Voltage amplifier (15mV signal on 120VAC offset) for shunt resistor/current sensing

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Hi,
I'm looking to create a simple circuit to convert current on a nominal 120VAC line into a voltage. I have a few designs but am having some difficulty keeping parts count low and maintaining accuracy.

What I have is as follows:
- I am using a shunt resistor, outputting 1mV per ampere of current (0-15mV total).
- I take the output from the shunt resistor (120VAC +/- 15mV) and use it as an input for two op-amps (running single supply [+3V] differential amplification).
- From the op-amps, I take the output and make it positive only via some diodes, then through a filtering capacitor.
- This gives me a voltage proportional to the current, or should..

The problems I am having have to deal with two things...
First is parts count. That is a hugely inefficient way to do things, having two entire amplification circuits, even with a dual op amp. I am wondering if there is a way to do this with a single amp.
Second, pin voltage is exceeded, as even though the voltage differential is 15mV max, it is on a 120V offset.

Do any of you have ideas on how to simplify this circuit? Op amps that might handle 120V on a pin, preferably at a low cost? That also take monopolar power? I tried a voltage divider, but if I divide the offset, I also divide the signal and lower my read resolution.

Thanks!
 

Answers and Replies

  • #2
Averagesupernova
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Two words: Current Transformer.
 
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That would be nice, and non-contact, too...
But how could I get 3V for 15A? What is the inductance of the primary wire? Orientation of the other coil? How do I calibrate it?

edit: forgot to specify, it would be ideal if this was non-disruptive (ie, could be installed as either a pass-through device or purely inductive, with no need for circuit disconnection/disassembly to install the unit).
 
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  • #5
MATLABdude
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Depending on how much current you're trying to measure, maybe a Hall Effect sensor? CUI makes some (sold at DigiKey), but it starts around 3A:
http://www.cui.com/srchresults.asp?catky=560054&subcatky=731029&subcatky2=825892 [Broken]

But if you only want to take one or two-off measurements (and you're not, say, designing instrumentation or making a long-term current monitor / logger), buy yourself a current clamp as berkeman suggests. Aside from Fluke, Extech and Meterman make more affordable (yet high-quality) clamps. If you want to splurge, get the ones that do AC AND DC current measurement (the cheaper ones only do AC--which might be all you need).
 
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Why don't you put the shunt on the low side of the load? That will give you a much smaller signal within your op-amps input range.
 
  • #7
jim hardy
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that Fluke clamp-on Berkeman suggests is a godsend to troubleshooters. I scored one at a flea matket.

Why don't you put the shunt on the low side of the load? That will give you a much smaller signal within your op-amps input range.

great suggestion !


you can put your shunt in the neutral wire but place the whole works in an insulated box with no exposed metal.
Reason: By Murphy's law, second time you use it you'll plug it into a recetpacle that was wired backwards, hot and neutral reversed, and the smoke will be let out of your opamps.

google "high side current sense" there's several other techniques. Isolating opamps is one of them.
 

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