Talk:Schmitt trigger

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Contents 1 Rewrite 2 Circuit is faulty. 3 regenerative comparator? 4 inadequate explanation

[edit] Rewrite

I can't see that the recent rewriting has improved the article all that much, if at all. The explanation of what the device is is now somewhat obscured by the way it has been reworded - for my money the previous explanation was a better overall summary and easier to understand. Also, the mention that the device has memory is erroneous - it does not. What it does have is positive feedback, which is something altogether different. Further, there is no requirement that a schmitt trigger circuit has to also invert the input - while a typical op-amp based one will do this, non-inverting designs are also possible. While the original text can no doubt be improved upon, I don't feel this effort has achieved that. We must always write for a lay audience and avoid bringing in extraneous details in an opening paragraph - the essence of an explanation is all that is necessary and desirable. Can we have another go please? Graham 04:19, 10 October 2005 (UTC)

That was my rewrite, and feel free to change it or completely trash it if you like. Regarding specific points:

• Obviously, I thought my changes made it easier to understand. If you disagree, please change.
• The positive feedback causes it to have memory. ("Memory" simply means that the present state/output can depend on past inputs as well as the present one. The Schmitt trigger has memory because when its input is between the two threshold voltages, its output depends on past values.)
• You're right about inverting vs. non-inverting Schmitt triggers. I screwed up a bit: my goal had been to explain what a Schmitt trigger does by explaining what an inverting one does, and then explaining that non-inverting ones exist as well. I then forgot to include the explanation that non-inverting ones exist as well. (And thinking about it, it would probably make more sense to explain non-inverting ones first, and then to mention that inverting ones exist and are more common nowadays.)
• I don't understand what you're getting at with your statement that "we must always [...] avoid bringing in extraneous details in an opening paragraph"; the first paragraph of both versions is the same, and the second paragraph of my version has no details that the second paragraph of the older version lacked.

I think what's really needed is a graph demonstrating a typical trigger's I/O characteristic; text is always rather opaque compared to a figure. But again, feel free to make any change to my text that you want to make, or to revert my edit if it really bothers you. Ruakh 10:51, 10 October 2005 (UTC)

I don't wish to trash it, or even revert it - it's better to move the article forward where we can. Overall it's not too bad. I accept your argument about memory, I suppose it's fair enough to consider it in that way, though I hope that this isn't causing a muddle (it isn't to electronics experts of course, but that's not who we are writing for - they already know this stuff and so can interpret accordingly. A lay reader might think this had something to do with digital type memory, in which case this might be a bit confusing!) I completely agree that a graph would speak a thousand words. The inverting/non-inverting is a minor issue that can be easily fixed up. I might have a go one of these days! Graham 11:58, 19 November 2005 (UTC)

[edit] Circuit is faulty.

The given circuit diagram isn't a Schmitt trigger, it's just a simple inverter, or digital inverter. Ie, it has no hysteresis. For a Schmitt trigger, you need the feedback to affect the reference voltage. A 'correct' example is this one: http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/schmitt.html

I'll have a look at fixing this myself sometime :) --Ch'marr 11:03, 22 December 2005 (UTC)

AFAICS, the diagram is OK. The circuit will exhibit hysteresis, and it is not an inverter (the input is connected to the non-inverting input of the comparator). A schmitt trigger does not require that the feedback affects the reference voltage, though that is certainly one way to build one. (By the way, that link appears to be dead). Graham 00:25, 23 December 2005 (UTC)

I respectfully disagree. That circuit will not exhibit hysteresis. As soon as Vin transcends the reference voltage (unmarked in the diagram), the output voltage will hit Vs+ or Vs-. (I'm assuming a comparitor is being used, but you can achieve the same effect by reversing the inputs of a standard op-amp). To have hysteresis, the reference voltage for a positive transition, and a negative transition, needs to differ. Eg: For L-H, Vin >3V, for H-L, Vin<2V. The circuit in my link (which does work) does that... and I realise it's a link to the same place as the 'External Links' :)

So, I ask that we change the diagram to something similar to that found in the link... unless we can find another implementation that shows it to be as simple as the one in the wiki page right now. --Ch'marr 00:16, 25 December 2005 (UTC)

Okay, colour me stupid, that will work if that's actually a comparitor rather than a standard op-amp. But it assumes that Vs- is lower than the reference voltage. --Ch'marr 00:25, 25 December 2005 (UTC)

I'm not sure what distinction you're drawing between a comparator and an op-amp - while circuits may be optimised for one application of the other, in essence there is no difference. So your remarks about 'reversing the inputs' don't mean anything sensible to me. The circuit pictured is probably not the most ideal design to be honest, but will exhibit hysteresis as long as it is driven from a low source impedance. We also need to choose our resistors so that the feedback resistor is somewhat larger than the input resistor. Assume the output of the circuit is in the low state. The resistor network forms a potential divider to the -ve supply rail. (and yes, I would assume that the -ve supply is more negative than ground, which is not unreasonable, as most op-amp basic circuits are shown with this arrangement in mind). The input needed to flip state will therefore be somewhat higher than Vref. Once it reaches this value, the comparator will flip, its output going high. The resistor network now forms a potential divider to the +ve supply rail, so to bring the + input back below the value of Vref will require that the circuit input is brought down much lower than the original L->H trip voltage. The only reason I say that a low source impedance is needed is because it must be able to sink or source the required current into the potential divider without a significant change in its own voltage, otherwise it won't perform very well. So we just need to choose the input resistor to be large enough to make sure this doesn't happen. Now I'm not saying this is the best circuit for the job - certainly by arranging the feedback to alter the reference voltage rather than the effective input voltage is probably a better approach, but there's no reason that this wouldn't work and it's easy to understand.Graham 08:47, 25 December 2005 (UTC)

Yep, I see all that now, thanks for the explaination. I think I'm just so used to seeing the schmitt trigger set up in the same way as the link above that anything else looks weird and makes my brain explode :) I still maintain that the other circuit is a little easier to see what's happening, and more flexible (not requiring a deviation from ground). But... that's just me being petulant. Feel free to zap this section of the discussion, since it's wrong/moot. --12.104.153.15 01:59, 28 December 2005 (UTC)

[edit] regenerative comparator?

should it also be noted that it is also a regenerative comparator? Excuse me if I am not correct, but it appears to be the same thing.

Yes, it's the same thing. Though a google search for "regenerative comparator" returns just 263 hits versus 364,000 for "schmitt trigger" and 25,400 for "schmidt trigger" so it's not a term used very much at all, and to me sounds like technobabble. I don't think we should encourage the use of this wordy term for such a common circuit by giving it a wikipedia endorsement. Graham 12:00, 15 March 2006 (UTC)

[edit] inadequate explanation

Could someone please better explain the purpose of resistor R4 in the figure at http://en.wikipedia.org/wiki/Image:Opampschmitt_realistic_xcircuit.png . The current explanation is "R4 has to limit the current due to the offset voltage on the input of the amplifier". I do not understand why R4 is needed at all. Thanks.