Clipping Circuit

Discussion in 'The Projects Forum' started by BurninBri, Nov 17, 2011.

  1. BurninBri

    Thread Starter New Member

    Aug 16, 2011
    27
    3
    I have a circuit that outputs 3v-17v for an enable pin. However, I need to limit this to be no more than 5V for many applications (but 2V-5V is fine). My preferred output is around 3v or so. I was thinking of putting 5 diodes together (5*0.65v = 3.25V -- but even 4 diodes which gives 2.6v should be fine). Otherwise, I would use a regulator. It's only about 40ma of current or so. This seems like the "easy way" to do it (I know there are other ways). My question: which way is better? which way is cheaper?

    Thanks!
     
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    A zener diode would seem to be an obvious solution. 40mA is not all that low though: at 5V this comes to 200mW dissiipation.

    Would it be possible to add some resistance in series before the limiter?
     
  3. crutschow

    Expert

    Mar 14, 2008
    13,016
    3,235
    What is the minimum load current you need? That will determine how much power the zener will need to dissipate.
     
  4. BurninBri

    Thread Starter New Member

    Aug 16, 2011
    27
    3
    This is only for an enable pin for a DC voltage regulator/converter (It drives a LM3150 and an LM22677. It may be used to drive a R1211N002B as well). All of these are enable pins only with very minimal current draw (and about 2V minimum to enable the chips, 5V maximum). Thanks again!
     
  5. crutschow

    Expert

    Mar 14, 2008
    13,016
    3,235
    Then a series resistor and a small zener diode to ground should work fine.
     
  6. BurninBri

    Thread Starter New Member

    Aug 16, 2011
    27
    3
    Thanks! Yes - I just put this in my circuit (and simulator with LTspice) and it seems I can use one diode instead of many. I added a resistor as well. Thanks again.
     
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