I hope you will find patience to read my beginner question.

I need to do some test on a diode using a simple setup: AC power supply + load + diode. The problem is that I must generate a sinusoidal or square waveform with 8 periods (the period duration varies from 300us up to 10ms), with amplitude varying between ±10V and ±20V,0V DC Offset (at least ±0.5V steps) and peak current up to ±250A.

These would be the ideal specifications. The minimum would be ±200A, ±10.5V and ±20V, 100Hz and 3Khz, sine or square waveform, 8 periods.

I'm not familiar with such kind of tests and I don't have any experience in this field. I'm asking here hoping someone having more experience in this field can recommend me some an alternative or some instruments.

I've searched on the internet for some instruments but I can't find something which is suitable for my setup. Or maybe I don't know exactly what to search. I've found some 4-Quadrant Amplifier and AC Sources but always are limited: if they have current capability then they have limited output voltage or vice versa.

For the load I think I can use something from http://www.frizlen.com/en/home/ but I'm not sure. I need a resistive load between 50 and 100 mOhm.

I've also thinked to create an DC-AC Pure Sine Wave inverter without a 220V transformer at the output, and power it from a High Current DC Voltage source. I've found this document:


in which they explain how to design a pure sine wave inverter and I want to ask the community if this idea might work, keeping in mind that my required output frequency are between 100Hz and 3KHz.

Is anyone who can advice me or suggest how to generate these waveforms? Thank you in advance :)

  • 4
    \$\begingroup\$ This does not appear to be a project suitable for beginners. Generating such large currents is not a simple task. Can you supply more information about this diode and what tests you hope to perform on it. \$\endgroup\$
    – Barry
    Commented Jul 13, 2016 at 15:33
  • \$\begingroup\$ It's a diode very similar with diodes used in the bridge rectifiers in car alternator . \$\endgroup\$
    – NumLock
    Commented Jul 13, 2016 at 16:16
  • \$\begingroup\$ "I need to do some test on a diode" - why? \$\endgroup\$ Commented Jul 13, 2016 at 18:37
  • \$\begingroup\$ I have seen some pretty big programmable power supplies on a popular auction site, HP/Agilent/Keysight made some pretty big units (the N8754A does 20V @ 250A), I can't vouch for their suitability for your particular application, but the ratings are right although I imagine they'd cost mega $$$. \$\endgroup\$
    – Sam
    Commented Jul 13, 2016 at 22:09
  • \$\begingroup\$ It's not me who pays for the instruments. If they cost under 30k,it's not a problem. \$\endgroup\$
    – NumLock
    Commented Jul 14, 2016 at 6:21

1 Answer 1


Look for programmable power supplies with 5000W (5kW) peak rating. Something might come up on Google.

Trying to make one on the cheap from random parts might work but may not give repeatable results for characterising parts.

A DC coupled 5kW servo/audio/lab power amplifier connected to a programmable arbitrary pulse/signal generator might be the DIY direction to investigate.

For the load resistor you can look up Shunt resistors, that should be the easy part. Make sure you use robust cables and bus-bars for the connections so you do not have volt drops in unexpected places. The frequency of 3kHz will not require you to have too much concern about inductance but you will need to keep the cables short. Check out welding and car starter motor cables for ideas and then decide if the volt drop is acceptable, your cabling may be enough of a resistor to dispense with the separate load.

Corrected my hasty calculations.

Continuous power likely not required to be at the peak rating as the tests are short in duration but impressive apparatus in any event.

Finding a used bit of gear from a Diode manufacturers lab would likely be the best if you want to keep it simple. Making a resonant capacitor discharge circuit that has only passive power components might be the cheapest DIY solution requiring only good quality measuring methodology.

  • 1
    \$\begingroup\$ You might want to check your math. 20 volts and 250 amps. \$\endgroup\$ Commented Jul 13, 2016 at 17:35

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