Multiple diode - transfer characteristic

Discussion in 'Homework Help' started by xxxyyyba, Mar 7, 2013.

  1. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    Hi!
    Here is my circuit:
    Code ( (Unknown Language)):
    1. 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? :)
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,751
    4,799
    Where is R4?

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

    Thread Starter Member

    Aug 7, 2012
    249
    2
    I made mistake, there is no R4. I will upload photos today so you will see what I have done.
     
  4. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    D2 and D3 on:
    [​IMG]

    D1 and D4 on:
    [​IMG]
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,751
    4,799
    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: Mar 8, 2013
  6. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    After analysis I concluded that Vout=-4.65V for -10<Vin<-4.65 and Vout=4.65 for 4.65<Vin<10. These results match with results from my first post (http://oi48.tinypic.com/2ym6ihe.jpg). I need now to sketch transfer characteristic. How should I do that? :)
     
    Last edited: Mar 9, 2013
  7. WBahn

    Moderator

    Mar 31, 2012
    17,751
    4,799
    What about from Vin from -4.65V (don't forget units!) and 4.65V?
     
  8. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    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:
     
  9. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    Any suggestion?
     
Loading...