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OR gate using diodes

  1. Feb 24, 2012 #1
    http://www.tutorvista.com/content/physics/physics-iv/semiconductor-devices/logic-gates.php
    I dont know how to post a pic so please go to this site and look at the OR gate

    why is resistance R connected to the point C. I dont understand its use? I think the OR gate would function without it too

    IF A=1 and B=1 both diodes are conducting and a potential is created at the point Y
    If A =0 and B = 0 both diodes do not conduct and no potential is created
    So whats the use of that resistance?
     
  2. jcsd
  3. Feb 24, 2012 #2

    wukunlin

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    Gold Member

    One reason is you need a R because your diodes would otherwise get fried.

    The other reason is that you typically use the output of a logic gate to drive another circuit. R represents this following circuit.


    edit: and I forgot to write the most important reason. In order to measure a output signal you've got to have a load (in this case R) to measure across. Without the R you are either measuring the ground, 0V with diodes blown to pieces, or if the end isn't connected to anything you simply wouldn't have a circuit
     
    Last edited: Feb 24, 2012
  4. Feb 24, 2012 #3
    Why would your diodes get fried and how does the resistance R prevent this?

    I don't really get your other answers as well.

    If you leave out the resistor, the output won't be connected to anything. This can be okay if use it to drive the coil of a relay, or a LED in series with a resistance, but:

    Some inputs, like an opamp will have a very large input resistance, and the positive charge at the output might stick around for a long time, and you're at the mercy of all kinds of electromagnetic noise.

    For some inputs you must be prepared to sink a current to get the output to 0V.
     
  5. Feb 24, 2012 #4

    cmb

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    The resistor pulls the output low if there are no inputs on the diode.

    Imagine you had a signal into one diode without the resistor. Then the signal drops off the diode; the output should go low too but the output will remain high - that part of the circuit is still charged up so it needs some path to ground to discharge whatever charge is on the output.

    This could be charge simply by virtue that it is a conductor, but also bear in mind that that output might be connected to the gate of a FET, which is capacitive. Also, the diodes will act as small capacitors so if one is 'on', it will charge the other up. That charge is dissipated by the pull-down resistor.
     
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