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The oscilloscope trace shows an 800 kHz square wave with some transition time (slope.) The way to reduce the transition time would be to use a high-speed comparator IC or op-amp.

Is it possible to reduce the transition time greatly (increase the steepness of the the slope) by using simple discrete components (such as transistors or resistors) without the use of any comparators or op-amps at high frequency?

ocrdu
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jessica smith
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    What is the frequency of the square wave and what transient time (risetime) are you looking for? These parameters will greatly affect the answer to your question. – Barry Dec 18 '22 at 13:13
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    What are the voltage requirements for the square wave? Must it be variable? – Elliot Alderson Dec 18 '22 at 13:31
  • @ElliotAlderson, no :) – jessica smith Dec 18 '22 at 13:33
  • @Barry, the frequency would be 800kHz and most op-amps and comparator will have these transient time which I try to reduce it so that it would look like perfect square wave :), so I believe the normal transistors such as npn 2n2222, will be able to reduce the transient time right? – jessica smith Dec 18 '22 at 13:34
  • @jessicasmith please answer Barry's question on what the rise time is that you **need**. Don't solve a problem before stating the thing you want! – Marcus Müller Dec 18 '22 at 13:48
  • What if the waveform is sharp in reality but your scope can't show it because it does not have enough bandwidth? You don't define any of these and you may not need any other than a better scope to see a sharper signal. – Justme Dec 18 '22 at 14:02
  • That waveform appears to have slew-rate-limited rise/fall time. Have you studied *speed-up-capacitor* methods for compensation? https://electronics.stackexchange.com/questions/378631/how-does-a-speed-up-capacitor-work-in-schmitt-trigger – glen_geek Dec 18 '22 at 14:51
  • @jessicasmith I asked two questions, and the first one is not a yes/no question. You still haven't given us an answer. – Elliot Alderson Dec 18 '22 at 15:42
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    @jessicasmith:"so that it would look like perfect square wave ". Perfection is extremely difficult. Chasing it is crazy making. Set reasonable boundaries. If square looks are important, just zoom out. – RussellH Dec 18 '22 at 17:23
  • Use digital circuit ... – Antonio51 Dec 20 '22 at 09:31

2 Answers2

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The transistors can be 2N3904. (2N2222 are also fine, but they are a bit less good in general for working with signals).

The assumed input signal is between -1V to +1V.

The input was a triangle wave, and the comparator almost made it a square wave.

It's just a simple comparator with a bit of positive feedback for stability against noise (part of he output goes back to the input through the 100k resistor). The larger the feedback resistor the less influential is the feedback, until it's an open circuit and then there's no feedback at all.

Without any feedback it would be even more able to convert to square wave, but it'd be less protected against power supply & breadboard noise. (To increase the feedback the 100k resistor can be reduced).

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ee_student
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    This is what I was thinking of. A simple differential pair. You may need something faster than a 2N2222 if you want to reduce the transition time below the 20 ns OP already has. – SteveSh Dec 20 '22 at 11:42
  • Yup, a transistor with a larger transition frequency perhaps.. – ee_student Dec 20 '22 at 21:17
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The functionality you need is "is the voltage above or below a threshold? If so, pull the output up or down very rapidly".

That's the functional description of a comparator. No matter what you'll build, it will have the functionality of a comparator.

without the use of any comparator

You can tell yourself you didn't use a comparator if you build it yourself. You could also buy and use a comparator, close your eyes and very loudly say "this is not a comparator"! (by the way, the comparators in this use case will be called "non-inverting buffer", typically, if this wave is some logic-typical level)

using simple discrete components circuit (such as transistor or resistor )

You would build a comparator out of transistors. That's not easier than using a comparator IC; and harder to make high-speed. Doing high-speed things in discrete components is harder, because suddenly your leads and board design become part of your circuit.

But…

800 kHz is not very fast. Looking at your scope output, it looks like the rise time is around 20 ns – which is pretty good, already! I'll need to think very hard to find an application where you need faster rise times at these frequencies, really. (And an application where the jitter of the wave is allowed to get worse, based on building discrete circuitry to make it steeper)

Also, look at the ripples at the edges: they are about half as wide as your rise time.

What you think is a slowly rising edge probably is much faster, it's just that your measurement setup has inherent low-pass behaviour: Your old CRT oscilloscope possibly doesn't have > 10 MHz bandwidth, you're driving it through band-limiting and loading cables … hence the ringing.

So, before optimizing anything, make sure you can actually see your edge properly. Then, figure out which edge steepness you actually need, because:

most op-amps and comparator will have these transient time which I try to reduce it so that it would look like perfect square wave

… already tells us that you're not winning anything if these edges were steeper than they are.

Marcus Müller
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