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diodes conducting in different cases?

Discussion in 'Electronics Homework Help' started by ArFa, Dec 7, 2010.

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  1. ArFa

    ArFa

    17
    0
    Dec 7, 2010
    Please, can you tell me why the diode D1 in case (a) is conducting but in case (b) it's not?
    And how to find all currents in this problem?

    notice that the only thing that are changing are resistors...

    the problem is in the link below...
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,412
    2,780
    Jan 21, 2010
    Pretend that the 2 resistors are connected together and between the supply rails.

    Find the voltage at the junction of these 2 resistors. It may give you some insight into the problem.
     
  3. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    The problem states that the diodes are ideal. That must mean that the diodes have no forward voltage drop and no reverse current. So in one circuit D1 will have no voltage across it and in the other D1 will have no current flowing through it.
     
  4. ljcox

    ljcox

    49
    0
    Jul 10, 2010
    As far as I can recall, you're right - an ideal diode has no voltage drop and no reverse current.

    Imagine that diode D1 was not connected and calculate the voltage at the anode of D2.

    If the result is positive, then D1 will conduct when it is re-connected.

    If the result is negative, then D1 will be off and so no current will flow through it when it is re-connected.

    Thus the currents can be calculated very easily.
     
  5. ArFa

    ArFa

    17
    0
    Dec 7, 2010
    Are you sure that this method is right?
    Becuase I proved this but in some cases doesn't give the right results...
     
  6. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    The general technique is to replace the diode with either a short circuit or an open circuit then look for a logical inconsistency. Does the short circuit current flow in the forward direction? Does the open circuit voltage reverse bias the diode? If one of these conditions is false, then that is the wrong answer! Therefore the other condition must be true.
     
  7. ljcox

    ljcox

    49
    0
    Jul 10, 2010
    It is correct for the example given.

    See the attachment.

    Each case must be considered on its merits. See Laplace's comments.
     

    Attached Files:

  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,412
    2,780
    Jan 21, 2010
    Read my very first suggestion and just do the calculations for a voltage divider using just the resistors.

    Calculate the voltage at the mid point.

    Now presume there is a diode between the junction of the resistors and ground. Will current flow through the diode in either case?

    If so, what will the voltage be at the bottom of the top resistor? And therefore what will the current through the top resistor be? Now add the second diode. Given that you know the voltage at the bottom of the top resistor in this case, what current will flow through the diode between the resistors?

    If the diode to ground does not conduct, then the circuit is effectively a voltage divider. Those calculations are trivial.

    Bonus question... What effect does the diode between the resistors have?
     
  9. lexroxas

    lexroxas

    37
    0
    Apr 15, 2011
    This is an old post.

    1. In case A, both diodes 1 and 2 are conducting due to the forward bias of +5 Volts. The -5 Volts supply cannot reverse bias diode 1 because its series resistance of 10k is more than the series resistance of the +5 Volts supply. Therefore, the output voltage is zero (0).

    2. In case B, diode 1 is reverse biased and diode 2 is forward biased for the same reason above. This makes all voltage sources and all resistances in series. Why? because you can just remove diode 1 in the circuit.
     
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