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Diodes in a circuit

  1. Mar 2, 2014 #1
    I wanted to post a similar looking circuit but I couldn't find any, so I'll have to describe it unfortunately.

    Imagine a circuit where you have two resistors and you want to measure the voltage across each of those resisistors so you put a voltmeter parallel to each one. Except, instead of voltmeters, imagine diodes with one facing the current and the other facing opposite.

    The source pd is 3V with negligable internal resistance, the forward facing diode is parallel to a 10Kohm resistor and the reverse facing diode is parallel to a 5k resistor.

    We want to know the pd and current for each resistor and diode, and then what the pds and currents would be if the diodes were reversed



    2. Relevant equations

    V= IR, Kirchoff 1st and 2nd law, forward diodes have pd of 0.6V...


    3. The attempt at a solution

    the pd is 3V for each diode and resistor? (incorrect)
    the pd is 0.6V for the forward facing diode (correct) so the pd for the parallel 10K resistor is 2.4V (incorrect)

    I know that voltmeters have a very high resistance so that they can read a pd as close to what the resitor's pd is and so the diode with its very high resistance in the reverse would read the same pd as the parallel 5k resistor. However I still can't work that out as I don't know the current for the circuit, And I can't can't work out the current because I don't what the total resistance is because I don't know what the resistance of the forward diode is.

    I've tried alot of algebra but can't get anywhere.
     
  2. jcsd
  3. Mar 2, 2014 #2

    ehild

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    Homework Helper
    Gold Member

    Does the circuit look as that in the picture?

    You can take that the resistance of the forward diode is zero, and the resistance of the reverse-biased diode is infinite.

    ehild
     

    Attached Files:

    Last edited: Mar 2, 2014
  4. Mar 2, 2014 #3
    The answer that my textbook gives for the current of the forward diode is 420 microA, so I calculate the resistance as 1429 ohms.
    Hmm I think my textbook has made a mistake in giving me too many unknowns with not enough knowns.
     
  5. Mar 2, 2014 #4
    Oh and yes the circuit does look like that except the diodes are facing each other.
     
  6. Mar 2, 2014 #5

    NascentOxygen

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    Staff: Mentor

    The textbook is okay, there is nothing wrong with the question.

    What characteristics are you going to assume about the diodes?

    Does any current flow, for the arrangement you describe? Trace the current path.
     
  7. Mar 2, 2014 #6
    Ok I've got it.

    I want to show my calculations because I'm not sure I did it the most efficient way.

    Firstly I can see from Kirchoff's laws that the pds will be 0.6V for the forward diode and parallel resistor.
    Therefore, the pd for the resistor and reverse diode will be 2.4V

    Now the tricky part was getting the current. The current for the 10K resistor was simple as 0.6/10,000 = 60 microAmps. I don't know the current for the forward diode because I don't know its resistance. However I do know that the current leaving the junction is equal to the current leaving the junction for the other diode-resistor pair. Because the other reverse diode has no current, the current leaving the junction is simply 2.4/5,000 = 0.48milliA which also what the current through the parallel resitor is.

    I can then use this equation V1/R1 = V2/R2 and plug in 0.48mA = 0.6/R2 . This then gives the value 1250 ohms.
    Becuase I now know the total resistance leaving the junction, I can use 1/1250 = (1/10,000) + (1 Rdiode) . The resistance comes out as 1429 ohms for the forward diode which then gives me the current 0.42mA if use the 0.6V information.

    it was fun getting there but I'm still not sure if I took the simplest route...
     
  8. Mar 2, 2014 #7

    NascentOxygen

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    Staff: Mentor

    Good, that's one way. But you didn't have to determne an equivalent resistance for the diode. 480uA - 60uA would have got it.
     
  9. Mar 3, 2014 #8
    Doh!! But thanks
     
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