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How do I use a 7 segment LED display to show steps ie fan is speed 1,2,3...8,9 etc?

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Discussion in 'LEDs and Optoelectronics' started by apples, Sep 21, 2012.

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  1. Sep 21, 2012 #1
    apples

    apples

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    Jul 1, 2012
    I have successfully used a 4511 IC to make numbers 1-9 appear on a 7 segment led display, cool.

    Now how can I use this to say have it display a number for a set voltage? Example a 12v fan, I make a circuit that gives 3v, 6v, 9v and 12v. Have 4 switches one for each of the voltage taps. So then how can I use that 7 segment LED display to show speed 1,2,3,4 which would correspond to the related voltages?
     
  2. Sep 21, 2012 #2
    CocaCola

    CocaCola

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    Apr 7, 2012
    Send the appropriate binary number to the 4511 as each circuit is powered up... Since you have varying 'switched' voltages it's probably best to send the binary number via transistors off a set logic rail, so that you know the logic levels are correct in all cases...
     
  3. Sep 21, 2012 #3
    gorgon

    gorgon

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    Jun 6, 2011
    Use one transistor for each voltage, and code the binary value with diodes.

    For this application the diodes could even drive the segments of the display directly, without the 4511 in between. You'll need a couple of diodes(5 really) more, but save the chip.

    TOK ;)
     
  4. Sep 21, 2012 #4
    Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    You can use a decimal to BCD encoder (e.g. 74xx147) to convert the (decimal) signals from the switches. A simple circuit along the lines of what CocaCola stated could look like this:
    [​IMG]
     

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  5. Sep 21, 2012 #5
    CocaCola

    CocaCola

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    Apr 7, 2012
    Yep, exactly what I had in mind...
     
  6. Sep 21, 2012 #6
    gorgon

    gorgon

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    Jun 6, 2011
    With a 74xx147 you also need to generate a 5V supply for the logic chips. It is possible to get 4511 versions that work on 12v, without the need for an extra 5V.

    BTW if you add a quad comparator (like LM339) it is possible to sense the voltage directly without the extra transistors.

    TOK ;)
     
    Last edited: Sep 21, 2012
  7. Sep 21, 2012 #7
    Harald Kapp

    Harald Kapp Moderator Moderator

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    But the 4511 is a BCD to 7-segment decoder. That would requirea an additional decimal to BCD encoder or a more complex circuit for generating the BCD code directly from the switch positions. Not impossible, just a bunch of diodes more:
    [​IMG]
    Since the inputs of the 4511 are active High, the BCD code needs to be inverted (not shown inmy schematic).
     

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  8. Sep 21, 2012 #8
    CocaCola

    CocaCola

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    Apr 7, 2012
    If you are only counting to 1-4 only 2 diodes are needed on the #3 circuit...

    1 ------0001----I<-- 3
    2 ------0010----I<-- 3
    4 ------0100
     
  9. Sep 21, 2012 #9
    Harald Kapp

    Harald Kapp Moderator Moderator

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    Right, so just place diodes where there are none in my schematic and leave out the ones I have drawn :).
    Good thinking, CocaCola
     
  10. Sep 22, 2012 #10
    gorgon

    gorgon

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    Jun 6, 2011
    Just a question. If I'm not totaly wrong, you count 2-4-6-8 in this example. Could it be that you have shifted one bit?

    TOK ;)
     
  11. Sep 22, 2012 #11
    Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Oh, yes. Shift up one bit an d it goes 1-2-3.

    Thanks for the hint.
     
  12. Sep 23, 2012 #12
    MrEE

    MrEE

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    Apr 13, 2012
    Here is another solution. since you already are using a 4000 series CMOS logic, then how about trying the 4532 priority encoder (if you have one handy) . With this you can have up to 8 inputs refer to the truth table of the datasheet at http://www.ti.com/lit/ds/symlink/cd4532b.pdf . In your case, you'll only need to use D1, D2 and D3. Tie all other inputs low. connect all 3 outputs to your 4511 and you have what you need. Note: make sure to use pulldown resistors on the 3 inputs. Also make sure the device is enabled (Ei pin is HI)
     
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