Help needed for diodes!

Discussion in 'Homework Help' started by Teee, Dec 4, 2012.

  1. Teee

    Thread Starter New Member

    Nov 11, 2012
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    Can I know how can I calculate the diode current and current flowing through each resistors?
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Try use thevenin's theory and reduce the circuit to a single Voltage source with series resistor.
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    If you haven't had Thevenin's Theorem yet (or even after you have replaced the circuit as seen by the diode by the Thevenin's equivalent), you want to first determine if the diode is forward biased or not. Perhaps the simplest way to do this is to remove the diode from the circuit and analyze the modified circuit to find the voltage across the nodes where the diode was. If this voltage would result not result in the diode being forward biased, then you are done since putting the diode back in the circuit would not significantly change anything. But if that voltage would result in the diode being forward biased, then replace the diode in the circuit with a 0.7V battery to hold the voltage between the two nodes equal to the forward diode drop (from the figure I am assuming that the diode model you are using assumes that, when forward biased, the diode will have 0.7V across it regardless of the current through it). Then reanalyze the circuit and whatever current is flowing in the battery is the same current that would be flowing in the diode if you put the diode back into the circuit (in place of the 0.7V battery).
     
  4. anhnha

    Active Member

    Apr 19, 2012
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    Thanks, how can I prove that method work? I think it will have big difference if we remove diode.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    I don't understand the question. Are you asking how to prove that the analysis method I described will lead to a valid result?

    If the diode is reverse-biased, then removing it will have no effect. If it is forward-biased, then removing it will have a significant effect. The point of analyzing the circuit with the diode removed is only to determine if the diode will be forward biased or not when it is inserted back into the circuit.
     
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  6. justtrying

    Active Member

    Mar 9, 2011
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    you need to determine whether the diode will turn on or not given the conditions. This is done by removing the diode and analyzing the circuit without it. Questions to answer are usually two - is it forward biased and is there enough voltage to turn it on (in this case 0.7V).
     
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  7. anhnha

    Active Member

    Apr 19, 2012
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    I mean, for example, if I remove the diode and then I recaculate the voltage between anode and cathode in the circuit. If the voltage is Vd=0.3 V, how can I know that if I insert the diode into circuit, the diode will be reverse-biased?
    I understand your point that If the diode is reverse-biased, then removing it will have no effect but what about the invert?
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    If you come up with Vd=0.3V, then when you put the diode back in the circuit, will it begin to conduct or not. Imagine the moment just as you put it between two points that have a voltage that will forward bias the diode by 300mV. Will current start to flow in the diode (and we mean enough current to have a noticeable impact on the circuit)? If it will, then the diode is forward biased and you have to reanalyze the circuit with that knowledge. If not, then you are done.
     
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