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Homework Help: Analysis of a circuit that contains two diodes

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data


    v8ei3q.jpg

    The diodes are ideal, so their internal resistance is zero. The zener diode has a 4V activation voltage. There is only one variable element in the circuit is V0. The question is what is Vab for every value of V0, and what are the directions of the current.


    2. Relevant equations

    Kirchoff laws

    The directed sum of the electrical potential differences around any closed circuit must be zero.

    And the properties of the diodes.


    3. The attempt at a solution

    Well, I have the solution but I don't understand some questions, that is the reason of this thread.

    The solution is this:

    -V0<0 implies that the current goes for the mesh of the mesh of the zener diode. The other diode cut the current in his mesh.

    -6V> V0 > 0 , is the same situation but with less current in the same mesh.

    -10V> V0> 6 V , there is no current in any mesh

    -14V> V0> 10 volts, zener polarized but we don't have sufficient voltage to polarize the other mesh of the diode.

    -V0>14 volts, both diodes are polarized and they carry the current.

    I don't understand something, in this analysis we consider the rises and drops of the voltage in the diodes and in the voltage sources to know if a mesh will carry current or not, but it seems to forget the resistances . But evidently if the resistances carry some current they will produce some potential drop.

    So ¿why do you need only V0=14 volts to polarize the zener diode?, ¿Should not be higher due to the presence of the resistance r0?.

    ¿What is the reasoning to arrive at that 14 volts?.

    And finally ¿what is the potential difference in ab due to the 8V source?¿How does the diode affect it?

    THanks in advance.
     
  2. jcsd
  3. Jan 19, 2010 #2

    Mapes

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    There's a voltage drop across a resistor only when current is flowing. When a diode begins to conduct, the current in its loop is zero, and so the voltage on either side of a resistor in that loop is zero.

    What are the requirements for conduction in a Zener diode?
     
  4. Jan 20, 2010 #3
    First of all, thanks for the answer mapes:smile:

    -Ok, so you can consider that the diode is conducting before carrying any current, and if the current is zero the voltage drop in the resistances is zero.

    -supply an inverse voltage of Vz volts

    So to secure ideas, i am going to consider another circuit, this one:

    2802rsn.jpg

    Imagine that we can change the value of the R resistance, and at the start is very high so there isn't conduction in the circuit , if we diminish its value ¿both branches will start to conduct at the same time (both zener with the same properties)?

    --------

    And another circuit:

    296jd6o.jpg

    With Vz= 5volts, and voltage drop of 0,5 in direct polarization.

    When i have seen the circuit i have thought that it was simple to know the answer to this question , ¿what are the values of V0 that activate zener 1 in any direction?, and the same for zener 2.

    I have made this reasoning:

    For V0 > 0:

    -the branch of the zener 1 needs more than 1V +0.5(of the zener)=1,5 volts to be activated, and you have 4volts -0.5= 3.5volts supplied by the other source, so it doesn't matter what is the value of the V0 source, zener 1 will be activated always due to the 4V source.

    But the teacher applies mesh and node analysis to arrive at the conclusion that if V0 is < 2,5 volts the zener 1 won't carry any current ¿what went wrong in my reasoning?

    For V0<0 i have similar problems.
     
    Last edited: Jan 20, 2010
  5. Jan 20, 2010 #4

    Mapes

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    Maybe it would be better to consider a simple circuit with a battery, resistor, and 4V Zener diode in series. What is the necessary battery voltage for Zener diode conduction? Does it depend on the resistor value? What about if the Zener diode is flipped? Does the voltage depend on the resistor value now?

    If the answers are clear, then it's time to move to two-loop circuits.
     
  6. Jan 20, 2010 #5
    -V supplied > = Vzener

    -not depend on the resistor

    -if the diode is flipped it will conduct if V supplied is > voltage drop across the diode (usually 0,5 volts) or in the ideal case it will conduct always.

    -not depend on the resistor
     
  7. Jan 20, 2010 #6

    Mapes

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    Looks great, you've got it. Do the answers in the first problem make sense now? It's an iterative approach: you assume no current flow, then check whether any of the diodes conduct. If they do, you have to revise your answer to incorporate the voltage drop across the resistors, then check again. It's generally not possible to just look at the circuit and see all the answers immediately; it takes some work.
     
  8. Jan 22, 2010 #7
    mapes thank you another time

    -for te first circuit I still don't udnerstand something:


    -V0<0 implies that the current goes for the mesh of the mesh of the zener diode. The other diode cut the current in his mesh.(clear)

    -6V> V0 > 0 , is the same situation but with less current in the same mesh. (In this case ¿do you consider that the 8V source increases the voltage on A?, i suppose that the answer is NO)

    -10V> V0> 6 V , there is no current in any mesh.(the same question for the 8V source)

    -14V> V0> 10 volts, zener polarized but we don't have sufficient voltage to polarize the other mesh of the diode.(the same question)

    -V0>14 volts, both diodes are polarized and they carry the current.(¿why 14 Votls?¿should not be 18 volts).


    I don't understand why you don't need 18 volts ( 8V + 6V from the sources + 4 V of the zener diode), I say this because if not it should be sufficient with 12V on V0 ( 8 to compensate the other 8V source and the 4V of the zener diode).

    Or 18 or 12 that is what i thought, but i am wrong ¿why?. IF you agree with the 14 V ¿why do you not consider the zener voltage drop?


    For the third circuit I continue to think the same but i am wrong, I don't know why.
     
  9. Jan 23, 2010 #8
    perhaps it is a little off topic, but I would like to ask you if you know what is the origin of this expression of the bjt:

    hie = (hfe + 1)rd

    hie= input impedance of the bjt with the output sorted, common-emmiter configuration

    hfe=forward current gain with the output sorted , common-emmiter configuration

    rd= dynamic resistance of a diode

    I have spend two days searching for it and I can't find what is the origin of this formula.

    THanks in advance.
     
  10. Jan 24, 2010 #9
    please HELP
     
  11. Jan 24, 2010 #10

    Mapes

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    Sorry, was out of town this weekend. In the first circuit, the regular diode will begin to conduct when point A is >8V (because the diode is forward biased). So there must be a 2V drop across the resistor next to point A. But the other resistor is in the same loop and has the same resistance, so there must be a 2V drop across it as well. The Zener diode also drops 4V because it is conducting in reverse. So V0 must be at 6+2+4+2=14V.
     
  12. Jan 24, 2010 #11

    Mapes

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    Horowitz and Hill derive this in The Art of Electronics, which is a great general reference.
     
  13. Jan 26, 2010 #12
    I see, you consider that the current in both resistances is the same because the other branch doesn't carry current yet, now it is clear, but as you said, you need some experience to see it.


    thanks, i will search that books.

    Mapes, i appreciate very much your comments, if you need something tell me with a mp.
     
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