Author Topic: Vpp from DSO in Square Wave  (Read 1794 times)

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Offline vinloveTopic starter

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Vpp from DSO in Square Wave
« on: July 05, 2019, 04:21:41 pm »
This is a real newbie question I guess, but I have a guess for the answer, but I just want to clarify first, if my guesswer is correct or not.

I got this square wave of 100Hz generated from a multi meter, and connected that output to DSO.
The DSO showed a nice square wave, and it said Vpp: 3.5V, and Freq: 100Hz

But when I measure the voltage from the source of the wave with another DMM, there is nothing read on DCV. It is 0.00 volt.
Why does the DSC says Vpp from the source is 3.5V?
 

Offline Jeroentjerad

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Re: Vpp from DSO in Square Wave
« Reply #1 on: July 05, 2019, 04:28:15 pm »
Because the squarewave is not DC. It is alternating anyway, like a sinewave.
 

Offline vinloveTopic starter

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Re: Vpp from DSO in Square Wave
« Reply #2 on: July 05, 2019, 04:59:21 pm »
Because the squarewave is not DC. It is alternating anyway, like a sinewave.

Then why does DSO reads its Voltages?
 

Offline mikerj

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Re: Vpp from DSO in Square Wave
« Reply #3 on: July 05, 2019, 05:07:48 pm »
Because the squarewave is not DC. It is alternating anyway, like a sinewave.

Then why does DSO reads its Voltages?

Because DSOs are designed to calculate the voltages and frequencies of waveforms whether they are pure AC or have a DC offset.  Measure the waveform with your DMM set to AC.
 
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Offline Jeroentjerad

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Re: Vpp from DSO in Square Wave
« Reply #4 on: July 05, 2019, 05:08:40 pm »
The Vpp value of the signal, the maximal difference between the two extremes of it, is called the peak-to-peak value. Of you are measuring a sinewave the so called RMS value of the sine wave comes in play. You can measure that with the meter on AC, most of them can measure those signals and are labeled as "true RMS".
I think the meter cannot measure other than sinusoidal signals (in the ACV setting) because the formula used to do so uses a rotational factor. I suppose you know the difference between alternating and direct current? There are very many topics here explaining that.
 
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Offline vinloveTopic starter

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Re: Vpp from DSO in Square Wave
« Reply #5 on: July 05, 2019, 06:30:11 pm »
Great answers and info.
Thank you.

I was thinking because 100hz square wave is too fast alternating voltages for DMMs to capture.
But DSO are made to capture these fast changing voltages with the bandwidths.

But you have given clear and right explanations for the point. Thanks.
 

Offline rcbuck

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Re: Vpp from DSO in Square Wave
« Reply #6 on: July 05, 2019, 08:47:41 pm »
Most DMMs, even the cheap ones, can measure AC signals up to a couple of hundred hertz. If you put the DMM in the AC mode you should be able to measure something. It probably won't be 3.5 volts but you should see something above 2 volts.
 
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Offline David Hess

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Re: Vpp from DSO in Square Wave
« Reply #7 on: July 05, 2019, 11:03:49 pm »
Most DMMs, even the cheap ones, can measure AC signals up to a couple of hundred hertz. If you put the DMM in the AC mode you should be able to measure something. It probably won't be 3.5 volts but you should see something above 2 volts.

For an AC RMS reading meter, it will measure one half of the peak-to-peak value or 1.75 volts.
For an AC average reading meter, it will measure 11% higher than the AC RMS value or 1.94 volts.  Note that this assumes that the peak values are correct.  An oscilloscope will calculate the peak-to-peak value as the difference between the maximum and minimum which includes overshoot on the edges.  What you really want is the different between the top and bottom as determined by a histogram which will be closer to what a multimeter sees.

The DC multimeter measurement is zero volts because the average (actually mean) DC value is zero.  In practice it may jump around depending on the frequency because the input may not average out to zero during the measurement time.  The DC oscilloscope measurement should also be zero for the same reason.
 
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