How do Diodes work?

Discussion in 'Homework Help' started by Tangent100, Nov 3, 2013.

  1. Tangent100

    Thread Starter New Member

    Nov 3, 2013
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    I am having a very first look at diodes in circuits, and can't find easy-to-understand information anywhere. Unfortunately, this happens often since I understand very little in any subject I attempt...

    So, if I have a diode in series with a resistor, and it is a 'forward' diode, is the pd always going to be 0.6V? Are there diodes that get smaller or bigger pd then that? Does the value of resistor or the emf affect the pd of the diode?

    'Reverse' Diode gives an infinite resistance, in my calculations, can I say this resistance is in fact 0?

    When the total emf is smaller than 0.6V, the pd of the diode will still be 0.6V when a current passes through it? Or will it be equal to the emf (if there is no r or R in the theoretical circuit)?

    I am attempting a question regarding diodes in parallel...
    I need to work out pd & current for each resistor and diodes.
    [​IMG]
    X has a pd of 0.6 and Y seems to be reverse so infinite resistance (whatever that means) but how am I suppose to know its voltage? Do reverse diodes have any voltage at all if they don't allow current to pass through? And then I don't know how am I suppose to work out the voltages for everything else in order to work out the current.

    Thanks, I know I am asking much...
     
  2. Dr.killjoy

    Well-Known Member

    Apr 28, 2013
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  3. Danm1

    Member

    Jul 19, 2010
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    your comment about infinite resistance, no this is not 0, you probably meant to say 1/0,

    But in practice it's actually not infinite, just very large.

    Think of a diode as a one way street.
     
  4. MaxHeadRoom

    Expert

    Jul 18, 2013
    10,553
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    Or as its symbol portrays, a Check Valve.
    (Although in terms of conventional current not Electron flow).
    Max.
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,757
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    Saying that the forward voltage drop (what is usually called Vf or Vd, not pd -- but that is a matter of convention and the author you are using may simply be using a different convention) is 0.6V is an approximation that generally gets answers that are "close enough" if you are using common silicon junction diodes. Many people use 0.7V.

    In reality, for a given diode, the more current that is flowing through it in the forward direction the greater the forward voltage drop will be. But the device physics are such that you have to increase the current by a factor of ten just to see a 60mV increase in Vd for most diodes.

    If you are using diodes other than silicon, such as germanium, then the voltage drop at their typical operating currents will be different than 0.6V or 0.7V. For germanium, for instance, it is typically about 0.3V.

    Yes, but indirectly. Like a resistor, the voltage that actually appears across the diode is whatever voltage corresponds to the current flowing through it. Or you can equivalently look at it the other way and say that whatever current flows through it is whatever happens to correspond to the voltage across it. The operating point (the particular combination of voltage and current that actually results) is determined by the interaction of the device (the diode) and the circuit to which it is connected.

    Consider a diode sitting on a table all by itself. Do you expect there to be a voltage of 0.6V across it? No. That would make no sense. Similarly, if you connect it to a power supply that is 0.4V you will not see 0.6V across the diode; you will only see 0.4V across it.

    Uhh...no. There is a HUGE difference between "infinite" and "zero".

    Thank about that for a minute. If the emf is less than 0.6V but the voltage across the diode is 0.6V, then there would be a net voltage that would push current the other way!

    As explain earlier, in this case the voltage across the diode would match the voltage apply by the source and virtually no current would flow.

    The rules are actually pretty simple (using the constant voltage model, which is usually good enough). If the diode is forward biased then the current is greater than zero (meaning that current is flowing in the forward direction) and it has a forward voltage drop of 0.6V (or whatever is appropriate for that type of diode). If the diode is not forward biased, then the current is zero and the forward voltage is less than 0.6V (and a voltage of -3V, meaning a reverse voltage of 3V, is less than 0.6V and so is consistent with then). Note that this is ignoring things such as reverse breakdown, but at the level you are working at this is reasonable to ignore -- you'll get there soon enough.

    So one way to analyze a circuit is to assume the state of each diode and then analyze the circuit under that assumption. Once the analysis is done, you then verify that the conditions that result are consistent with the assumptions made. If they aren't, make a different set of assumptions and try again.

    In your circuit, you have two diodes. That means you have four possible sets of assumptions you can make: {X,Y}={ON,ON},{ON,OFF},{OFF,ON},{OFF.OFF}.

    If the diode is "ON", replace it in the circuit with a 0.6V battery oriented such that the anode is 0.6V greater than the cathode.

    If the diode is "OFF", replace it with an open circuit (the infamous "infinite" resistor).

    Do the analysis and then verify that for each diode that is "ON" that there is actually a current flowing from anode to cathode and for each diode that is "OFF" that the voltage at the anode is no more than 0.6V greater than the voltage at the cathode.
     
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