Impedance Measurement of a DC-DC power supply 1

[EngApprentice]

How do you measure the input and output impedance of a dc-dc power supply? I need to characterize it from DC to 1MHz and I am wondering if an external circuitry has to be setup to measure it? I have tried the website and the FAQ and some suggested that input impedance is as simple as Z=V/I and the output impedance is the internal voltage drop of the supply between no load and full load, divide that by the full load current ((V1-V2)/Io). Could someone clarify to what seems to be a basic and fundamental problem?

 

[Skogsgurra]

Such measurements are usually done by injecting a capacitor coupled load signal and reading current and voltage for different frequencies. You need a signal source capable of delivering about 10 percent of the supply's rated voltage and current from low frequencies up to 1 MHz. You also need a dual channel oscilloscope or two wide-band meters to measure voltage and current. Z = U/I over the frequency range. There are probably specialized instruments for this. But they will cost.

Gunnar Englund

http://www.gke.org

 

[EngApprentice]

Thanks skogsgurra! I'm a fresh grad so please bare with me. For the current measurement, do I measure the peak current injecting from the signal source through the capacitor? Could I perform the test using a simple signal generator? Same setup both for input and output?

Yeah, impedance analyzers are really out of the question.

 

[Skogsgurra]

You shall measure RMS current and voltage. If your signal is a pure sine (which it shall be) then you can measure peak-peak and divide by 2.828 (two times sqrt(2)).

You may need some kind of current transducer. A clamp-on current probe usually has a limited frequency range, but there are Fluke DC - several hundred kHz clamps. And Chauvin-Arnoux and Tektronix going well beyond 1 MHz.

An ordinary signal generator doesn't produce more than about 20 mA (10 V and 50 ohms internal impedance) so it may be insufficient if you are checking power supplies with more than about twenty times that current (bad accuracy).


There is an entirely different way of doing it. Just connect your generator via a blocking capacitor and see how much you can influence the output voltage. Measure AC voltage and calculate internal impedance from the measured AC component. Example: You have set the generator to 5 V RMS at no load. Connect to the PSU and measure AC signal. Say that is 0.2 V (in this example). You then know that the impedance reduces your original 5 V to 1/25th or 0.2 V. This means that the PSU has roughly 1/25 of the generator's 50 ohms or 2 ohms. You should set up the proper equation for the voltage divider created by generator 50 ohms and PSU internal impedance to get the correct values.

Gunnar Englund

http://www.gke.org

 

[fsmyth]

The impedances are as simple as that -- at DC.
I would suggest using shunts, and being careful about
scope grounding.
If you use Gunnar's second method, calibrate the output
of the cap using some low-value resistors; the output
impedance will no longer be 50 ohms. If the generator
output jack has a maximum voltage rating, chances are
that it is capacitively coupled internally, and can be
directly connected (in which case it IS 50 ohms, if so
labeled).

<als>

 

[EngApprentice]

I was suggested to inject the ac signal through an isolated transformer, elimnates grounding issues and additional impedance might have introduced using caps and resistors! But skogsgurra was right about the insufficient current sourcing out of the generator, I see a voltage attenuation going through the transformer (nothing else connected just source and scope on other end) just changing the frequency!!! Or is the transformer acting like a filter now? What's going on?

 

[Skogsgurra]

Yes, the transformer is very much a filter. There is stray inductance and there are eddy currents in the core. Both reduce the signal as frequency gets higher.

There is yet another way of doing it. Use a switch transistor and feed a square wave to the DUT. Take FFT of current and voltage. Then calculate impedance by dividing U and I at different frequencies in the spectra. With a switch transistor, you can easiy have frequency components up to 1 MHz.

If you do not need to go all the way to 1 MHz, you can use an audio amplifier to have almost any signal power you want. At least up to hundreds of watts.

Gunnar Englund

http://www.gke.org

 

[EngApprentice]

Thanks Gunnar, I'll probably find the transformer with the least attenuation with respect to my frequency range before venturing using a switch transistor. They are all very good suggestions.