Current to voltage converter analysis, help needed

[romhunter]

I've not done much with opamp so don't really have the experience. Recently I have to do a project and it require me to use a current transformer to form a sort of overcurrent protection for AC load switch (more detail: protect the relay contact from overcurrent using the signal from the current transformer, it was what I thought though). After searching the internet the following image came up. I think it was some kind of current to voltage converter? (minus the series RC). So can anyone tell me why they put it in the first place? And what value should I pick? (Some mathematics is fine). Thanks in advance

P/s: if you can tell me better stuff to do than this then it would be even more awesome.

[Dave]

Using an opamp there means that the secondary (sensing) side of the transformer is effectively shorted and therefore your current transformer will have less effect (lower inductance) in the circuit you're measuring.

[romhunter]

Using an opamp there means that the secondary (sensing) side of the transformer is effectively shorted and therefore your current transformer will have less effect (lower inductance) in the circuit you're measuring.

I mean the series CR. Looks like some kind of differentiator or something. I don't even know what's that used for so that's why I'm asking

[Dave]

It reduces the gain of the amplifier at higher frequencies (the capacitor) but only by a certain amount (transimpedance of R||r at higher frequencies).

If you really want to understand how various feedback elements behave in these sorts of circuits, you should study up on control systems theory. This channel has pretty good videos on the topic.

[romhunter]

It reduces the gain of the amplifier at higher frequencies (the capacitor) but only by a certain amount (transimpedance of R||r at higher frequencies).

Yeah now I see, but still why would you need it? The current transformer is working at 50Hz anyway (main voltage frequency over here)

[MrAl]

Hello there,

Without the RC string we just have one resistor R1 in the feedback path, so the output is:
Vout=Iin*R1

where Iin is the current from the output of the current transformer.

With the RC string intact, the circuit acts the same except also like a low pass filter, which filters out some of the higher frequency noise that might be present on the signal.

Probably start with R2=R1/10, and C=1uf and go from there.

[Dave]

Yeah now I see, but still why would you need it? The current transformer is working at 50Hz anyway (main voltage frequency over here)

The fundamental frequency is 50Hz, yes, but most loads are not going to draw a sinusoidal current waveform, which means that there are higher frequency components present.

A horrible load, for example, would be a universal motor. Reasonably large current draw and abrupt switching, which causes large current spikes. This is why you want to limit the bandwidth of the measurement system, to avoid such disturbances from affecting your system.

You could simply omit the series resistor and just use a capacitor. Setting the corner frequency at 1kHz or so should be alright, I think.

[David Hess]

Look up "transimpedance amplifier" for details on that configuration of the operational amplifier.

The series RC circuit not only limits bandwidth limiting noise but I think it may also have been included to cancel out parasitic capacitance across the transformer which could otherwise make the operational amplifier unstable.  Transimpedance amplifiers often have a small capacitance across the feedback resistor for this reason.

The output of a current transformer should never be allowed to go open as would happen if the operational amplifier were to lose power.  The OP-07 has back to back diodes across its inputs however they will not be sufficient if the current from the transformer is more than a few milliamps so some external diodes or other protection may be called for.

Using an opamp there means that the secondary (sensing) side of the transformer is effectively shorted and therefore your current transformer will have less effect (lower inductance) in the circuit you're measuring.

Just in case this was not clear, this is a benefit to using a transimpedance amplifier with a current transformer.  It also extends the low frequency response which probably does not matter here.

[romhunter]

Thanks everyone, I got the hang of it now. Still quite bad at analyzing opamps circuit though, can somebody give me something for reference? I've done with reading a lot but still those books can't cover everything.

[MrAl]

Hello again,

I found a useful approximation for the values in this circuit, R1, R2, and C.

First:
w=1/(R1*C)

and since w=2*pi*f we have:
f=1/(2*pi*R1*C)

then:
R2=(sqrt(2)+1)*R1

R1 is somewhat arbitrary here because that is chosen based on your original circuit requirements.
f is the cutoff frequency in Hertz.

As an example, when i used R1=100 and calculated C from the above and then R2 also, i got a response of 0.735 at 500Hz instead of the exact target of 0.707 at 500Hz.  So that's close enough for most purposes like this one.  Of course testing is always required.