The Miller Theorem and the Frequency Response of the Common ...

The Miller Theorem
and the
Frequency Response of the
Common Emitter Amplifier
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 1

A useful approximation: The Miller Theorem
• Consider a shunt admittance connected between the input and output of an
inverting voltage amplifier of gain G.
Y
In Out
V V In
I F
-I F Out
-G
-G
I in
I (1+G)Y
1 I 2 I (1+1/G)Y
out
• Assuming that Y does not affect the gain, KCL at the input says that:
( ) ⎡⎣( 1 ) ⎤⎦
( ) ⎡( 1 1/ )
i = i + i = i + v − v Y = i + + G Y v
in 1 F 1 in out 1
in
i = i − i = i − v − v Y = i + ⎣ + G Y⎤⎦
v
out 2 F 2 in out 2
out
• This is the same current as we would get if we connected Y(1+G) in parallel to
the input and Y(1+1/G) in parallel to the output (on the right)
• Note that this is an approximation; Y always changes the gain,
unless Z out
// Z L
= 0. Nonetheless, very often it is a good approximation.
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 2

An application of the Miller theorem
Input impedance of the inverting amplifier (if op-amp has zero Y in ):
R1
R2
v in 2
G V out
Zin
= R1
+
( 1+
G)
R
We can even use the Miller theorem to calculate the gain of this circuit
for finite op-amp gain G:
( G )
/( 1)
v R / + 1)
out
−GR −R
Av
= =− G = , lim A =
v R R G RG R R R
in
2 2 2
v
1+ 2
+
1
+ G→∞
1+
2 1
This is the correct answer for the closed loop gain as we shall later see!
SF1.6p34
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 3

The small signal model the of BJT
• The model below is useful to frequencies of up to 100s of MHz
• The model is strictly valid for B-E voltage variations smaller than V T
=25mV.
• To analyse a circuit:
– Each transistor in a circuit is replaced by this model
– The “base spreading resistance” R BB
is lumped with the thevenin
impedance of the signal source. It is not shown explicitly in these
notes.
What does R BB
do:
• R BB
is responsible for most of the noise of transistor amplifiers
• R BB
is responsible for the power gain rolling-off at high frequencies
• R BB
is often ignored at “low” (f < MHz) frequencies
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 4

BJT model: impedances and transconductance
• The large signal BJT equation gives the transconductance:
Vbe ( / V
⎛ V ⎞
th CE Vbe )
/ Vth
IC
= I0 e − 1 ⎜1+
⎟
I0e
⎝ VA
⎠
kT
B
Vth
= 26 mV at room temperature
e
∂IC
IC
gm
= ∂ V V
be
th
• However, since I E
=(1+1/β)I C
,
∂VBE
Rπ = = βV / I = β / g
∂I
T C m
• Same equation gives output conductance:
G
CE
B
∂I
= R = =
C C
1/
CE
∂VCE
VA
I
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 5

BJT model Capacitances
• BC junction is reverse biased:
– depletion capacitance
– overlap capacitance
• BE junction is forward biased:
– depletion capacitance
– storage (diffusion capacitance)
– overlap capacitance
• CE capacitance is small
– overlap capacitance
• Fringing/Overlap capacitances:
C = C + C
BC BC 0 j
C = C + C + C
C
BE BE0
j D
C
CE CE0
C , C , C
BC0 BE0 CE0
• Junction capacitance:
– Reverse biased diode
– M=1/2-1/3, from doping
– Φ approx 0.8V
– M, Φ usually determined from two
bias points
C
j
0
=
M
⎛ VBE
⎞
⎜1−
⎝
C
Φ
⎟
⎠
• Diffusion capacitance
– from transit time
C = g τ , τ = 1/ 2 π f
D m T T T
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 6

An important figure of merit:
the BJT Current gain and the transit frequency f T
The transistor is connected as a current amplifier:
• Driven by a current source
• Driving a current meter
f T
is the frequency at which the current gain drops to unity.
f T
is usually specified by the transistor manufacturer as a figure of merit.
f T
is a rough estimate of the highest frequency at which the transistor
can be used as an amplifier. Typically fT / 10 is this highest frequency.
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 7

An important figure of merit:
the BJT Current gain and the transit frequency f T
( ( ) 1/
π )
iB = vBE s CBE + CBC
+ R ⎫⎪ ⎬ ⇒
iC = gmvBE
⎪⎭
i g g R
β
= " h " = = =
i s C C R sR C C s f
C m m π
0
fe
B ( BE
+
BC) + 1/
π
1+ π ( BE
+
BC)
1 + β0
/2π
T
β
g g
β
where fT
= = , since: R =
2 πR C + C π C + C πC g
the current gain is
π
0 m
m
0
π
BE BC
2
BE BC
2
BE m
( ) ( )
h
fe
( f )
β
= ⇒
1+
j f f
lim
0
β f > f
0 T
/ T
/ β
0
h
fe
fT
= with fT
=
f
1
2πτ
T
f T
allows the easy calculation of C BE
+ C BC
since g m
only depends on the
collector bias current! Note that C BE
>>C BC
so knowing f T
defines C BE
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 8

Load
The Common Emitter amplifier
Source
Amplifier
Model parameters:
⎧ ∂IC
IC
⎪gm
= =
⎪ ∂VBE
VT
−1 −1
−1
⎪ ⎛ ∂I
⎞ ⎛∂
( IC
/ β
B
) ⎞ ⎛ ∂I
⎞
C
β
⎨Rπ
= ⎜ ⎟ = ⎜ ⎟ = β ⎜ ⎟ =
⎪ ⎝∂VBE ⎠ ⎝ ∂VBE ⎠ ⎝∂VBE ⎠ gm
⎪
−1
⎪ ⎛ ∂I
⎞
C
VA
RCE
= ⎜ ⎟
⎪
⎩ ⎝∂VCE
⎠ IC
V = kT / q = 25mV at 290K=17C
T
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 9

The Common Emitter amplifier (2)
Combine , , into a Thevenin circuit:
S
R
= =
S
π
S
VS1
VS
R
π
+ Z
S
1+
Z
S Y
in
RZ
π S
ZS
RS1
= Rπ
// ZS
= =
R + Z 1+
Z Y
π
V
S S in
V Z R π
2
RCEZL ZL
Consider RCE and ZL together: R = RCE / / ZL
= =
R + Z 1+
Z Y
and note that
Z = R + R
S S BB
CE L L out
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 10

The Common Emitter amplifier (3)
Gain:
A
v
v ∂V
V 1 −g Z
= ≡ =− g R =
v V V RY Z Y
out 2 s1
m L
m 2
s
∂
s s
+
s in
+
L out
If Y in
= 0 and Y out
= 0 this is the familiar:
A
v
= ≡−g Z
out
v m L
vs
( 1 ) ( 1 )
Note that lowercase letters represent small signal variations, i.e. differentials
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 11

The Common Emitter amplifier (4):
Input and output impedance
Input Impedance:
Output Impedance:
Z
Z
in
out
∂V
∂I
−1
BE
B
= = =
B
∂V
∂I
BE
−1
C
C
= = =
out
⎛ ∂I
⎜
⎝∂V
⎛ ∂I
⎜
⎝∂V
CE
⎞
⎟
⎠
⎞
⎟
⎠
R
π
R
CE
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 12

Frequency response of
the Common Emitter amplifier
If we momentarily neglect C BC this would be the same as before but with:
Input Impedance:
Output Impedance:
We still have:
A
V
1
Rπ
Zin
= = Rπ
// CBE
=
Y 1 + jωC R
in
1
RCE
Zout = = RCE // CCE
=
Y 1 + jωC R
vout 1 −gmZL
v
= =
and = G =
v ( 1+ RY ) ( 1+
Z Y ) v
BE
out CE CE
s s in L out
π
−g Z
out m L
( 1+
Z Y )
B L out
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 13

Dealing with C BC
I CBC
In the absence of C BC
we know that:
the current through C BC
is:
v
v
= G =
−g Z
out m L
( 1+
Z Y )
B L out
( ) ( 1 ) ( 1/ 1)
I = jωC V − V = jωC − G V = jωC G−
V
CBC BC B C BC B BC C
We can account for C BC
by introducing an additional input “Miller” admittance:
Min ,
ω
BC
1
( )
Y = j C −G
and an additional output “Miller” admittance:
( )
YMout
,
= jωCBC
1−1/
G
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 14

Frequency response of the CE amplifier
V S1
and R S1
is the Τhevenin equivalent of R S
V S
and R π
Use the Miller Theorem to get:
( ) ( ) ( )
with C = C + 1+ G C 1 + G C and C = C + 1+
1/ G C C
2 BE BC BC 3 CE BC BC
The low requency gain from V1 to V2 is: G = −gmR2
−gmZL
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 15

Frequency response of the CE amplifier (2)
the gain of this amplifier is:
dV2 −gm
R2
G
= =
dVS1 1+ sRS1C2 1+ sRC
2 3 ( 1+ sRS1C2)( 1+
sRC
2 3)
input pole
If the gain is large the input pole can be at a much lower frequency than the output pole:
τ = R C = R // R C + 1+ G C R 1 + G C since R R
τ
if τ
in
> τ
out The gain-bandwidth product of this amplifier is:
G = gmR2
gmZL
⎫
⎪ gmR2
G
1
1 1 ⎬ ⇒ GBW = . lim GBW
.
BW =
2πRC G →∞
S 2
2 πRC S BC(1 + G) 2πRC
S BC
2πRS1C2 2πRSC
⎪
2⎭
This is independent of G!
The tiny C and the s ource impedance can define the gain-bandwidth product of the CE amp.
BC
( π ) ( )
( ( ) )( )
This suggests that feedback is at work!
Note that the input pole will be at a lower frequency than the output pole if
> ⇒ > ⇒ = / > 1
G RS1 R2 gmRS1R2 R2
gmRS β RS
R π
( ) ( ) ( π )
IN S1 2
S BE BC S BC S
= RC = C 1+ 1/ G + C R // Z C Z (since C C
)
OUT 2 3 BC CE CE L BC L BC CE
output pole
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 16

Importance of the gain-bandwidth product
the gain of the CE amplifier is:
dV2 −gm
R2
G
= =
dVS1 1+ sRS1C2 1+ sRC
2 3 ( 1+ sRS1C2)( 1+
sRC
2 3)
input pole
output pole
One of the two pole frequencies is usually much lower than the other.
We then say that we have a dominant pole amplifier whose frequency response is:
A
v
G G
1+ sτ
1+
jωτ
0 0
( s) = =
for a DC gain G 0
and a bandwidth ω = 1/
We have already seen that in the limit of large G 0
the Gain-Bandwidth product:
G = G = G
is independent of the amplifier gain.
B
0ω
/
B 0
We will see later in the course that this is a very general result which applies to
all one pole feedback amplifiers regardless of how they are constructed.
τ
B
τ
L2 Autumn 2009 E2.2 Analogue Electronics
Imperial College London – EEE 17