I am trying to test a filter I have designed on a PCB which aims to minimize the noise of the input power supply. To do this, I need to have a noisy power supply. I currently have access to a benchtop power supply and an oscilloscope which can generate any waveform, but is there a way to combine the two?

I initially thought I could set a voltage on the power supply, then use the positive lead as the GND reference for the Oscilloscope signal generator, then take the output between the positive of the signal generator and the GND of the power supply, but this unfortunately doesn't work since the GND of the oscilloscope is common, so when I connect oscilloscope GND to power supply GND there is a short.

Thanks for the help!

  • \$\begingroup\$ Maybe you don't need the PSU. What kind of filter? Is the DC bias important for it? Can the FG provide a DC bias? Or are your concern about DC current saturating chokes? \$\endgroup\$
    – tobalt
    Sep 5, 2023 at 5:25
  • \$\begingroup\$ Yes, sorry - the filter is a low pass filter, and DC bias is important. The FG can provide 3V of bias but the circuit is 24V so I would like it to operate with the noisy input. \$\endgroup\$
    – JohnFahey
    Sep 5, 2023 at 5:26
  • \$\begingroup\$ What spectrum range of noise are you interested in? Or, what frequency range do you want to inject and, what amplitude? What power supply currents are present? Is the current drawn under normal circumstances DC or, does it have a lot of AC components? \$\endgroup\$
    – Andy aka
    Sep 5, 2023 at 7:40

2 Answers 2


Usually you can make something like this work. Also read here: Bias-tee


simulate this circuit – Schematic created using CircuitLab

But part selection is a bit critical. In particular, the inductor L1 must be a proper high (higher than Rsrc+C1) impedance at your noise frequencies of interest. If you wish to cover a wide spectral range, you might need to use several inductors in series here each being good at blocking different parts of the spectrum. Moreover, the inductors should be able to carry the DC current without saturation.

Typically the function generator V1 has an output impedance of 50 Ω. If you need it to be higher, you can place an additional Rsrc. Oppositely, if you need a lower noise source impedance, you might need a step-down transformer after V1.

The voltmeter (e.g. oscilloscope) can be AC-coupled.

  • \$\begingroup\$ Thankyou for this answer tobalt, just to clarify: the combination of L1, Rsrc and C1 form the bias tee, and the idea is that L1 only allows DC to pass whereas C1 and Rsrc only allow AC? Also, do you have any references or suggestions on how to select values? or does L1 need to be high impedance whereas C1 should be low impedance for the frequency of interest? \$\endgroup\$
    – JohnFahey
    Sep 7, 2023 at 0:29
  • \$\begingroup\$ Yes exactly, the inductance branch should have highest possible impedance (at your frequencies) whereas the capacitance (+ resistor) branch should be much lower. At least 10x smaller. A typical impedance for the high freq branch is 50 ohm (i.e. only the capacitor), so the inductance branch should be a couple of kohms at least. \$\endgroup\$
    – tobalt
    Sep 7, 2023 at 14:15

There are options to float the scope by using an isolation transformer, and disconnect the earth, however this carries a shock risk

The better option is to use two probes and the A-B maths function to measure the PSU in differential mode. Probe shield are either left floating, connected together, or to PSUetalwork/ ground.

  • \$\begingroup\$ Thankyou for your answer colin, but I would like a little more clarification please: Should I connect channel 1 to the PSU, and then channel 2 to the function generator? If that is the case, where does the differential signal come from if I only have 2 channels? Thanks again! \$\endgroup\$
    – JohnFahey
    Sep 5, 2023 at 6:19
  • 3
    \$\begingroup\$ Do not float the scope. If it goes wrong, the best case is damaged equipment. The worst case is a dead engineer. Use an isolation transformer on the equipment you are testing. \$\endgroup\$
    – JRE
    Sep 5, 2023 at 6:46
  • \$\begingroup\$ Channel 1 to signal, Channel 2 to gnd. Shield clips connected together. Ch1-Ch2 is the differential signal that I think you are trying to measure \$\endgroup\$
    – colintd
    Sep 8, 2023 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.