# How to estimate the analog bandwidth?

I have created the low cost oscilloscope and need to estimate the analog bandwidth.

I do not have any specialized equipment and just trying to estimate it by watching the response to the square signal.

Here is the 0.5MHz square wave signal:

I think I can estimate is as 6-8MHz. Am I right?

• You might want to tell us about your thought process to verify it. And share some measurements. Rise time measure, rise time of the actual square wave etc. – PlasmaHH Oct 8 '18 at 10:39
• I think it's less, assuming your input is really a square wave. Which btw I would verify using a better scope connected at the same time as your device. – Dmitry Grigoryev Oct 8 '18 at 10:41
• The problem is that my front end opamp has a slew rate of 20V/us and it limits the rise time . I do not have a decent sine wave generator to measure it properly. – P__J__ Oct 8 '18 at 10:46
• Weird ripples causing odd step-type distortions – pipe Oct 8 '18 at 15:21

If your scope's input amplifier has a frequency response of a first-order RC-filter, you can roughly estimate the bandwidth from the rise time:

$$BW ≈ 0.35 / t_R$$

To clarify, bandwidth is defined by the frequency which is attenuated by -3dB, and the rise time corresponds to the input signal going from 10% to 90% of its amplitude.

Of course, this only applies when you're sure that the observed rise time is due to your scope delaying the signal which originally is (close to) an ideal square wave. If your input signal has a known rise time itself, it should be subtracted from the measured rise time before applying the formula. At 500kHz however, I expect your square wave to be very close to ideal, compared to the rise time you observe with your scope.

• Taking into account the OPAMPs slew rate (which will no affect the sine wave) my approximation is about 6.5MHz so my first guess was not bad :). Thank you very much for your help. Tomorrow I will test it using proper sine wave generator. – P__J__ Oct 8 '18 at 12:57

It's very hard to say whether your estimation is right without knowing more about the system and the input signal. Looking at the rise and fall times it seems reasonable by eye, but if you want a good estimation of bandwidth, it makes much more sense to use a sinusoid waveform rather than a square one. With the square input your effectively checking the slew rate, but you can't be sure how much of the slew rate limiting is happening because of your source and how much is happening because of the scope.

By sweeping the frequency of a sine input, you should be able to monitor the frequency at which the displayed amplitude drops by 3dB (voltage amplitude becomes $$\1/\sqrt{2}\$$ ), which will give your -3dB corner frequency. You will also be able to measure the rolloff if you're so inclined, this will depend on how you have designed the input stage of your scope.

You say you have no specialized equipment, so assuming you don't have a sinusoid function generator, I would suggest building something like a Wien Bridge Oscillator. This will give you a neat sinusoid source, with only a few components. By changing the resistor values, you can get different frequencies for your sweep. If you don't have a small low voltage bulb, there are other designs which don't need it (you lose a bit of sine linearity though).

• "you can't be sure how much of the slew rate limiting is happening because of your source and how much is happening because of the scope" - wouldn't that also be the case with a sine generator? – Dmitry Grigoryev Oct 8 '18 at 11:00
• @DmitryGrigoryev In theory yes, but by building the wien bridge osc with a fairly high speed amp, you can more or less guarantee that the source isn't slew-rate/bandwidth limited, so you're just measuring the limitations of the frontend. It's a bit rough for sure, but with no bench equipment it's not a bad solution. Edit: plus seeing that P__J__ used a 20V/us amp in the frontend, that makes it easier to know how fast the oscillator amp should be. By picking something with at least 40V/us slew rate, that should cover it. – Matt S. Oct 8 '18 at 11:05