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as far I know, the schematic works also with a simple TTL NOT gate.

Is there any special reason to use the Schmitt Triggers?

enter image description here

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  • \$\begingroup\$ I have always imagined that it's to avoid a metastable edge at the transition point. You don't want the whole thing stuck stationary at some accidental fixed point. I think it would need some more formal analysis (for example, why doesn't it just mean you get twice as many accidental fixed points), but generlly I'd have more confidence that as there's hysteresis, there's probably no good metastable states. \$\endgroup\$
    – Dan
    Commented Sep 10, 2022 at 17:14
  • \$\begingroup\$ Thanks all for the answer; it is hard to choose one. After a while, I think that Mattman944 has the most effective answer. \$\endgroup\$
    – ozw1z5rd
    Commented Sep 11, 2022 at 14:39

3 Answers 3

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Schmitt triggers have hysteresis. Without hysteresis, the inverter will behave more like a linear amplifier. The frequency will be less predictable because it will depend more on the internal characteristics of the IC, than on the RC time constant. Also, the output may reverse before it reaches legal logic high and low.

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The reason is hysteresis.

It keeps the output in a stable state until input voltage crosses the threshold voltage to toggle output state.

It should not work with a standard inverter.

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This is a riff on the answer by Mattman944:

If it's really TTL (or 4000 or 74C series CMOS) but not 74HCxx or a more advanced CMOS, then the inverter with just a feedback resistor will act like a linear amplifier -- if I recall correctly, these are often even stable. There's more internal circuitry in CMOS parts starting with the 74HC series, but even there the circuit will still sorta-kinda act like a linear amplifier.

This amplifier will have some gain, which will be highly dependent on the series logic it is, and also on temperature, process variation, the actual process used to fabricate it (i.e., date of manufacture and brand), etc..

This gain will be limited in bandwidth, much like an op-amp is, but the limiting will not be strictly intentional -- instead, it'll be a byproduct of what the gate is designed for, which is to be a logic part. So this limiting will be all over the map, and will also depend on temperature, process variation, where you got the part, etc..

When you hook that up with a capacitor to ground, it'll act like a unity-gain op-amp does in the same configuration: it'll oscillate. It'll oscillate at a frequency that's determined by that totally uncontrolled gain and bandwidth of the gate-as-amplifier.

The amplitude of the output will be indeterminate (except that it'll be there). The frequency of the output will be indeterminate (except that it'll be there). How close the output is to a sine wave or a square wave will be indeterminate. All of these things will vary from part to part.

If, on the other hand, you use a Schmitt trigger in that circuit, the following things will be true:

  • If you design the circuit to work at a frequency well below the chip's design frequency, the output will be nice and square.
  • If you do some analysis based on the output voltage and the input switching points, then you can predict:
    • The highest possible frequency
    • The lowest possible frequency
    • The highest possible duty cycle
    • The lowest possible duty cycle

You will, in short, be able to state to a roomful of skeptical engineering peers and your manager, exactly what the predicted performance of the circuit will be, and the only assumption you'll have to make is that the part behaves the way the data sheet says it will. Then you can show that either the circuit will work as intended and the design can go forward, or that the circuit won't work as intended, and some other clock source should be sought.

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