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Calculating Voltages in a circuit (With Diodes)

  1. Oct 2, 2011 #1
    Say D1 is a silicon diode( .7 volts) and D is a Germanium diode (.3 volts)

    Determine the voltages at A,B,C and D with respect to ground

    You can use hypothetical values

    Any help would be greatly appreciated

    Thanks
     

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  2. jcsd
  3. Oct 2, 2011 #2
    Remember KVL? ƩV=0 around a loop. Apply this fact to solve for unknown voltages.
     
  4. Oct 2, 2011 #3
    But the voltage in each point should be the same right even with a resultant voltage?

    What about the diodes?
     
  5. Oct 3, 2011 #4

    gneill

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    Staff: Mentor

    This is one of those questions where the answers will depend upon the level of detail and the assumptions that you make; a simply posed question that could take pages to answer!

    For example, is Vcc > Vc or is Vc > Vcc? (are the diodes forward or reverse biased by the supply voltages?). What about the region where the diodes are just beginning to turn on or off and their I vs V curves are nonlinear? What sort of model are you going to use for the diodes? Real diodes have small reverse bias leakage currents, will that be taken into account?
     
  6. Oct 3, 2011 #5
    Well lets just say its not that complicated. Lets put Vc as 12 V and Vcc as 24. The diodes are forward basis where D1 is Silicon and D is Germanium. Thats all the details given. Just straightforward.
     
  7. Oct 3, 2011 #6

    gneill

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    Oh sure, spoil ALL the fun! :devil:
     
  8. Oct 3, 2011 #7
    lol not for me. But still any ideas on the problem?
     
  9. Oct 3, 2011 #8

    gneill

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    Sure. Since they're forward biased, pencil in the forward voltage drops for the diodes. What does that leave for the voltage drop across the resistor? (KVL)
     
  10. Oct 3, 2011 #9

    ok Im going to use the total voltage for the voltage drops in each diode. So for D1, the voltage drop would be Vd1 = VR - .7. Which is Vd1 = 36-.7 = 35.3

    For D, the voltage drop would be Vd = VR - .3. Which is Vd = 36-.3 = 35.7

    Does that help?
     
  11. Oct 3, 2011 #10

    gneill

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    No, in fact it's rather bizarre :eek: The voltage drop across the silicon diode is 0.7V. That's it, that's all. The voltage drop across the germanium diode is 0.3V. That's it, that's all. So what's left for the resistor?
     
  12. Oct 3, 2011 #11

    the total for all voltages?
     
  13. Oct 3, 2011 #12

    gneill

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    Are you familiar with KVL? Can you write the KVL equation for the loop?

    EDIT: Here's your circuit with the known voltages penciled in:

    attachment.php?attachmentid=39562&stc=1&d=1317673487.gif
     

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    Last edited: Oct 3, 2011
  14. Oct 3, 2011 #13

    VR = Vcc -.7 -.3 + Vc ?
     
  15. Oct 3, 2011 #14

    gneill

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    Almost. Watch your polarity on Vc; when you pass through it on your way around the loop, you're going from its + terminal to its - terminal.
     
  16. Oct 3, 2011 #15
    So VR= Vcc - .7- .3- Vc?
     
  17. Oct 3, 2011 #16

    gneill

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    Is that a guess? Why don't you explain your reasoning? Take me on a "round the loop tour" of the circuit.
     
  18. Oct 3, 2011 #17
    ok well the voltage for the resistor is the Vcc added to the silicon diode. It is in forward basis, so its 12-.7 and then past the resistor into the second diode, again with forward basis, and then the last terminal where it goes from the + terminal to its - terminal into the ground.

    Hence reasoning: VR= Vcc- .7-.3- Vc
     
  19. Oct 3, 2011 #18

    gneill

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    Well, that's more or less it. For KVL work it's better to deal with individual potential differences across the components rather than the potential at various nodes around the circuit with respect to ground.

    So I would say: beginning the loop at point C (at the right end of the resistor) and proceeding clockwise, There is a voltage drop of 0.3V for the germanium diode, followed by a drop of 12V for Vc, followed by a rise of 24V for Vcc, and then a drop of 0.7V for the silicon diode. The sum is then: -0.3V - 12V + 24V - 0.7V = 11V. This remaining 11V will be the potential drop across the final component in our loop, the resistor, bringing us back to point C where we started. Thus the potential drop across R is 11V.
     
  20. Oct 3, 2011 #19
    ok fair enough but what about the voltages ( A,B,C and D) across the different points or is it the same for the voltage drop across R?

    Thanks again for all the help
     
  21. Oct 3, 2011 #20

    gneill

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    After you've established all the individual potential differences, then you can go back and sum them from the ground node to the individual nodes (you can work in either direction around the loop).
     
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