Help using a operational amplifer as a 2 position latch circuit

Discussion in 'Digital Circuit Design' started by sornjs, Dec 29, 2017.

  1. sghioto

    Member

    Dec 31, 2017
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    ak, I would have to disagree with your second paragraph. Parallelling inverters is a common way to increase output. I have used such a circuit to control a small dc motor for years without a problem. The circuit looks clean on a scope and most likely a single pair of inverters could drive the motor if they only run at 4 ma. I do agree the CD40106 would be a better choice, I didn't have one on hand to breadboard so went with the 4069 to test.
    Steve G
     
    Last edited: Jan 12, 2018 at 8:05 AM
  2. ScottWang

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    Aug 23, 2012
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    I was redrew your board as the normal circuit below.

    [​IMG]

    DETAILED CKT_sornjs_ScottWang-n2.png
     
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  3. BobaMosfet

    Senior Member

    Jul 1, 2009
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    This is a clever little circuit. The entire operation of the circuit relies upon R3 to override the non-inverting input. Without which it would not sustain an output signal one-way or the other. The inverting input, set at a reference voltage of 500mV, gives the OpAmp a reference to climb toward upon start, otherwise it wouldn't work at all.
     
  4. AnalogKid

    AAC Fanatic!

    Aug 1, 2013
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    Referring to the circuit in post #15:
    1. R3 does not override anything. As a resistor connected to a low impedance voltage source, it *sets* the non-inverting input voltage level as equal to the output voltage but at a much higher impedance. It is the switches, with their vastly lower resistance, that override it.
    2. The inverting input reverence level is 50% of Vcc, not 500 mV.
    3. The start-up state of the circuit is random. This is addressed in later posts.

    ak
     
  5. AnalogKid

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    Here is a first pass at the 555 version mentioned in post #28. It powers up with both outputs low. Each 555 needs a decoupling capacitor.

    Note - CMOS 555's are much better at these long time delay periods than the original bipolar parts. Also, the CMOS output stage does not have the bipolar part's cross conduction current spike.

    ak
    OpAmp-Latch-4.gif
     
    Last edited: Jan 12, 2018 at 6:10 PM
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  6. sghioto

    Member

    Dec 31, 2017
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    Revised schematic using a CD40106 inverter as suggested by ak. Output : 10.5 volts at 8ma., 5 second drive time.
    Steve G

    EEE CD40106 motor driver.jpg
     
  7. AnalogKid

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    What software are you using for schematics?

    ak
     
  8. sghioto

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    Dec 31, 2017
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    Windows Paint.

    Steve G
     
  9. MrAl

    Distinguished Member

    Jun 17, 2014
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    Hi,

    Some cute ideas in this thread i think. I have not been able to read them all yet though.
    But what is striking about the op amp circuit is that a more typical latch is made from two cross coupled inverters, but all thta is is a high gain amp with a gain of 1 or more with the feedback connected to the input, which means we have a gain of at least 2 and that constitutes an amplifier that oscillates as a sine modulated by an increasing ramp. That leads the output to saturate either high or low, and that stops the oscillation plus keeps the output in that one state. So it does make up a latch. We just have to force the input to be another value to get the output state to change, which is similar to the way a cross coupled pair of inverters works as a latch.
    The only drawback i guess is that we have to use the whole op amp which could be made from 20 transistors. There are multiple transistor IC chips that can be turned into latches also but we have to use at least 2 transistors per latch anyway.
     
  10. AnalogKid

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    The need for a full H bridge eliminates a lot of transistor array parts. My fav is the ULN2003/4. Two sections make an excellent set-reset ff that can sink 1/2 A at 50 V. But no source...

    Actually, there's a thought. If the DC source were 24 V for a 12 V motor, then 1.5 K collector loads would source 8 mA at 12 V. A single ULN2804 would get you 4 circuits in one chip.
    Hmmm ...

    ak
     
  11. BobaMosfet

    Senior Member

    Jul 1, 2009
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    1. I see what you're saying. I misused the term 'override'. To be correct, R3 causes a voltage to be continuously fed into the non-inverting input whether or not a button is pressed, otherwise the circuit would not be able to sustain a steady output one way or the other. This can be proven by simply removing R3 altogether, in which case the circuit won't even work until a button is pressed-- and held down.
    2. Given your example, it appeared that the non-inverting input was 1VDC, and the inverting input was 0VDC (ground). In which case your voltage divider set the voltage at 50%-- ie. 500mV.
    3. The startup state of the circuit can be argued as random because neither button is pressed, however given the fact that the output is not connected to the input in an OpAmp, and that your inverting input is at ground, due to quiescent current it will output a positive signal which is then fedback into the non-inverting input, which sustains the circuit.

    Please don't think I'm arguing with you, I'm not. I agree with what you said. I'm just noting specific particulars I see and tested by building the circuit.
     
    Last edited: Jan 15, 2018 at 10:49 AM
  12. AnalogKid

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    In your response to #2, I'm not sure which schematic you are referring to. None of my schematics have a voltage divider that sets either input to 500 mV.

    In the schematics of posts 15 and 29, the inverting input reverence potential is 50% of Vcc and is independent of the non-inverting input or the output state. For a 12 V circuit, the - input sits at 6 V. The + input voltage sits at the saturation voltages of the opamp.

    The schematic in post #37 is essentially the same circuit. With bipolar supplies, GND is conveniently at 50% of the total power supply potential without needing any resistors to establish it. I don't see any condition where the + input is at 1 V. The current limiting resistor in series with the switches forms a voltage divider with the feedback resistor, so the + input should be at either +8 V or -8 V (-ish).

    Note that in #37 the negative rail is -12 V, so the output pulls the + input way below ground.

    ak
     
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