Different voltage measure while probe connected to 50/5A CT

[Nikhil N]

I have connected the probe to secondary of ring type CT and placed a wire as primary with one multi-meter connected at both ends. When I am checking the probe ends, its 7.1V AC, but it reading 45mV AC when connected to CT. Why this is happening?

 

[davenn]

I have connected the prob to secondary of ring type CT and placed a wire as primary with one multi-meter connected at both ends. When I am checking the prob ends, its 7.1V AC, but it reading 45mV AC when connected to CT. Why this is happening?

so if you disconnect the CT coil from the function generator and you measure the output of the generator ... what is it's output ?

 

[Baluncore]

The 10 turn secondary of a CT, (designed to conduct 5 amp), will have a very low impedance. It will effectively short circuit the output of a signal generator.

It would be necessary to buffer or amplify the signal generator with something like an automotive audio power amplifier, rated to drive a 2 ohm speaker.

 

[Nikhil N]

so if you disconnect the CT coil from the function generator and you measure the output of the generator ... what is it's output ?

7.1 V

 

[Nikhil N]

The 10 turn secondary of a CT, (designed to conduct 5 amp), will have a very low impedance. It will effectively short circuit the output of a signal generator.

It would be necessary to buffer or amplify the signal generator with something like an automotive audio power amplifier, rated to drive a 2 ohm speaker.

Yah. The secondary has only 0.2 Ohm resistance. So what you are suggesting is to amplify the signal from signal generator and feeding to secondary of CT right? Can you give me the rating of amplifier to be used? Can I use Op-amp for the purpose?

 

[Baluncore]

Can you give me the rating of amplifier to be used? Can I use Op-amp for the purpose?

I really think the use of a laminated CT to inject audio current will pose major problems. The 0R2 resistance ignores the inductance so impedance is probably closer to 2R.
But you can give it a try. There is plenty of signal voltage available from the generator so you do not need an op-amp. If you can find an old mains powered stereo amplifier with low impedance speaker output that could do the job. If not look for a power booster amp designed for automotive 12V operation it will be able to drive a low resistance coil on a CT. It will probably be rated at about 25W or more.

 

[Nikhil N]

I really think the use of a laminated CT to inject audio current will pose major problems. The 0R2 resistance ignores the inductance so impedance is probably closer to 2R.
But you can give it a try. There is plenty of signal voltage available from the generator so you do not need an op-amp. If you can find an old mains powered stereo amplifier with low impedance speaker output that could do the job. If not look for a power booster amp designed for automotive 12V operation it will be able to drive a low resistance coil on a CT. It will probably be rated at about 25W or more.

If I am using a audio amplifier which will work in 12V, I can expect a max voltage of 12V right? I need to amplify 7.1V, AC signal to 1000times. Beacuse when I used 7.1V, the voltage came to primary was only 5.3mV, which is not prominent to act as noise into the cable. Please suggest me a amplifier which will be suitable for this purpose.

 

[Baluncore]

Please suggest me a amplifier which will be suitable for this purpose.

This is the sort of amplifier I would consider using between your sig-gen and the 10 turn CT.
http://www.ebay.com.au/itm/191952191816?_trksid=p2055119.m1438.l2649&ssPageName=STRK:MEBIDX:IT

I need to amplify 7.1V, AC signal to 1000times.

No. That would produce 7kV. Your measurement was wrong.

If I am using a audio amplifier which will work in 12V, I can expect a max voltage of 12V right?

Yes, but that is more than you will need.

You are using a current transformer backwards to inject current into a common mode bundle of wires. The CT turns ratio is then 10 : 1.
Since power in = power out, the output voltage will be 1/10 of the input voltage. The current induced in the output will be 10 times the input current.
You measured 5.3mV output voltage, so you probably had 53mV input voltage.
Your signal generator was unable to do better than that when driving the 10 turn CT.

We do not know the circuit or circuit resistance you used to measure the 5.3mV.
We do not know the output current capabilities of your signal generator.

You do not have to use the 10 turn CT secondary.
You could wind more secondary turns onto the CT core to change the ratio.
I would try 50 or 100 turns, driven directly by the sig-gen.

 

[Nikhil N]

This is the sort of amplifier I would consider using between your sig-gen and the 10 turn CT.
http://www.ebay.com.au/itm/191952191816?_trksid=p2055119.m1438.l2649&ssPageName=STRK:MEBIDX:ITNo. That would produce 7kV. Your measurement was wrong.Yes, but that is more than you will need.

You are using a current transformer backwards to inject current into a common mode bundle of wires. The CT turns ratio is then 10 : 1.
Since power in = power out, the output voltage will be 1/10 of the input voltage. The current induced in the output will be 10 times the input current.
You measured 5.3mV output voltage, so you probably had 53mV input voltage.
Your signal generator was unable to do better than that when driving the 10 turn CT.

We do not know the circuit or circuit resistance you used to measure the 5.3mV.
We do not know the output current capabilities of your signal generator.

You do not have to use the 10 turn CT secondary.
You could wind more secondary turns onto the CT core to change the ratio.
I would try 50 or 100 turns, driven directly by the sig-gen.

thanks for the replay