Measuring current using current transformer ring type

[Esteban]

Hello everybody, well I'm working on project that consists on monitoring temperature, current intensity and rpm of an ac motor.

By now I'm using a current transformer ring type that provides 1 ampere every 50 amperes, so my idea is to rectify the current provided by the current transformer and then pass it on a current to voltage converter, the voltage will be used on a PIC 18F4550 ADC channel.

The circuit that I want to use for rectifying is the one that I've uploaded on this post. The opamps are used for compesate the voltage lost by the diodes.

I know the circuit may not be the best way to rectify a signal... but opamps are cheaper than the modules used for this applications, and I don't think that I can easily get an specific IC for this task on the place I live.

Could this opamp configuration help me? if not is there some other configuration that can help?
Another question is that I'm using 741 opamp for current to voltage converter, does 741 opamp can afford currents between 500mA and 1 amp?

 

[Harald Kapp]

There are a few issues with your circuit:

1. An opamp as current-to voltage converter does not need an input resistor (R1, R3). In fact, these are detrimental here.
2. No off the shelf opamp will be able to deliver 1A, which would be required to compensate the current flowing into the feedback resistor (R2,R4) to keep the input voltage diffferential ~0V.
3. The opamp doesn't compensate the voltage drop across D1, D2.

My suggestion:

1. convert the output current of the current transformer into a voltage using a burden resistor. Uisng e.g. 1Ω will give you 1V/1A (secondary units) or 1V/50A (primary units).
2. Use a precision full wave rectifier to convert AC to DC. Or simply add an offset (also called DC bias, again using an opamp) such that the voltage has a DC component at 1/2 Vref of the PIC's ADC and gain such that the AC component controls the PIC's ADC over the full input range. You can then do any necessary calculations (incl. "rectifying") in software.

A similar project using an Arduino is discussed here.

 

[Esteban]

Thanks a lot Harald Kapp!

Does the dc bias have to be done with an opamp or could be a voltage divisor as the circuit I've uploaded on this post?

I'm using 5V as Vref I will use 2.5V as DC bias, I want to get a relation of 5V:50A on my ADC channel (I will not get more than 40A on the motor I'm using), what value does my burden resistor needs for that Vref and so the relation I want? Do you have any reference about this AC and DC components dynamic for varying the ADC signal? What do you mean with software rectifying?

All of this sounds really interesting and also new for me, thanks

 

[Harald Kapp]

Sorry, your circuit won't work. As there is (almost) no input current into the ADC pin of the arduino, any change in voltage across the burden will not show on the ADC's pin.
This configuration can work:

 

[GPG]

. As there is (almost) no input current into the ADC pin of the arduino, any change in voltage across the burden will not show on the ADC's pin.

The ADC pin should be a high impedance? Sim works with 1M, 10K superfluous.

 

[Harald Kapp]

Of cour

The ADC pin should be a high impedance?

Of course it should, therefore "almost no input current".

Sim works with 1M, 10K superfluous.

Which sim? Which 10k?

Am I missing something?

 

[GPG]

Used 1M as input impedance. 10K in estebans diagram not needed
50Hz 1A RMS.!Ω Load..Simulator Simetrix

 

[duke37]

I do not follow all the aswers here.
I would use a bridge rectifier feeding the burden resistor to give a floating rectified output that could be connected to the computer. A capacitor may be necessary if the computer cannot log fast enough.

The transformer has a ratio of 1:50 so the diode drop of 1.2V would be reflected into the primary as 24mV.

 

[Harald Kapp]

I would use a bridge rectifier

That gives you 1.2V voltage drop on the secondary and a non-sinusoidal signal. As the secondary voltage is in the range of 0V...5V (the op wants to digitize the signal using a PIC), 1.2V is a intolerably high error component.
At low currents the measurement error will be high. Professionally it is as far as I know nerver done this way. As

 

[duke37]

If the rectifier comes before the burden resistor, it will affect the measured current by adding 24mV plus 20mΩ (with a 1Ω resistor) in the supply. There will be no 1.2V drop in the measured voltage. This is used as a current transformer, not a voltage transformer.
Any rectifier will give a non sinusoidal signal, in this case rectified AC but a capacitor across the burden resistor will smooth things out.

 

[Harald Kapp]

If the rectifier comes before the burden resistor, it will affect the measured current by adding 24mV plus 20mΩ (with a 1Ω resistor) in the supply. There will be no 1.2V drop in the measured voltage. This is used as a current transformer, not a voltage transformer.

Sorry, seems I've been a bit slow in taking up the order of components in your post. But: as the voltage is floating anyway, you don't need the rectifier at all, a simple offset circuit using 2 resistors is sufficient.

A rectifier can be used to increase the resolution of the digital signal by one bit, but one needs an active rectifier (opamp based):

• feed the rectified signal to the ADC which will see only positive signals and can digitize the amplitude at full scale resolution
• from the rectifier generate a "sign" signal, indicating whether the input signal is pos. or neg.
• combine amplitude from the ADC and sign (via an IO pin) to a sign-magnitude number which has effectively one bit more resolution than the ADC signal alone.

I also have to correct my post #4: The circuit as shown in post #3 will work likewise.

Sorry for any inconvenience I may have brought into this discussion.

 

[duke37]

I think that a bridge rectifier before the burden resistor will do all that an active rectifier will do with much less complication and no power supply required.
If the signal from the burden resistor is fed through a 100k resistor into a 10μF capacitor to get the mean, then the ADC can be quite slow. I do not see why an offset would then be required.

 

[Harald Kapp]

At least in a simple simulation this is not so:

The secondary currrent (I(L2) is heavily distorted during zero crossing which shows as clipping in the output voltage (V(n002) = voltage across R1).
Depending on the accuracy required this may be tolerable - or not.

 

[duke37]

It is some years since I used LTspice and it is not on my present computer.
The inductance of the transformer looks very low but whether this would have an effect in an ideal situation, I know not.
If I remember rightly, the source is 500A peak, the transformer has a turns ratio of 32 so that the secondary should produce 15.6A peak, giving 156V across the resistor.
I or the simulation is wrong. Could you please try it with mH instead of μH.

 

[Harald Kapp]

Using mH instead of µH works better. I didn't think this would make a difference since the sim deals with ideal components.

 

[Esteban]

Sorry, your circuit won't work. As there is (almost) no input current into the ADC pin of the arduino, any change in voltage across the burden will not show on the ADC's pin.
This configuration can work:

Thanks Harald Kapp, I've done your circuit on my spice sim. So can I do software rectifying using a MCU? does anybody has a reference about how to rectify signals with any mcu? I'm using a pic 18f4550 but any example could help.

Thanks everybody for replying

 

[Harald Kapp]

So can I do software rectifying using a MCU?

Use the abs() function:

if input >= 0 then
   output = input
else
   output = -input
end if