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Opinion Please: Which of These LED Drive Circuit is Best?

  1. Jul 30, 2008 #1
    Your opinion, please. Does one of the two variations of this circuit offer a clear advantage, or is it a wash (see attached)?

    This indicator circuit turns on the LED below 11V (set by the zener). The LED cathode must be at ground (it is actually part of a tri-color LED with a common cathode), so it cannot be in the collector circuit of an NPN.

    Any suggestions on the transistor biasing?

    Thanks for your comments.
     

    Attached Files:

  2. jcsd
  3. Jul 30, 2008 #2

    berkeman

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    Staff: Mentor

    The 2nd circuit is closest, but still not quite right. Thematically, you use a PNP BJT for high-side switches, as the 2nd circuit does. But the first NPN transistor Q1 should have its collector connected to Q3's base through the resistor, so that when Q1 turns on, it pulls down on Q3's base to turn it on. Re-draw the 2nd circuit that way, and re-simulate. Should work a lot better.

    Welcome to the PF, BTW!
     
  4. Jul 30, 2008 #3
    Thanks for the suggestion, Mike.
     
  5. Jul 31, 2008 #4
    Thinking further about what you said, I'm not sure that your proposed action is what I need functionally (or maybe I don't understand). To reiterate, the LED is to be OFF above 11V.

    I'm using Q1 as an inverter; when it is on, I want Q3 off (and the reciprocal). So I don't want to pull the Q3 base low when Q1 is saturated.
     
    Last edited: Jul 31, 2008
  6. Jul 31, 2008 #5

    berkeman

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    Ah, my bad, I missed where you mentioned the polarity. But circuit #2 still won't work, because Q1 can't pull its emitter up high enough to turn off Q3, given its base biasing. I'd suggest using an additional transistor instead -- use the configuration I mentioned in my previous post, and add an extra NPN stage in the middle to do the inversion.
     
  7. Jul 31, 2008 #6
    Mike,

    Thanks very much! This may not be exactly what you envisioned, but I'm sure it is close. Regardless, it now functions perfectly (my original design would never turn off the red LED).

    I've attached the full schematic. It is a voltage monitor using a single LED indicator that lights red at 10-11.2V, amber at 11.2-12.2V, and green above 12.2V.

    I learned an important lesson from you that high side switching is trickier to design than low side switching.

    Jim
     

    Attached Files:

  8. Jul 31, 2008 #7

    berkeman

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    Nice work, Jim. I'd say you have a future in circuit design! :biggrin:
     
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