Electronic Circuit Analysis

262 Electronic Circuit Analysis
The voltage drop across 7805 is 2V
.. The minimum input voltage required is Yin = Vo + Dropout voltage
Vin=12V
So a current source circuit using a voltage regulator can be designed for a desired value of IL'
by choosing an appropriate value of R.
Example : 8.2
Using 7805C, voltage regulator, design a current source that will deliv~r 0.25 A current to the
480 lOW load.
Neglecting
8.2 Current Limiting
VO=VR+VL
= 5V + (48 0) (0.25) = 17V
V in = V 0 + drop across Ie
V in = 17 + 2 = 19V Ans.
This means short circuit protection i.e. Even if the output terminals of the voltage regulator are
shorted, the current should be limited and should not exceed a particular value. This can be done
in two ways.
1. Simple limiting circuit
2. Foldback limiting circuit
8.2.1 Simple Limiting Circuit
As the drop across R 3 , V C2 B2 increases, 'V B2 E2 decreases.
! Q2 goes off.
R4 is called current sensing resistor.
",.. If IL < 600 rnA, voltage drop across R4 is < 0.6V (... R4 = 1 0,-1 x 600 rnA = 0.6 V).
The drop across R4 is V BE for transistor Q3'
:. Q 3
is cut off. So the regulator circuit works as a normal circuit, without any
limiting provision. When IL is between 600 and 700 rnA, the voltage across R4 is between 0.6 V and
0.7 V. Therefore Q 3
will turn on. So the collector current of Q 3
will flow through R 3
. So the base
voltage V BE
to Q 2 will decrease. Therefore output voltage Vo will decrease and hence load current
will decrease.