Current limiting resistor for 4511 and 7 segment?

Discussion in 'General Electronics Chat' started by jaygatsby, Jan 9, 2012.

  1. jaygatsby

    Thread Starter New Member

    Nov 23, 2011
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    I was runing some 7 segment displays with a CD4511 chip. As I understand it the chips provide the correct amount of current to the LED 7 segments. I was having problems so a friend of mine looked at the circuit and said that he thinks that current limiting resistors are need at the common cathodes of the 7 segments. Is this the case, or do the chips take care of that?

    Thank you
     
  2. Robert.Adams

    Active Member

    Feb 16, 2010
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    It doesn't look to include any current limiting. You should look in the datasheet at the output drive voltage and figure out what you are getting on the outputs.

    Find the current you want to pull for the supply voltage you've provided and you'll get the voltage that you should get on your output pin. If this is higher than the required forward voltage of your display you'll need to use current limiting resistors.
     
  3. elec_mech

    Senior Member

    Nov 12, 2008
    1,513
    193
    You should always use current-limiting resistors for LEDs, regardless of the output of the IC. This protects them and allows for proper, reliable operation. The resistors need to go onto each anode of the 7-segment display, not the common cathode. Here's why:

    If you connect a resistor to the common cathode and a 1 is displayed, two segments will be on. If an 8 is displayed, seven segments will be on. So:

    2 segments = 2x current_led
    7 segments = 7x current_led

    With only one resistor for all seven segments, the 1 will be brighter than the 8. To boot, LEDs rarely share current evenly, so some of the seven segments will be brighter than others when 8 is displayed.

    If you use a resistor for every LED segment, then every LED segment will get the same amount of current (approximately anyway) and shine with the same amount of brightness regardless of how many segments are on.

    Let's step back. Can you provide the following:
    1) A schematic of your circuit.
    2) Input voltage to circuit.
    3) LED ratings, voltage (typ.) and current (typ.) or part number.
    4) Description of the problem you are having.
     
  4. PaulEE

    Member

    Dec 23, 2011
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    If your circuit is running on 5v, the equation is as follows:

    (+5 - Vled)/Iled = Rlimit PER SEGMENT (above post 100% correct, sometimes one LEDs effective "on" resistance is low enough that it hogs all current for other segments...I've seen it!)
     
  5. PaulEE

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    Dec 23, 2011
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  6. jaygatsby

    Thread Starter New Member

    Nov 23, 2011
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    Unless I misunderstand, it looks like I need a resistor per segment. I have 8 7 segments. So I need 56 resistors for this project? One per annode (common cathode 7 segments)?

    I really thought the 7 segment 'drivers' were 'drivers' because they took care of this stuff. 56 resistors is a lot.

    Thank you
     
  7. PaulEE

    Member

    Dec 23, 2011
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    Driver = buffer = more output current than a regular old gate; typical drivers can do 10 to 30 mA (that is an estimate, just so you get the idea). Imagine if you put +5v on an LED with a series on resistance of 15 ohms (say, one segment) and it dropped 1.8v out of the 5v you applied...that's over 200mA. The LED would promptly be destroyed.

    In your case, it appears that 56 resistors is indeed what you need. That is a lot.

    They make DIP resistor packages that have 8 x 1k values in them for this purpose, as well as other values.
     
  8. PaulEE

    Member

    Dec 23, 2011
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    After looking at the datasheet, it appears that you can slave all the CD4511 chips to a pulsing source to modulate the LED current. This basically eliminated all the resistors we were talking about, since you can use a variable duty cycle square wave that'll just be enough to turn the elements on without burning them. Also, this chip can supply 25 mA/segment, so you can probably put 25-30 mA LEDs on the segment outputs without an issue. Whether this is an issue is dependant on the total power dissipation that the CD4511 can handle...
     
  9. PaulEE

    Member

    Dec 23, 2011
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    Looking once more, the CD4511 can dissipate 700 mW in the DIP package. I think you're okay to connect directly with a modulation input for the segment outputs. That's 87.5 mW/segment, including decimal,...
     
  10. PaulEE

    Member

    Dec 23, 2011
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    p.s. - that's a neat chip. I'm ordering me a few of those.
     
  11. jaygatsby

    Thread Starter New Member

    Nov 23, 2011
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    Thanks for all the help paul
     
  12. PaulEE

    Member

    Dec 23, 2011
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    No problem
     
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