# Dumb "current transformer" questions

• posted

This confused me for a while because I was thinking of conventional transformers.

I think it is better stated that the secondary of a current transformed must never be open circuited. If you do, the secondary voltage will be very high and destroy the transformer.

So if you use one meter and switch it between current transformers (for 3 phase applications) the secondaries not connected to the meter must be shorted. ie make before break switch.

chuck

• posted

Here's my plan for how to use the CT. From a post to sci.electronics.design.

Current xfmr 1 ----_ / \ + - \ / Current xfmr 2 ----~

(continued from the above diagram)

• ----------+------+----- R C=== to voltmeter

- ----------+------+-----

The leads marked "to voltmeter" can be always safely disconnected. No problem.

i
• posted

While this is correct it's not the best way to use a current transformer as a measurement device. This because most meters capable of reading a few volts AC FSD are messed up by the forward drop of diodes used to rectifiy the AC.

The trick is to use it as a true current transformer without any intermediate voltage transformation. Feed the output of the current transformer directly into a full wave rectifier - silicon diodes are OK because forward voltage drop is not important.

Short circuit the rectifier output directly through a DC AMMETER. Because a moving coil meter responds to the average value of the current it will read directly the MEAN value (0.90 x RMS) of the secondary current. The voltage drop of the rectifier diodes will slightly increase the voltage drop at the current transformer primary but will not affect the current transformation ratio. For sine wave input waveform the equivalent RMS current will accurately read as 1.11 x the indicated DC value.

Jim

P.S This is a repeat post as the original seems to have got lost. As others have commented a current transformer must never drive an open circuit.

If you're switching the meter between current transformers with this scheme use a relatively low FSD current meter -perhaps 1 to

10 ma, Make up a set of separate 5A shunts. Terminate each current transformer with a full wave rectifier loaded with one of these 5A shunts. This keeps the current transformer loaded while the meter is switched across one of the other shunts.
• posted

• posted

Yes.

I agree.

I think that what I could easily do, is parallel a higher resistance diode so that it is parallel to the rectifier bridge. Then, while the diodes in the bridge are recovering, this higher resistor would momentarily take on current. (which is low since we are talking about a transition of sinewave AC)

i

• posted

Are DC amp meters moving coil or a Darsenal movement with a shunt? If they are just a volt meter with a shunt, I think it will experience the same problem as iggys design.

Also the meter cannot be removed from your circuit or the CT will see an open circuit.

chuck

• posted

• posted

Yes, that is my plan, and I will make sure that the resistor is securely attached to the C.T. terminals.

i

• posted