# Help with using DSO

#### vaishya9

Joined Jul 9, 2023
3
Hello,
I am working on a project where I need to measure the peak voltage of a signal. The signal is usually at a constant voltage, but there is a small peak that appears when I press a button. Initially, I set my DSO to 1 second per division(time/division) and 5 millivolts(voltage/division) per division. I was able to capture the signal with a small peak of 20 millivolts and used the stop function in DSO to capture it. However, the peak was too small to be seen in the stop mode with the initial settings. To observe the signal properly, I decreased the time per division setting to 250 milliseconds per division. After doing so, I could see the signal clearly, but the peak value changed to 35 millivolts. I am confused about how this is possible since I kept the voltage per division setting the same at 5 millivolts per division. I am unsure what the true peak of the signal is as the peak voltage keeps increasing as I decrease the time per division value. Can you please advise if my approach is correct? Note that I used some initial values as I do not know the frequency of the signal.

#### ErnieM

Joined Apr 24, 2011
8,362
Your approach is correct. As you decrease the time/division you are increasing the resolution for fast events.

I would set the mode to single sweep, trigger set to start on this small signal. Then you can narrow down on it and see the whole event and get the best voltage and time readings.

#### Ya’akov

Joined Jan 27, 2019
8,510
Welcome to AAC.

Your thread has been moved to the Homework Help forum. AAC rules limit the sort of hep that members can provide with coursework-related questions to guidance using your own research and experimentation as a foundation. For this reason, all such questions must be posted in the Homework Help Forum to make their nature clear.

It’s good to have you join us, I am sure you will get the help you need.

#### KeithWalker

Joined Jul 10, 2017
2,927
The reason is that as you reduce the the time/div, the time for each sample also decreases. Each time sample contains the average value of the voltage during the sample. If the pulse being measured has a very sharp peak, the shorter the sampling time, the higher the average measured voltage will be at that peak. I hope this little diagram helps you to understand what is happening:

#### vaishya9

Joined Jul 9, 2023
3
The reason is that as you reduce the the time/div, the time for each sample also decreases. Each time sample contains the average value of the voltage during the sample. If the pulse being measured has a very sharp peak, the shorter the sampling time, the higher the average measured voltage will be at that peak. I hope this little diagram helps you to understand what is happening:

View attachment 298006
Firstly, thanks for sharing the information. I understood how it work. But now I am confused, how to decide what is the actual peak value and at what time/div it should be set to have the right value?
for example, I have tried to check a peak voltage of the same signal generated at 3 different time/div settings of 500ms/div, 200ms/div, and 100ms/div and I got all the 3 peak values different. As per the information you provided, it should have worked like that only. But in that case, what is the true peak value?
Please refer to the attached image(volt/div is constant to 2mv/div in all the 3 pictures)

#### KeithWalker

Joined Jul 10, 2017
2,927
If the peak is very sharp, you will probably never be able to measure its amplitude accurately using that sampling method. The result is the average voltage measured during the time of one sample. The time would need to be shorter than the width of the peak .

#### vaishya9

Joined Jul 9, 2023
3
If the peak is very sharp, you will probably never be able to measure its amplitude accurately using that sampling method. The result is the average voltage measured during the time of one sample. The time would need to be shorter than the width of the peak .
So isn't there any way to measure the peak value, actually this setup is where I am measuring the current consumed by the hardware by seeing the voltage drop across a 1ohm resistor in series to the GND line.
Also, do you think that reducing my series resistor can achieve a better resolution?

#### KeithWalker

Joined Jul 10, 2017
2,927
Reducing the series resistor will just give you a smaller vertical deflection. Decreasing the time/div will give you better resolution. The complete waveform you are interested in is only about 150mS long. Change the horizontal deflection to 20mS/div.
If you want to get the best value you can for the amplitude of the narrow positive spike, increase the trigger level to trigger on the positive slope of the spike, and reduce the time/div until you get the best results.

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