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current transformer circuit for measuring current

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The problem isn't clear. You have been asking about a ct circuit for an energy meter in post #1.

What's the expected input signal of your energy meter? Usual energy meter ICs are connecting directly to a standard current transformer, there's no specific "circuit" involved.

Do you already have a current transformer or a transformer core?
 

can I know the ratio of the CT?
precision rectifier is best one for converting AC to DC.
what is your i/p range, & what is your requirement in o/p?
 

dear that is as according to circuit but however thanks to all for kind suggestions and advices. i was hesitating while making this circuit and today i m trying and will be solved.
 

thanx to all designed my own ct and its working fine and giving me approx accurate values.but please a request to all my friends plz if some one is asking some thing then this mean he is starting from zero so guide him as to beginners.howerver i m very thankful to u all.
 

Hi all.

I am designing a circuit to measure the current through each of three phases via Current Transformer. Each phase consists of separate CT.

The ratio of the CT is 500:5 and my maximum primary current for each phase is nearly the same (500 Amps). I have done some calculation for the burden resistance, that will be used to convert current to voltage.

Max ADC i/p (for the PIC)= (Current at primary xRb)/Transformer Ratio

the above formula gives me approx 1 ohm, and the same formula gives me approx 10mV output/amp accross the burden.

Now the question arises that what should be the power rating of this 1 ohm resistor?



Thanks in Advance.
 

the question is simple. At maximum rating of CT you have 5amps. So the resistor must be P=I^2*r=25*1 =25watt. But i suppose the nominal current which have will be about 20 -100 amps. So the transformer will give you 1amp at burden resistor. So the wattage will be drop to P=1^2*1=1watt.
 

sabouras said:
the question is simple. At maximum rating of CT you have 5amps. So the resistor must be P=I^2*r=25*1 =25watt. But i suppose the nominal current which have will be about 20 -100 amps. So the transformer will give you 1amp at burden resistor. So the wattage will be drop to P=1^2*1=1watt.
Click to expand...

No that is not the case.:p The primary is literally gonna see around 410 amps (It's a 500kVA genset). I was wondering if there is a 25 watt resistor available>

I have once clamped the CT on one of the phase and measured the output voltage without any burden resistor with the multimeter. It sees around 7V AC. I think these voltages are going to drop down as i will put a resistor across the secondary.
 

Current transformers have a power rating specification, which defines the maximum burden resistor. It's unlikely that a standard 500A CT has more than 5 or 10 VA rating. You get better accuracy by using a lower burden value.

In any case, you can't give the AC output voltage of the burden directly to a PIC ADC, in so far the voltage calculation isn't right. And you have to consider that 5V rms is about 14 Vpp for sine voltage.
 

FvM said:
Current transformers have a power rating specification, which defines the maximum burden resistor. It's unlikely that a standard 500A CT has more than 5 or 10 VA rating. You get better accuracy by using a lower burden value.
Click to expand...

Yes mine one has a 10VA. So what is the exact formula to calculate burden resistance? I only need to know how much AC volts will the CT generate accross the burden per amp.

FvM said:
In any case, you can't give the AC output voltage of the burden directly to a PIC ADC, in so far the voltage calculation isn't right. And you have to consider that 5V rms is about 14 Vpp for sine voltage.
Click to expand...

That means the volatges i measured accross the secondry without the burden were false?

Yes i will use a presision rectifier once i am done with the AC side.

- - - Updated - - -

jayanth.devarayanadurga said:
Have you seen CT's by ABB. They have a broad range. You can measure current passing through a wire by contactless method. As it is done in clamp meters.
Click to expand...

No. I have purshased it from local market :-D. The usage is same, it clamps on the phase and generates current in the secondry that needs to be converted into voltages via a burden resistor.
 

Hi to all above - some serious points:

1) SSNQ You can't run a CT with no burden resistor, if you do the volts on the o/p will fly up until the insulation breaks down and you get the 7Vac you were talking of. A 500:5 CT (100:1) has a ratio of 1:100, there is 230Vac (or 400Vac) available on the 1 turn primary to drive current through the very high inductance of an open sec CT, hence the volts can go to 400 00, 40kV ac rms, enough to break down the winding insulation. In practice in the field CT's are installed shorted, and left shorted when not needed, i.e. burden = 0 ohms, therefore power inthe CT sec is very low (5A x very little voltage = near zero watts).

2) CT accuracy is affected by the burden resistor, if the nominal CT power is 5VA, for 5A out, the burden is 1/5 ohm, 5W, although you would use a 10W or higher resistor to keep the temp rise down to some thing reasonable, for best accuracy you might run a burden less than 1/5 ohm.
 
1. Sorry i did not understand how the 230 Vac can rise upto 40kVac:?:. It means the CT on which i did the experiment can no more be used?


2. So i have to use a 10/25= 0.4ohm 10W resistor? And is there some formula to know how much volatages would the burden resistor meausres for each ampere passing through the primary?
 

So i will use 0.2ohm 10W resistor.

At 450 Amps running through primary secondry will read 4.5Amps.

V=IR

That gives me 0.9Vac (theoretically, but near to that practicaly). Is that it?
 

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