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Diode circuit

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to find the output voltage mathematically or graphically


    2. Relevant equations



    3. The attempt at a solution
    My approach :
    as long as the Vi is < 9.3 V , the diode D1 will be ON because in that case VA will be < 10 V and current can flow through 10k resistor. And if this is true then D2 will also be 0N and Vo = Vi.
    Now second interval i.e Vi > 9.3 V, same reasoning for D3 i.e D3 will be ON as long as Vi > -9.3 V & D2 is ON & D4 is also ON. But in this case if we track from input to the output through D3 & D4 , then i get Vo = Vi and this is making D1 ON which i assumed to be OFF. Summarizing my results :

    0<Vi<9.3 all diodes are ON & Vo = Vi
    Vi>9.3 all diodes are ON & Vo=Vi..... but this is not possible because this makes VA > 10 which shows current can't flow from +10V battery. So in short i know i am doing but the question is where ????
     

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  3. Oct 29, 2012 #2

    NascentOxygen

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    Staff: Mentor

    If, for a diode to be forward biased, you need a voltage of a level which can't exist, then your assumption of that diode conducting is wrong.
     
  4. Oct 29, 2012 #3
    can you tell me what will be the output voltage Vo for Vi = 5V ?
    According to me, Vi=5V so all diodes can be ON & in that case VA = Vi+0.7 = 5.7V and Vo = 5.7-0.7 = 5V, this will make Vc = 4.3 V again possible. So is this the correct output for Vi=5V ?
     
  5. Oct 29, 2012 #4

    NascentOxygen

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    EDIT: I've changed to Vi=4V for your example.
    For Vi=4V that sounds right, but you need to check the currents through the diodes to be sure — do this by determining the currents through all 3 resistors since you know the [expected] voltages across them.

    If you go on increasing Vi much beyond 4V, the currents through D1 and D4 will be decreasing and at some point these currents will become zero. At that stage, Vo is no longer directly determined by Vi.

    Evaluate the currents for Vi of 5V and you'll see what I'm talking about.
     
    Last edited: Oct 29, 2012
  6. Oct 29, 2012 #5
    i got it nascent
    is there a way to find that which diodes will be ON and which will be OFF for a particular input voltage without calculating the currents ?
     
  7. Oct 29, 2012 #6

    CWatters

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    That's not what I got..

    If Vi=-10V ....

    Point A will be at Vi+0.7 = -9.3V D2 will be off.

    Current will flow from ground, through the 10K near the output, through D4 and the 10K to -10V. Vo will be about -9.3/2 = -4.65V
     
  8. Oct 29, 2012 #7

    NascentOxygen

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    Staff: Mentor

    If you imagine D1 and D4 not conducting, you can calculate the standing potential at point A (and B). This defines the changeover point which, if Vi falls further, will forward bias D1.
     
  9. Oct 29, 2012 #8
    this is the solution given in the textbook. Still i have a doubt that how do we come to know that we have to observe the state of D1 & D4 ?
     
  10. Oct 29, 2012 #9
    It's always harder to see what is happening in nonlinear circuits.

    What I do is follow the shorts in the diodes (pretend they conduct) and see if some conditions for conduction arise.

    For this circuit, you will see if Va > Vc (plus some diode drops), all the diodes conduct. So Va=Vc (plus some diode drops) is one transition point that will see one or more diodes shut off. If the diodes are on then there is a relationship between Vi, Vo, Va and Vc.

    (This circuit is a variation of the diode-bridge modulator / diode-bridge switch. One use is in AM radio modulation).
     
    Last edited: Oct 29, 2012
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