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Can a diode light up when it's connected backwards?

  1. Mar 15, 2012 #1

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 15, 2012 #2

    Averagesupernova

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    The only way that LED will light up is if the battery voltage exceeds the breakdown voltage of the LED and it destroys itself in a bright flash. :)
    -
    The short answer to your question would be NO!
     
  4. Mar 15, 2012 #3

    cepheid

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    Just to complement what was already said: the whole point of a diode is that it is a device that conducts current in only one direction and not the other.
     
  5. Mar 15, 2012 #4

    AlephZero

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    Actually the question of whether you could physically make a light emitting zener diode is interesting (and I don't know the answer).

    But conventioal diodes usually self-destruct very fast if you try to operate them in the Zener region, because the Zener voltage is high (typically hundreds or thousands of volts) so the heat generated in the diode (volts x amps) when the Zener current starts to flow is also high.

    So the answer is "no" for any "mass produced" LED.
     
  6. Mar 15, 2012 #5
  7. Mar 15, 2012 #6
    Well I had one that did, but I think that the manufacture connected the LED backwards when it was manufactured.
     
  8. Mar 15, 2012 #7
  9. Mar 15, 2012 #8

    cepheid

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    But this can be regarded as a set of two LEDs wired in "inverse parallel" (i.e. cathode-to-anode and anode-to-cathode) correct?

    Either way, whichever one is active is forward-biased
     
  10. Mar 15, 2012 #9

    Femme_physics

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    So if I have this decoder ->

    http://img33.imageshack.us/img33/5601/decoderd.jpg [Broken]

    Current can only flow backwards? I thought the point of a decoder is that signals flow based on the signals defined in A0, A1, and A2. But in this case no signal can be received because the LEDS are blocking the way
     
    Last edited by a moderator: May 5, 2017
  11. Mar 15, 2012 #10

    I like Serena

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    Not quite.
    At the Q connections the voltage is either high or low.
    If the voltage is low, the corresponding LED lights up, with current flowing from Vcc to the low voltage Q connection.
     
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  12. Mar 15, 2012 #11

    Femme_physics

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    Ah, so the current flows FROM Vcc, into the LEDS, to the decoder, and eventually leaves as possibly A0, A1, A2 or possibly neither depends on the combination chosen?
     
  13. Mar 15, 2012 #12

    I like Serena

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    No. A0, A1, and A2 are inputs.
    The decoder will have separate connections to a voltage supply that is not shown in your drawing (just like an op-amp).
     
  14. Mar 15, 2012 #13
    If you change your dc supply to an ac one, it will ;)
     
  15. Mar 16, 2012 #14

    Femme_physics

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  16. Mar 16, 2012 #15

    I like Serena

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    Yes, but the Decoder with have extra 2 connections instead of just one.
    One to Vcc, and one to -Vcc or earth.

    Btw, I have a little trouble with you naming them Vin and Vout.
     
  17. Mar 16, 2012 #16

    Femme_physics

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    Heh, do you? :wink:


    Well, current has to come in of somewhere and come out to somewhere, so I figured Vin and Vout are the best choices.
     
  18. Mar 16, 2012 #17

    I like Serena

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    Yeah well, my problems are:
    1. Vin and Vout are not currents but voltages.
    2. Vin is usually the input signal (in your case you have 3 input signals: A0, A1, and A2) and not the power supply.
    3. Vout is usually the output signal (in your case you have 8 output signals: Q0, Q1, ..., Q7) and not the power supply.
     
  19. Mar 16, 2012 #18

    vk6kro

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    It is like this:

    http://dl.dropbox.com/u/4222062/common%20anode%202.PNG [Broken]
     
    Last edited by a moderator: May 5, 2017
  20. Mar 22, 2012 #19

    Femme_physics

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    I completely understand now, thank you :)
     
  21. Apr 2, 2012 #20

    Femme_physics

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    If I have a decoder (same one from page 1) and I am told that the decoder doesn't have a need in a choice signal - CS. What does it mean?
     
  22. Apr 2, 2012 #21

    vk6kro

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    CS usually means "chip select". So, you have pull this pin high (or low) to make the chip operate.

    This is used when several chips are connected but you only want to use one of them.

    If you have only one chip and you want to use it alone, you may have to tie this pin high (or low.. read the data sheet) to make the chip work.
     
  23. Apr 2, 2012 #22

    Femme_physics

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    According to the data the decoder acts on low. So telling me that the decoder has no need for choice signal - CS, seems like an irrelevant bit of information to solve for which code let's out which exit. I mean, it doesn't change the truth table right?
     
  24. Apr 2, 2012 #23

    vk6kro

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    It just means that the chip has to be wired with that pin grounded all the time, if you want it to work all the time.
     
  25. Apr 2, 2012 #24

    Femme_physics

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    Oh, ok, so when told to neglect it it's like in mechanics exercises where you're told to neglect friction. :)
     
  26. Apr 2, 2012 #25

    psparky

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    From what I remember from LED's from senior design....

    When you send them a "1".....they turn off.

    When you send them a "0"....they turn on.
     
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