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How to power 7 segment display?

  1. Nov 27, 2007 #1
    I have a 7 segment display that is rather large. It requires 20vdc and about 200mA to light up.

    Right now I have a circuit that outputs to a BDC-7segment decoder, however, the decoder isn't able to provide enough power to light up the segments on the display. What I was thinking was to use a transistor for each segment to handle the additional power. The transistor will be triggered by the 5vdc signal from the decoder.

    My question is, how do I wire each of 7 transistors into the 7-segment display? The display is CC (Common Cathode), and thus has me stumped. If it were common anode, I could just use an NPN and put the display between the +20vdc and the collector. Also, what would be better..a mosfet or a BJT? Any suggestions?

    I have attached an image of how the display is wired.

    Thanks
     

    Attached Files:

  2. jcsd
  3. Nov 27, 2007 #2

    rbj

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    in my opinion, you should tie the tops (anodes) of the LEDs together to a common +20v and the bottom (cathodes) of each LED would be tied to the collector of a good medium power NPN transistor (that might be driven by another NPN in a Darlington circuit). the emitters of all of the NPN transistors would be tied together and connected to ground. for each NPN, there would be a pull-up resistor tied between the base and +5v or +3.3v or whatever is the B+ for your logic. you use a pull-up because most logic can "sink" more current than it can "source" (i.e. they make pretty good switches to ground), so the resistance of the pull-up resistor would be determined by how much current the logic can sink when it goes to logic "0".
     
  4. Nov 27, 2007 #3

    berkeman

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    Staff: Mentor

    To drive a CC LED display, you use high-side switch transistors. Use PNP BJTs with enough current and power capability. You will need to do level translation for the 5V decoder signals, but that just amounts to driving low-side NPN transistors that sink collector current to pull down on the PNP bases to turn on the high-side switches. Depending on the output level of the decoder, you may need an inverter between each decoder output and the associated NPN low-side drive BJT.
     
  5. Nov 27, 2007 #4

    rbj

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    whoops. i missed in the original post that it was common cathode.

    sorry, but berkeman is correct, instead of the easy NPN transistors that live close to ground, all of your LEDs are already tied to ground, so it's PNP transistors on the "high side". now how to connect those transistors that are connected to +20v to a bunch of logic that live in a 0 to 5v environment, is more of a problem. i think, instead of the Darlington thing, you will have NPN transistors connected with the collectors connected to the bases of the high-side PNPs. dunno if there need be a pull-up, pull-down, or no resistor.

    sorry for the bad advice. i need to read things more carefully. i did look at the graphic, and thought it was a common anode, but nooo, it's CC. ooops, sorry.
     
  6. Nov 28, 2007 #5
    Triden,

    Take a look at the ckt I have attached. I just used a NPN to drive the LEDs.
    The voltage at the base of the transistor has to be greater than 0.7v+forward voltage drop across the LED.
    I think 5V will be more than enough.
    LEDs are usually connected on the collector side of transistors, but I connected them on the emitter side.
     

    Attached Files:

  7. Nov 28, 2007 #6
    likephysics: Thanks for your reply, but I can guarantee that the ckt you atatched will not work. I thought this originally, but after thinking about it, the 20vcd on the collector will not allow the bjt to turn on with a 5vdc base. A BJT will allow you to turn on a 2amp device using only 10mA, but not 20vdc using only 5vdc if you see what I'm saying.

    Berkeman: awesome, I understand what you said and it makes some sense now to me. I am going to try using that method and see if I can get it to work. I was thinking, since I have an open-collector decoder laying around, I can use a pullup resistor on it with 20vdc and then I can drive a PNP from the output of the decoder...that way I won't need to use an NPN for level shifting..atleast I think anyhow. Thanks for the advice
     
  8. Nov 28, 2007 #7

    berkeman

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    Staff: Mentor

    Be sure to check the BVceo rating for the open-collector decoder outputs -- 20V might be kind of high for them. If they can take it, that's a neat simplification, though. Also check their current drive, and the beta of the PNP high-side drive transistors that you choose, to be sure that you don't need a current gain stage in between. Have fun!
     
  9. Nov 29, 2007 #8
    triden, I tried the ckt in the simulator. It worked.
    If you don't mind can you explain why it doesn't work.
    thanks.
     
  10. Nov 29, 2007 #9

    berkeman

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    Staff: Mentor

    The issue is that you are switching just 5V (-Vbe) across the LED with your configuration, instead of a full 20V (-Vcesat).
     
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