# Analysis of a comparator circuit

I am struggling to analyze this (admittedly homework) Op-Amp circuit.

## Voltage divider approach

Now if we treat the input into the negative terminal as a voltage divider we get

$$V_- = 10V* \frac{5}{20}$$

It has been kindly pointed out in the comments (thank you Andy aka), that this is a comparator circuit.

A comparator circuit is going to output a $$\V_{cc+}\$$ if $$\V_{+} > V_{-}\$$ and $$\V_{cc-}\$$ otherwise.

$$\V_o\$$ is therefore going to roughly be (ignoring slew rate)

Is this a sensible analysis, or am I missing something here?

• Your schematic does not show an Rf Jan 9 at 17:03
• Furthermore, there is no feedback. Any formula you have for an op amp circuit with negative feedback will not apply here. Jan 9 at 17:11
• It's not an op-amp; it's a comparator. When you drew the circuit using the tool, you made a mistake in assuming an op-amp and that has compounded your thinking difficulty. Show a picture of the actual homework question. Jan 9 at 17:14
• I think you should go talk to your instructor. I think you have some very fundamental confusion about op amp circuits, which is too broad to address here. Jan 9 at 17:18
• It doesn't say it's an op-amp does it. So, all your thoughts about it being similar to a differential op-amp are misconceptions. So, fix your question to remove those misconceptions and google what a comparator does. Jan 9 at 17:21

Is this a sensible analysis, or am I missing something here?

You are no longer missing anything and your waveform looks nearly correct (more below).

• Always look for negative feedback when assuming an op-amp circuit.
• For comparator circuits either no feedback is used or a little bit of positive feedback.
• The positive feedback is useful to avoid the output toggling rapidly as both inputs get close. It would slightly skew your waveforms and is a little bit more complex to analyse but, nevertheless, it might be important in more complex homework problems.

However, you should really analyse the input waveform a little more carefully should you want a better answer. I estimate that the input waveform rises to 9 volts in about 1 second and, that means it crosses the 5 volt level at about five-ninths of a second (555.6 ms). If you look at your waveform, the output toggles negative to positive at a little less than 0.3 seconds so, strictly speaking your answer isn't as accurate as it should be.

So, after much thought, the ramping input crosses the 2.5 volts (the threshold) at 277.7 ms and your graph looks good.

• hmm, why are we looking at 5V? Is my V- computation incorrect? (10*5/20=2.5V). I used the 2.5V as the reference voltage Jan 9 at 19:27
• @sjaustirni you are absolutely correct and I shall amend my answer accordingly. I don't know where I'd got 5 volts from!!! Jan 9 at 20:05