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Current transformers?

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seven_segment

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Hi all. Just recently I have purchased a current transformer for a project I'm working on where I require a low voltage circuit to sense current on a 240v supply. Before I carry on with this post, the specs of the transformer are outlined below:

40A:5A ratio

VA 5

Ith = 60 I 1n

Saturation Coefficient <6

I should point out that this is the first time I have worked with current transformers. I've read up on them as much as possible, although there's still some confusion.

What I really want is a variable voltage between 0-5v DC in proportion to the primary current that I can feed straight into a microcontroller. I had hoped that I would be able to select a suitable burden resistor to get 0-5v AC on the secondary, and then convert this to AC. In theory this would mean I would need a 5v/5A = 1Ohm burden resistor. However, the specs on the transformer state 'VA 5'. I can only assume means that the maximum capacity of the secondary coil is 5VA, which limits me to 5VA/5A = 1v maximum voltage on the secondary.

I have no idea if I am applying the above calculations correctly, and would appreciate it if someone could shed some light on this area. Furthermore, could anyone explain to me the best way to get 5v DC when the secondary current is 5A AC (and values inbetween in proportion to the primary current) while not exceeding the specs of the transformer?

All help very gratefully appreciated.
 

A burden resistor connected across the secondary produces an output voltage proportional to the resistor value, based on the amount of current flowing through it. With our 1:10 turns ratio transformer that produces a 10:1 current ratio, a burden resistor can be selected to produce the voltage we want. If 1A on the primary produces 0.1A on the secondary, then by Ohm's law, 0.1 times the burden resistor will result in an output voltage per amp.
More info here: **broken link removed**

So, 5A on the secondary and burden resistor of 1Ω gives you 5V, and this is exactly what you need ..

Regards,
IanP
 
Hi Ian, thanks for the reply. I understand why a 1 Ohm resistor will generate 5v due to V=IR, however I'm still confused about the VA rating. As I understand it, into a resistive load VA is pretty much the same thing as power in watts. Since P = I * V surely if I try to get 5v @ 5A from the transformer that would be 25w, which is greater than the 5VA rating?

Thanks for the help so far. I have attached the specifications for the transformer, hopefully it might be a bit of help.
 

If you are concern about VA rating of the secondary then you will need to use 0.2Ω burden resistor, amplify signal by 5 times and using peak detector convert it to DC. An example of simple peak detector based on opamps is below ..

Regards,
IanP
 
Again thanks Ian, you confirmed what I thought about the VA capacity of the transformer. I assume the peak detector circuit will give the peak voltage, i.e. 1v AC will produce around 1.41v on the output of the detector?

Ideally I would have liked a 1:1 ratio where 1v AC on the input would produce 1v DC on the output. It isn't a major problem if the circuit you posted doesn't work that way, I can easily generate a 1:1 ratio by adding another OpAmp and adjusting the gain to get the desired affect, it's just useful to know how the circuit should perform before I go ahead and breadboard it.
 

Peak detectors give you the level of 1.41, but because the voltage has to amplified to get the 0-5V range, the amplification factor can be corrected to 5/1.41≈3.55 .
Amplification of 3.55 you can achieve with non-iverting amplifier (picture below) with Rin=10kΩ, Rf=25.5kΩ (22kΩ+5kΩ variable) and Rs≈10k ..
As you operate with AC currents the supply to this amplifier and to peak detector should be symmetric (+/-12V or similar) ..
Regards,
IanP
 

One final question ( honest :D )...

You mention that I need a symmetric supply to all the OpAmps in the circuit. However, the output I require is only 0-5v DC, so can I can away with just a single rail supply here? I am of course assuming here that the peak detector circuit does give a DC output.
 

The output from peak detector is dc voltage.
However, if you don't have +/- supply for peak detector you will have problems with low voltages close to 0V.
What you can do is to generate negative voltage from positive using a simple dc-dc voltage converter (example below) ..
Regards,
IanP
 

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