Author Topic: Probe compensation question  (Read 3737 times)

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Offline fubar.grTopic starter

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Probe compensation question
« on: November 16, 2015, 04:51:26 pm »
I've watched the excellent video made by w2aew here:



I have a question that I don't think was properly addressed: At around 12:30, when adjusting the trimmer cap and the square wave becomes rounded, that's because higher frequencies are attenuated more than lower ones.

Ok, that's pretty straightforward.  But what exactly happens when he turns the cap the other way and the square wave becomes pointy? The voltage overshoots above and below the original 0 to 5 Volts of the test signal.

What exactly causes this overshoot?

Offline Stray Electron

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Re: Probe compensation question
« Reply #1 on: November 16, 2015, 05:14:28 pm »
  The center of a square wave is made of the low frequency, the corners are made up of the high frequencies. (Odd harmonics). The misadjusted probe that shows overshoot in the corners is over attenuating the low frequencies.
 

Offline tggzzz

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Re: Probe compensation question
« Reply #2 on: November 16, 2015, 05:20:21 pm »
Instead of the higher frequencies being subjected to positive attenuation they are subjected to negative attenuation, also known as gain. In other words the higher frequencies are emphasised.
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Offline fubar.grTopic starter

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Re: Probe compensation question
« Reply #3 on: November 16, 2015, 05:25:08 pm »
Instead of the higher frequencies being subjected to positive attenuation they are subjected to negative attenuation, also known as gain. In other words the higher frequencies are emphasised.

But how can a probe made of entirely passive parts add gain to a signal? Is it due to some resonant effect?

Offline grumpydoc

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Re: Probe compensation question
« Reply #4 on: November 16, 2015, 06:07:11 pm »
Quote
But how can a probe made of entirely passive parts add gain to a signal? Is it due to some resonant effect?

All components are attenuated, but if the low frequencies are attenuated more than the high frequencies then there is relative "gain" and you will get the peaking on the rising edge that you see in the video.
 

Offline Someone

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Re: Probe compensation question
« Reply #5 on: November 17, 2015, 04:53:19 am »
Instead of the higher frequencies being subjected to positive attenuation they are subjected to negative attenuation, also known as gain. In other words the higher frequencies are emphasised.

But how can a probe made of entirely passive parts add gain to a signal? Is it due to some resonant effect?
The overall impedance of the probe provides a 10:1 attenuation when attached to an oscilloscope, so the adjustment is a relative amount of loss. It needs a trimmer as each input on a scope (and different scopes) have slightly different capacitance, to make it easier the range of adjustment is small and each probe has only a range of oscilloscope input capacitances that it can be adjusted to match.
 


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