⇒ ↓ V doi to ↑ i d 



410/454

When does change over take place ? 

• Current order is given to both the converters 

I ref(C) > I ref(I) 

I ref(C) > I ref(I) -I margin ⇒ +ve (assume) 

I margin = 0.1 – 0.15 I rated 



411/454

• Assume that i/p AC has dipped due to fault, 

I dc ↓ , 

‣ α conv ⇒ α min 

and with this new value 

of ‘α’, I dc is ↓ 

• If I dc < (I ref(C) -I mar ), inverter takes over the 

current control & converter is working under 

C.I.A, after some time tap changer changes the tap 



412/454

Review 

Rectifier 

characteristics 

Constant current 

by ‘α’ control 

Constant ignition 

angle control 

C.C 

Can have a –ve slope 

Can be parallel to Y-axis 



413/454

Contd.. 

• Current control is given to both converters 

But I ref(R) > I ref(I) 

I ref(R) -I ref(I) = I margin ≈ 0.1I rated 

• Current control loop of inverter is inactive 

when current ≈ I ref(R) 



414/454

• ‘e’ is –ve, ‘K’ is +ve ⇒ I act should be ↓ 

I act > (I ref –I mar ) 

⇒ ‘γ’ should be decreased 

⇒ o/p of PI is zero 

⇒ selector switch selects γ min 



415/454

• ‘e’ is +ve, I act < (I ref –I mar ) 

⇒ ‘γ, should be ↑ , so that I act ↑, ‘K’ is +ve, 

o/p of PI starts increasing 

⇒ 

Selector switch selects maximum of two inputs 



416/454

• Due to line fault or during low i/p AC voltage 

condition V dco(R) will drop 

⇒ Assume V dco(R) Cosα min 

< V dco(I) Cosγ 

• If there is no current control by the inverter , 

i d will ↓ and eventually becomes zero 



417/454

• In order to avoid this situation inverter is also 

provided with current control 

• Operate at E ' till tap changer changes the tap 

What happen If I mar is –ve ? 

⇒ Rectifier is trying to control I ref(R) 

⇒ Inverter is trying to control I ref(R) + I mar 



418/454

Inverter side : 

• I d can be ↑ by ↑ ‘γ’ 

• As γ↑ , I d ↑, but rectifier 

controller tries to ↓ the current (I ref(R) 

• Since I d is ↑ due to increase in γ , 

rectifier controller ↑ αto reduce I d 

α ⇒ towards 90 o 

γ ⇒ towards 90 o 



419/454

⇒ New operating point could be ‘D ' ’ 

⇒ Correct sign to I mar is 

very important 

• I mar should not be too small 

because there could be 

measurement error 



420/454

Mode stabilization : 

• Intersection of α min characteristics of converter 

and inverter CEA may not be well defined 

• There could be multiple crossings 

• Instead change the slope of the 

inverter characteristics 

near the crossing 



421/454

Alternative inverter γ control 

• Instead of regulating ‘γ’ (CEA) 

• Maintain a constant DC voltage at a desired 

point 

• Could be sending end 

• Required inverter voltage to maintain the above 

voltage is estimated by computing I.R drop 



422/454

• ‘V’ profile is flat 

• Constant ‘γ’ characteristics has drooping 


γ≈18 o in voltage 

control mode 



423/454

Constant ‘β’ control : 

β = μ + γ 

μ ⇒ function of i d & V ac 

⇒ Choose ‘β’ for worst case 

⇒ At low loads additional security against 

commutation failure 

⇒ As i d ↑, minimum ‘γ’ may be encountered 



424/454

• V dcoi Cosβ remains constant 

• As i d ↑, V d = V doi Cosβ + (R L +R ci )I d also ↑ 



425/454

• Use either constant V dc or constant β control 



426/454

Current limit 

Maximum current limit : 

Max. short term current = (1.2 -1.3) I rated 

Minimum current limit : if i d ↓ below a 

certain limit due to finite ripple in I, 

current will become discontinuous 

• 12-pulse converter 

• 12 times in one cycle current become zero 

(current interruption) 



427/454

• There could be lightly damped oscillations 

(smoothing L & line C) 

• Over voltage across the device 

• Simulation study is required 

• Ensure I min in DC link 



428/454

Voltage depend current-order limit (VDCOL) 

• Under L.V condition it may not be desirable 

or possible to maintain rated current 

• Commutation failure 

• At one converter end V ac has ↓ 

∴V α dco 

Cos ↓ 



429/454

• To maintain the current, voltage at the other 

end of the line is adjusted 

• Either ‘α’ or γ↑ 

• Reactive power demand ↑ 

• V ac has ↓, ‘Q’ supplied by ‘C’ or filter also ↓ 

• Above problems can be addressed using 

voltage dependent current order limit 



430/454

• VDCOL characteristics could be a function of 

AC voltage or DC voltage 



431/454

Review 

Rectifier 


Constant current 

by ‘α’ control 

Constant ignition 

angle control 

• Inverter ⇒ Constant extinction angle control 

• Current control is given to both converters 



432/454

Contd.. 

But I ref(R) > I ref(I) 

I ref(R) -I ref(I) = I margin ≈ 0.1I rated 

• Current control loop of inverter is inactive 

when current ≈ I ref(R) 



433/454

I mar should +ve : 

Contd.. 

• If I mar is –ve, reversal of power takes place 

(only academic interest) 



434/454

Contd.. 

Mode stabilization : 

• Intersection is not well defined 

⇒ Change the slope 

Constant V dc 

Constant ‘β’ 



435/454

Current limit : 

Contd.. 

⇒ I max = (1.2 -1.3) I rated 

⇒ I min 

⇒ Should not be allowed to go into 

discontinuous 

• There could be lightly damped oscillations 



436/454

Voltage depend current-order limit (VDCOL) 

• Under L.V condition it may not be desirable 

or possible to maintain rated current 

• Commutation failure 

• At one converter end V ac has ↓ 

∴V α dco 

Cos ↓ 



437/454

• To maintain the current, voltage at the other 

end of the line is adjusted 

• Either ‘α’ or γ↑ 

• Reactive power demand ↑ 

• V ac has ↓, ‘Q’ supplied by ‘C’ or filter also ↓ 

• Above problems can be addressed using 

voltage dependent current order limit 



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• VDCOL characteristics could be a function of 

AC voltage or DC voltage 



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Rectifier inverter V-I characteristics 

• Power transfer over the line can be controlled 

by varying I mar 

• Signals are transmitted through 

telecommunication lines 

• Communication may fail or DC line fault 

⇒ Reverse power flow may occur 

⇒ Inverter is provided with min. α limit 

≈ 95- 110 o 



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Summary of basic control principle : 

• HVDC system is basically current control 

⇒ To limit over current 

⇒ To prevent the system from running down 

due to fluctuations in AC voltage 



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Significant aspects of basic control : 

Rectifier 

Current control 

‘α’ limit 

• In current control mode closed loop regulator 

controls the firing angle to regulate I d at I ord 

• Tap changer control of the converter brings ‘α’ 

within 10-20 o 



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• Inverter is functioned with CEA control and a 

current control 

• In CEA mode, γ is regulated at around 15 o 

• Inverter control could have constant ‘β’ control 

• Under normal operation rectifier is in current 

control & inverter is on CEA control mode 



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• If there is a ↓ in AC voltage, 

‘α’ of rectifier ⇒ α min (CIA mode) 

• If current falls to a certain limit, inverter 

will assume C.C 

Valve blocking & by passing : 

• If one bridge is to be taken out of service 

⇒ Only blocking will not extinguish the current 

that was flowing through the thyristor pair 



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⇒ Inject AC voltage in the link 

⇒ There could be ‘V’ & ‘I’ oscillations due to 

lightly damped circuit 

⇒ Transformer feeding the bridge is also subjected 

to DC magnetization 

⇒ By pass the bridge when the devices (valves) 

are blocked 



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⇒ Achieved using by pass valve and by pass switch 

⇒ Assume T 2 & T 3 are conducting & blocking 

command is given 



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⇒ Commutation for T 2 to T 4 is in usual manner 

⇒ But incoming device T 5 is prevented by not 

triggering T 5 . When T 1 get F.B (V AB +ve ) 

trigger T 1 

⇒ Current by pass pair is shunted by closing S 1 

& open S 



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• For energization of blocked bridge 

⇒ Current is first diverted from S 1 to bypass pair 

⇒ S 1 will generate arc voltage 

⇒ Trigger bypass pair 



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Modern techniques 

• HVDC using line commutated converters 

• Requires AC voltage for commutation 

• Requires reactive power 

• DC link is equivalent to a current source 

• ‘V’ can reverse but ‘I’ can not reverse 

• Devices should be able to block –ve voltage 

• Not suitable for weak grid 



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• Instead use VSI 

• ‘I’ could be in phase with ‘V i ’ 

• Inverter devices are self commutated 



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• No AC voltage is required for commutation 

• Conversion at UPF is possible 

• DC link is voltage source 



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• ‘V’ can not reverse, but ‘I’ can reverse 

• Devices should be able to carry ‘I’ in 

both directions 



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Thank you 



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BG Fernandes Department of Electrical Engineering II T - Power ...

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