Measure AC current through Shunt Resistor

Thread Starter

Tankesh singh

Joined Jan 21, 2017
13
Hi,
I am Trying to Build a current monitoring device through shunt resistor. I want to measure 0- 3 amp AC current.
i use schematic which mention in attached file. when i connect the oscilloscope i got 11 volt some noisy square wave signal. i don't understand what i do wrong.image.jpeg
 

Dodgydave

Joined Jun 22, 2012
9,438
Make all the resistors 330K, this will give you a gain of 1, so at 3Amp you should have 3V across the 1 Ohm resistor and 3V out of the op amp,.
 

ericgibbs

Joined Jan 29, 2010
10,776
hi T,
How is the +12v supply for the OPA connected to the mains supply.
I expect your local mains supply input has an Earth line.??
What does the Gnd symbol for the OPA 0V on the circuit diagram signify, is it 0V common on the PSU or Earth.?

Another point, ensure that the 330K resistors are suitable for 325Vpk AC operation.!!!!!

A complete circuit diagram would be helpful.
E
 

bertus

Joined Apr 5, 2008
20,754
Hello,

The use of a current transformer or a dedicated hall effect current measurement would be much safer as the shown direct to mains connected circuit.

Bertus
 

ericgibbs

Joined Jan 29, 2010
10,776
hi MrC,
This is the point I trying to dig out from the TS, Is the PSU 'floating' ie: fully isolated from the mains.Earth/Ground.
E
 

ronsimpson

Joined Oct 7, 2019
917
I think most all of us see that 330k and 1meg resistors reduce the line voltage only slightly. So much of the line voltage could make it to the amplifier. So the output of the amp is a " square wave signal". The input common mode range of the amp can not take that kind of signal. Reduce the 1 meg to 5k to get the input voltages down. (still other problems but …)

Here is a IC that isolates from the power line, and hands you 0.4V/amp up to 5A. It is a good place to start looking for safe ways to measure power line current.
1572013632828.png
 

MrChips

Joined Oct 2, 2009
21,853
I have done a similar project with an electronic circuit breaker.
I cannot recall how I did it. I will have a look later.
 

DickCappels

Joined Aug 21, 2008
6,587
The current transformers I have tried have poor frequency response, even when driving a summing node. The Hall sensors I looked at did not have the sensitivity I need for a particular project but this one might work. Thank you all :)
 

atferrari

Joined Jan 6, 2004
4,092
I think most all of us see that 330k and 1meg resistors reduce the line voltage only slightly. So much of the line voltage could make it to the amplifier. So the output of the amp is a " square wave signal". The input common mode range of the amp can not take that kind of signal. Reduce the 1 meg to 5k to get the input voltages down. (still other problems but …)

Here is a IC that isolates from the power line, and hands you 0.4V/amp up to 5A. It is a good place to start looking for safe ways to measure power line current.
View attachment 189675
Can measure DC as well. Wow!!
 

SLK001

Joined Nov 29, 2011
1,548
Your 1 ohm resistor is likely to blow up after a couple of seconds. At 3 amps, and P=I^2 x R, you'll have about twice the power dissipation as your resistor is rated for. Also, what op amp are you using? Well, it doesn't matter, because none will work with your setup. With AC current, your output will be AC voltage, something that you cannot get with your setup. You will need a bipolar supply. Reduce your resistor to 0.1 ohms and give your amp a gain of 10 and put in a bipolar supply and your circuit just might work.
 

ronsimpson

Joined Oct 7, 2019
917
your circuit just might work. …. give your amp a gain of 10
The signal you want to see is getting smaller, with the 0.1 ohm resistor, and the voltage you substract is hundreds of volts. The gain is already so high the output of the amp is a square wave. The input to the amp is over driving the amp.
bipolar supply
I agree with that.

Used the ACS758 a while back. Measured 50A, DC to 100khz, isolated, 100uOhms of loss.
1572209653619.png
 
Last edited:

SLK001

Joined Nov 29, 2011
1,548
The signal you want to see is getting smaller, with the 0.1 ohm resistor, and the voltage you substract is hundreds of volts. The gain is already so high the output of the amp is a square wave. The input to the amp is over driving the amp.
On retrospect, I believe that you are correct. Even though the difference between the inputs is small, the individual inputs are way too high for your common variety, everyday opamp, with peak voltages around 165V. I'm now surprised that the opamp didn't die immediately. The OP is going to have to use a specialized opamp that is designed for these voltages, or implement a discrete version. It's possible that a resistive voltage divide prior to the opamp may work, say twin divide by 10.
 
Top