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I am measuring the differential voltage across a resistor connected between the supply (16.5V) and the ground. Below is my Circuit.

Circuit

Using differential probe (Probe setting 9.96:1), when I measure, I am getting a Pk-Pk voltage across the resistor as 8V. Using a passive probe (Probe setting 10:1), Probe Ch-1 at Supply and Prboe Ch-1 at Collector of NPN Transistor. Applying MATH Function=Ch1-Ch2, I am getting a Pk-Pk voltage across the resistor as 15.78V. What do I need to check? Where am I going wrong?

Please help. Thanks

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  • \$\begingroup\$ Your first step is to THINK, what do you think the voltage should be? Now you have a clue which measurement gives a reasonable result and which one gives an unexpected result. Then analyze / debug the measurement with the unexpected result. \$\endgroup\$
    – Bimpelrekkie
    Sep 11, 2019 at 8:34
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    \$\begingroup\$ What kind of differential probe do you use? \$\endgroup\$
    – jusaca
    Sep 11, 2019 at 8:41
  • \$\begingroup\$ Photos of your probe setup are usually helpful, and you should specify model numbers and links to datasheets of your probes and test equipment, and full details of what you're using to drive the transistor base. \$\endgroup\$
    – pericynthion
    Sep 11, 2019 at 8:49
  • \$\begingroup\$ keysight.com/en/pd-2156263-pn-N2750A/…. This is my differential probe \$\endgroup\$
    – user220456
    Sep 11, 2019 at 8:52

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Where am I going wrong?

From looking at the Keysight N2750A Differential Probe User's Guide, your test appears to be exceeding the allowed input range of that probe. That would explain the unexpected readings.

Here is a copy of part of page 42 from that User's Guide, showing some of the relevant specifications:

Screenshot from page 42 of Keysight N2750A Differential Probe User's Guide

Using the 10:1 attenuation probe setting you described:

  • The maximum differential voltage is 10 Vpp, but you are trying to measure approximately 15.8Vpp (according to the single-ended probes + maths measurement).

  • The maximum (positive) common mode voltage is +2.5V above 100Hz, but your 2kHz signal has a common-mode voltage of around 8.6V (from my calculations and based on that 15.8V value).

If you haven't damaged the probe by exceeding its limits, and if this is a test where you can change the parameters, then you might get the expected measurements if you keep within its limits. For example, slow down the clock to < 100Hz and use a 5V Vcc supply to that transistor. Then try again.

However if that partial schematic you supplied is from a piece of equipment and no changes can be made, then this probe appears to be incompatible with performing that measurement. You could contact Keysight for more advice.

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  • \$\begingroup\$ How did you calculate my Common-mode voltage(of the 2kHz signal) to be around 8.6V? Could you please explain? \$\endgroup\$
    – user220456
    Sep 12, 2019 at 5:26
  • \$\begingroup\$ @Newbie - Hi, You explained the voltage across the resistor was 15.8Vpp (to 1 decimal place). That sounds like the voltage to Gnd at the transistor's collector is between 0.7V and 16.5V. In other words, the average voltage at the collector, assuming a symmetrical waveform, is 8.6V with (15.8Vpp / 2 =) 7.9Vp positive and negative swings (8.6V - 7.9V = 0.7V and 8.6V + 7.9V = 16.5V). I haven't seen the waveforms from the single-ended probes, but that's my interpretation of your description. \$\endgroup\$
    – SamGibson ♦
    Sep 12, 2019 at 22:35
  • \$\begingroup\$ Another version of the calculation (same result) is: \$V_{cm} = \frac{V_{1} + V_{2}}{2}\$ where in your case \$V_{1} = 16.5 \textrm{V}\$ and \$V_{2} = 0.7\textrm{V}\$ showing that, again, \$V_{cm} = 8.6\textrm{V}\$ \$\endgroup\$
    – SamGibson ♦
    Sep 14, 2019 at 3:35

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