Download Find the dc transfer characteristic of the circuit shown. Given that

Find the dc transfer characteristic of the
circuit shown.
Given that (VZ1 = 3V and VZ2 = 6V).
Solution:
We'll first try to find the regions where the
diodes operate as (forward, reverse or
breakdown).
-
Since there's always a –ve feedback
by the lower (20K) resistance, then V- = V+ = 0
Note that V+ = 0 as there's no current flowing in RC.
-
Since this is a non-inverting amplifier configuration, then Vo = -K * Vi
Where K is constant.
1) When (Vo) is +ve, current will be flowing in the diodes from P to N, then
they'll be both forward biased.
Then, @ Vi < 0 the diodes are forward.
The circuit will have two (20K) parallel with (10K) resistance in the feedback,
and the equation will be:
Vo = - Vi
2) When (Vo) is –ve, current will flow from N to P, then both diodes are reversed
and current will flow only in the lower (20K) resistance.
Then @ Vi > 0 the diodes are forward.
This will be valid as long as the diodes reverse voltage didn't exceed the break
down voltage
i.e: Vo > -3 & -6
=> Vo > -3.
The circuit will have (20K) resistance in the feedback, and the equation will
be:
Vo = - 4 Vi
3) When -6< (Vo) < -3 , VD1 will be in break down, and VD2 will remain in
reverse.
The circuit will have (20K) resistance parallel with (20K) and a battery of 3v
of the zener diode …. You'll make the analysis of the circuit by making KCL
@ V- and get:
Vo = - 2 Vi + const
4) When (Vo) < -6 , both VD1 and VD2 will be in break down.
The circuit will have (20K) resistance parallel with (20K) and a battery of 3v
of the zener diode, parallel with (10K) and a battery of 6v of the zener diode.
You'll make the analysis of the circuit by making KCL @ V- and get:
Vo = - Vi + const
The final VTC will be like this: