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Schmidt Trigger?

  1. Dec 11, 2008 #1
    Hey guys.
    I was just wondering if the image in the file attached is a Schmidt Trigger?

    Attached Files:

  2. jcsd
  3. Dec 11, 2008 #2


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    Staff: Mentor

    Is this homework/coursework? If so, I can move the thread to the Homework Help forums.

    Pretty goofy circuit, BTW. What are the first and 3rd stages for? What are your thoughts about any Schmidt trigger functionality? What kind of function is necessary to provide hysteresis that is part of a Schmidt trigger?
  4. Dec 11, 2008 #3
    well dont take this as gospel but i think the middle stage provides hysteresis due to the two resistors, I can't work out what the purpose of the other things are. if say the op amps has voltage rails of +10V and -10V, and both resistors are 5K. if the output of the op amp is +10V then the voltage at the -ve input will be (5K/ 10K * +10V) = +5v, now say the +ve input became less than the -ve and the output flipped to -10V, due to the two resistors the voltage at the -ve input will now be (5K/10K * -10V) = -5V.

    This is hysteresis and therefore a schmitt trigger, i don't know about the rest of the circuits though.
  5. Dec 11, 2008 #4


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    The middle stage has negative feedback. It is NOT hysteresis. The first stage has gain in the negative feedback path. The output of the first op-amp will never fall below about .7 volts. Strange circuit to say the least. However, if there are parts the OP has omitted for simplicity, and considering we don't know much else about it, it may not be that strange at all.
  6. Dec 12, 2008 #5
    yes i stand corrected, i just modelled the middle stage in LTspice and there is no hysteresis, oops.
  7. Dec 12, 2008 #6


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    I must be the only one here old enough to instantly recognize that (type of) circuit. The first and last stage perform an analog "log" and "anit-log" function repectively. [BTW. Though essentually just a forward biased diode, the BE junction of an active transistor provides a much better approximation to a true exponential VI relationship than does an actual diode].

    Anyway what do you get when you take the log, then multiply by 2 then take the anti-log. It's a simple enough maths problem. ;)
    Last edited: Dec 12, 2008
  8. Dec 12, 2008 #7


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    Anyway if those two transistors are matched then the transfer characteristic is :

    y = R2/(R1^2 I0) times x^2

    where R1 is the input stage resistor, R2 the output stage resistor and I0 is the diode reverse saturation current (typically very small)
  9. Dec 12, 2008 #8


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    Hmmmmm. Squaring circuit. I thought the first stage looked like a log amp before posting but wouldn't have gotten the 3rd stage correct. Thinking about it now though, I should have gotten it.
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