Multiple diode - transfer characteristic

Thread Starter

xxxyyyba

Joined Aug 7, 2012
289
Hi!
Here is my circuit:
Rich (BB code):
http://oi48.tinypic.com/2ym6ihe.jpg
Ed=0.7V, R1=R2=R3=R4=R, Vcc=Vee=10V. My task is to calculate and sketch transfer characteristic Vout=f(Vin) for -10V<Vin<10V.
My first assumption was that D2 and D3 are on while D1 and D4 are off. I calculated that it's true for Vin>-9.3V, Vout=4.65V.
My second assumption was that D1 and D4 are on while D2 and D3 are off. I calculated that it's true for Vin<9.3V, Vout=-4.65V.
What should I do next? Are my calculations ok? :)
 

WBahn

Joined Mar 31, 2012
26,398
Where is R4?

You computations don't include any mention of the resistors? Are you saying those don't affect the results?
 

WBahn

Joined Mar 31, 2012
26,398
In your first case, it would seem that the input voltage doesn't matter (other than needing to satsify any conditions to be consistent with the diode states you are assuming) and that the output voltage is dependent only on Vcc and two of the resistors (which your diagram fails to distinguish, although since you are saying they are all equal that's not a problem).

But now I see where you are getting your division by two from. The total voltage available to drive current through the resistors is (Vcc-Vd), where Vd is the diode drop. The total current is then (Vcc-Vd)/(2R) and the voltage across the output resistor is then (Vcc-Vd)/2.

It looks like you are on the right track. I will sit down and look at your answers in detail a bit later. But you do have a typo in your first post that makes things a bit confusing.
 
Last edited:

Thread Starter

xxxyyyba

Joined Aug 7, 2012
289
It seems that for -4.65V<Vin<4.65V all diodes are on. I replaced them with voltage generators (0.7V) but I don't know what to do next :confused:
 
Top