0
\$\begingroup\$

I have a device that is powered from a current source. The current source supplies 10A at 60Hz - the magnitude is roughly 5-10VPP.

I would like to inject another signal on top of that to simulate noise or data at a high frequency.

I am powering my device from the power line current through a CT, and I am also measuring the current through another CT.

Ideally I would like to be able to measure the frequency response of my power input circuitry to see what my passbands are at the input to my power supply, and I would also like to see what my passbands are at my ADC input.

So here is my harebrained Idea:

enter image description here

As you can see, the point between the two diodes should be DC Biased to 12V. There should be no DC current though. But because that section is biased up, the diodes should block any of the signal from flowing back into the current source or back into the source on my spectrum analyzer. Then I would strip the DC off using some arbitrary low valued power capacitor. I would effectively be floating my current source to a virtual ground 12V above my system ground. To do this I would need to use a cheater plug. If I tie the negative terminal of the battery to system ground and to the output tab of my UUT and tie the positive tab of the battery to the tab on the cheater plug I think this effectively makes the battery part of my AC current path which may be a problem. Anyone think this will work?

The only other thing I can think of is building some LC PI filters - one a high pass from the analyzer source, and then a low pass from the current source. I need to pass several amps of current though so I would need to keep the impedance down on the low pass. But that could be an option.

EDIT: Here is the type of current waveform I want to drive into my system. Note that I want to sweep the frequency on the higher frequency component.

enter image description here enter image description here

As of right now I leaning towards using a high current buffer amp w/ summing amp input to do this and ditching my existing current source. I can't figure out how I would wire up the above circuit to float my current source without burning the building down.

\$\endgroup\$
4
  • \$\begingroup\$ What are you actually trying to achieve - it's a lot easier to make suggestions on what you want rather than try and understand what you want from a "hair brained" circuit. \$\endgroup\$
    – Andy aka
    Jul 23, 2014 at 17:54
  • \$\begingroup\$ I explained that above. I am trying to measure the frequency response of my power system input and also my analog front end. I need to do that through my complete system including my CTs. To do that I need to energize them enough to overcome the excitation current - so I need to pass several amps of current. \$\endgroup\$
    – Nick
    Jul 23, 2014 at 17:58
  • \$\begingroup\$ I had another idea... I could build a summing amp with a buffer that is capable of driving 10-15A. I'll probably need an aluminum plate + a box fan to cool the sucker though. \$\endgroup\$
    – Nick
    Jul 23, 2014 at 18:00
  • \$\begingroup\$ My device is connected to mains power and measures current through a CT. I am trying to see what affect different frequency content coupled into mains power (such from BPL/PLC systems or any other weird noise sources) has on my system. \$\endgroup\$
    – Nick
    Jul 23, 2014 at 18:06

1 Answer 1

Reset to default
1
\$\begingroup\$

Could you make a little transformer, send the main signal through it and couple your high frequency signal in? (I don't know where to get the transformer core that will work, but perhaps you do.) I should add that I've used little caps (a few pF solder on, or a few turns of insulated wire wrapped around a lead.) to inject signals into higher impedance circuits. I didn't think that would work here.

\$\endgroup\$
6
  • \$\begingroup\$ Can you elaborate on how you propose wiring it up? I can think of a few ways to couple in my HF signal using an instrumentation transformer perhaps. But my real goal is to essentially remove the current source from the measurement. Some form of current source is a necessary evil for the test, but isn't a part of my system when its installed. \$\endgroup\$
    – Nick
    Jul 28, 2014 at 16:45
  • \$\begingroup\$ Hi @Nick, OK this idea may only be half baked, (or even worse!) But I was just picturing something like this,dropbox.com/s/dehcnnpxpt1p1ad/DSCF0045.JPG If there is something fundamentally wrong with that let me know and I'll take my answer down. (Does the voltage source have to float?) \$\endgroup\$
    – George Herold
    Jul 28, 2014 at 17:06
  • \$\begingroup\$ Okay, that is more or less what I thought you were suggesting. I don't think this solves my problem though. It may isolate and protect my signal generator's output, but (correct me if I am wrong) it doesn't change the fact that my impedance/response measurement will inherently be measuring the impedance of two parallel paths - one going through my current source and one going through my UUT. \$\endgroup\$
    – Nick
    Jul 28, 2014 at 18:19
  • \$\begingroup\$ My voltage + current sources don't need to float. I was just exploring that idea as a possible means to an end. In general I prefer grounding my test equipment - especially power supplies. I don't like getting shocked! \$\endgroup\$
    – Nick
    Jul 28, 2014 at 18:22
  • \$\begingroup\$ So I think you are trying to put different loads on your device. Some common one at 60/50Hz and then a high frequency signal to see how the device responds to that? (Is that correct?) So in my half baked idea I think you would have to float the voltage source that is driving the high frequency part. (Because the AC will be inducing a big common mode voltage across the HF voltage source.) Now that may be hard to do... (some batteries and an opto-isolator?) Hey, how about you just bang on the output with a square wave and look at the response? \$\endgroup\$
    – George Herold
    Jul 28, 2014 at 18:44

Your Answer

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

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