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Diode Logic Gates circuit analysis?

  1. Mar 5, 2008 #1
    I need to understand how diode-resistor gates work. I've read a lot about what's supposed to happen based on inputs, but I need to see some numbers & analysis to understand what's going on. Where do you start in terms of trying to analyze the following AND gate?

    [​IMG]

    How I would attempt it is by assuming all diodes are off and representing each diode's voltage in terms of Va, Vb, and Vr. That'll tell me what is required to make the diode turn on (forward-biased). But then I get mixed up because if I tried to do a KVL starting @ Vdd and downwards through R2, R1, and Va, I need to know what the current is through R2 and R1, which leaves me stuck. If they're all off I know i would be 0, but what if they weren't?

    Any suggestions on how to analyze this circuit? Thanks.

    PS: I suppose this should be moved to Homework help, sorry for posting in the wrong forum.

    What I've come up with so far, and I doubt it's correct:

    0) If all diodes off, Vout = 1k/(10k+1k) * 5 = 0.45V
    1) Vr < 5.7 for D3 to be off [Vd3 = -5 + Vr]
    2) Va < 4.75 for D1 to be off [Vd1 = 0.45 - Va + 5]
    3) Vb < 4.75 for D2 to be off [Vd2 = 0.45 - Vb + 5]

    Note: Von for forward-biased diode = 0.7
     
    Last edited: Mar 5, 2008
  2. jcsd
  3. Mar 6, 2008 #2

    chroot

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    If you turn on either D1 or D2, by applying a logic-low signal to Va or Vb, you get a current path through the resistor divider formed by R1 and R2. The division is between 5V and 0.7V, the forward-biased voltage across the conducting diode. Since the resistor divider has a ratio of 10:1, the output voltage will be 1/11th of (5V - 0.7V), or about 0.4V.

    If both D1 and D2 are off, because you've applied a logic-high signal to both Va and Vb, then there is no current path, and R1 and R2 no longer drop any voltage. The voltage at the output will then drift back up to Vdd.

    - Warren
     
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