1A current measurement using microcontroller

Thread Starter

ygs.mehta

Joined Sep 5, 2011
1


Hello sir I am new in hardware design. i have read your discussions. I think you guys help me in my project. I want to measure 1A AC current from Current transformer.
???

plz give some idea… in which direction I have to go.. I am confused about this ??? block ..
plz find attachment. basic and 1st daigram... i want replace ??? with proper circuit ...
 

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#12

Joined Nov 30, 2010
18,159
You need to use a rectifier circuit, then a filter circuit. Turn the AC into DC and smooth it out with capacitor then ADC. For best accuracy, use operational amplifier precision rectifier circuit.
 

NLightNMe

Joined Apr 20, 2011
16
At the most basic level, you only need a load resistor. The CT will be a current source whose full-scale value depends on the turns ration to the primary. You have 1A in the primary. So if you have 100 turns in the CT, you should get 0.01A full scale out of the CT. Then, just choose a resistor that will give you a voltage across it equal to the full-scale of you uC ADC when 0.01A flows through it.
 

#12

Joined Nov 30, 2010
18,159
Sorry if I was wrong. I don't do microcontrollers and so I don't know if they can ADC an AC voltage.
 

joeyd999

Joined Jun 6, 2011
4,270
Sorry if I was wrong. I don't do microcontrollers and so I don't know if they can ADC an AC voltage.
It is absolutely possible to sample an A/C signal and compute an RMS voltage (or current if you know the transfer function from current to voltage).

You need to sample faster than twice the highest frequency component of the AC waveform (Nyquist Criterion) and then do the brute force calculations to compute RMS.

This is *more* accurate than rectifying/averaging, but more difficult. Also, you need a A/D and CPU fast enough for the signals in question.

I do this all the time.

Edit: I have tricks to do this even with a slow A/D or CPU wrt the input frequency.
 

#12

Joined Nov 30, 2010
18,159
OK. Your answer is that the thing to put between the current transformer and the microcontroller is...a resistor.
 

joeyd999

Joined Jun 6, 2011
4,270
OK. Your answer is that the thing to put between the current transformer and the microcontroller is...a resistor.
You're probably going to need to do some level shifting also...you must keep the entire A/C waveform within the allowable a/d input voltage range of the CPU...usually 0V to +VRef, whatever that is.
 

#12

Joined Nov 30, 2010
18,159
You're probably going to need to do some level shifting also...you must keep the entire A/C waveform within the allowable a/d input voltage range of the CPU...usually 0V to +VRef, whatever that is.
Still simpler than a rectifier and a capacitor :)
 

joeyd999

Joined Jun 6, 2011
4,270
This is a neat circuit that I developed that works really well. I eliminated the burden resistor and replaced it with active feedback. There is always 0V across the secondary, and U408B transimpedance amp drives a current that cancels the flux generated by the primary.

This has the benefit of removing non-linearities due to changing magnetic flux density in the core (it is always 0) and magnetic hysteresis.

Vout = 1/2 Vdd + R423 * Isecondary, where Isecondary = Iprimary/n, n being the turns ratio of the transformer.

If I remember correctly, this particular circuit had a 1:500 transformer, and measured from 0 to 100mA AC RMS.
 

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If I remember correctly, this particular circuit had a 1:500 transformer, and measured from 0 to 100mA AC RMS.
I don't think U408B's negative input would stay within its operational voltage limits if T401's primary current were interrupted. Protecting its inputs is left as an exercise to the designer?
 

joeyd999

Joined Jun 6, 2011
4,270
I don't think U408B's negative input would stay within its operational voltage limits if T401's primary current were interrupted. Protecting its inputs is left as an exercise to the designer?
Talk about a blast from the past, this thread is almost 3 years old.

In any case, consider that, due to active feedback, the net flux in the core is always zero.
 
Understood, but unless your elegant circuit has a low impedance when it's powered off, I'd play it safe and snub any unexpected secondary voltages.
 

joeyd999

Joined Jun 6, 2011
4,270
Understood, but unless your elegant circuit has a low impedance when it's powered off, I'd play it safe and snub any unexpected secondary voltages.
You may do what you wish.

With the 500:1 transformer as described, the circuit as-is should withstand a 1A transient on the primary, which is well beyond the 100mA AC the circuit was originally designed for.
 

joeyd999

Joined Jun 6, 2011
4,270
BTW, Fred:

It seems odd that you joined AAC to argue the merits of a circuit posted in a long defunct thread.

Just sayin'.
 
As the best hit in three pages of google results for "current transformer burden resistor calculation", this thread is something of a public resource. It's a clever circuit and has wider application with different CTs. For example, but eliminating the magnetic flux with active feedback, it seems to me much smaller CTs could be used than in burden resistor designs - not?
 

joeyd999

Joined Jun 6, 2011
4,270
As the best hit in three pages of google results for "current transformer burden resistor calculation", this thread is something of a public resource. It's a clever circuit and has wider application with different CTs. For example, but eliminating the magnetic flux with active feedback, it seems to me much smaller CTs could be used than in burden resistor designs - not?
Ha! I am famous!

I appreciate you recognizing the value of the concept. I mistook your original query as implying that I was doing it wrong -- which flummoxed me, as the circuit has been in production for many years.

The advantage to me is the increase in accuracy and dynamic range over a standard CT configuration. I am sure there are other benefits.

I am an inventor (with patents). I try lots of things that others wouldn't consider. Sometimes I come up with something valuable -- maybe this is one of them. Let me know if/where you go with it.
 
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