Author Topic: Bandwidth Calculation: Caveats!  (Read 1584 times)

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Online T3sl4co1lTopic starter

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Bandwidth Calculation: Caveats!
« on: November 25, 2015, 04:45:08 am »
Some threads ago, there was confusion over what the proper way to measure oscilloscope bandwidth is.

There's a rule of thumb:
BW = 0.35 / t_r

If t_r is in us, then BW is in MHz, etc.

But this only applies when the applied waveform is an ideal step, and only when the passband characteristic is nearly linear phase!

When the response is peaked (rather than falling off gradually, as a Bessel filter does), the risetime doesn't actually change much (though there is much more ringing present, obscuring it), but the bandwidth can be extended more than double.

Here's a recent example I cooked up (in simulation, at the moment).  The circuit which generated the waveform is a crude model of a distributed amplifier (4 stages).  This circuit uses a chain of inductors to harness the amplifying devices' capacitance as a lumped transmission line: the devices act in parallel at low frequency, but additively at high frequency.  Whereas a single stage might do 10MHz (or any number of devices wired directly in parallel), the total achieves over 30MHz.

Top: AC small signal response (frequency)



Bottom: Time domain step response

The passband is +/-1dB up to 34MHz, and cutoff (-3dB) at 36MHz.

But the rise time is only 15ns, which implies a bandwidth of 0.35 / 0.015 = 23MHz.

If I tested this circuit with an impulse, I would observe slightly lower risetime.  A simpler circuit (with not as sharp a cutoff) would show a much more exaggerated response.

So don't do step response testing with waveforms other than step, because that's a necessary part of it being a step response test!  And if the time domain response isn't perfectly gentle and well-behaved, expect to get dubious results using that old rule of thumb.

That's why rules of thumb are not theorems or laws!

But
« Last Edit: November 25, 2015, 04:47:06 am by T3sl4co1l »
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Offline nbritton

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