Current transformer

The function of a 'current transformer' is as a near-perfect conductor on the primary circuit (it doesn't dissipate energy, ideally). If you leave the secondary open, the primary circuit is OPEN, just like you'd turned a switch off. You aren't going to measure the current that way, I hope!

Actually, the voltage would never get anywhere near that high, although = it=20 could get high enough to damage the insulation. For one thing, a 1000:1=20 ratio is a 5000:5 CT (usually), which is fairly rare. Most systems run=20 currents of 100 to 500 amps or so. If you do run current through a CT = with=20 an open secondary, it will just function as an inductor, and it will = soon=20 saturate at 60 Hz so the primary voltage drop will be only several = volts.=20 Thus the secondary will probably produce only several hundred to a = thousand=20 volts or so. However, this is often beyond the voltage rating of the = control=20 wiring and could cause breakdown of insulation, arcing, and other = damage.=20 And much higher spikes will be created when the power is switched on and =

off. This could be especially bad if the CT is in a PWM motor control = line=20 where the current may be switched at 20 kHz or so.

Most CTs are shipped with a shorting jumper or spring clip on the = secondary.=20 And some are protected by a pair of high current anti-parallel diodes. = But=20 sometimes you may need more than 0.6V peak to drive the load, and since = CTs=20 sometimes operate at 2x to 5x nominal rating, that can be a lot of power = to=20 dissipate in the diodes.

Paul=20

Yup.

John

On a sunny day (Mon, 23 Jan 2012 22:29:51 -0500) it happened DJ Delorie wrote in :

And even connect the sense leads below the connection of the transformer, just in case of a bad contact:

----- | |---------- CT [ ] electronics |---------- | ----

On a sunny day (Mon, 23 Jan 2012 21:17:58 -0600) it happened "vkj" wrote in :

Yes, simple! What impedance will you see primary? The secondary divided by n^2 (turns ratio square). But the secundary is now infinite. So you would get a large primary impedance only limited by saturation effects of the core material. There are no infinities in nature (my opinion), so that would be the limiting factor in the voltage you would see secundary.

Only as long as the primary current doesn't drive the core into saturation.

Sylvia.

the

integral)

then

ratio square).

effects of the core material.

limiting factor in the voltage

What?? Heres a little "thought experiment": SUppose you had a small solenoid (air core) of say 10 turns. Apply a 6V 50Hz voltage across it. Im sure you will agree that it would hardly present any impedance, and the current would be limited almost entirely by the resistance. So are you saying that if I now wind a secondary winding over the coil, I would magically get "infinite" impedence on the primary? Remember there is no iron here, and it would take a lot to saturate vacuum.

Answer: Think inductance. Transformer voltage equations dont directly apply in this case. The voltage across the coil is very small because of the small inducatance (ignroing resistance). If you have a secondary, the voltage across it would be gverned by the mutual inductance and the primary current. Vsec = M d(prim.ct.)/dt; M = mutual inductance.

vkj

--------------------------------------- Posted through

saturation.

Lets not confuse the focus of the issue and consider air cores only, thus eliminating saturation.

Lets say your current transformer presents an impedance of 1 milli-Ohm. Lets say your current is 20A. The prim. voltage is then 20mV. As someone has pointed out the turns ratio of a typical CT could be 1000:1, so the sec. open circuit volt is still just 20V. Hardly disastrous.

vkj

--------------------------------------- Posted through

Sorry, I dont think thats true at all. As I pointed out the fundamantal mechanisms are: Current causes flux. Flux is coupled to secondary. change of flux causes secondary voltage, Vsec = M d(iprim)/dt. In the same way, the change of flux in the primary itself causes a "back emf" and results in self inductance, V = L di/dt. The notion of a transformer follows from this. Assuming the primary inductance is "large", then you can equivalently talk of voltages. It is well known that ferro-magnetics have extremely non-linear permeabilities, and hence current )or voltage) transformers can be very non-linear.

vkj

winding

ammeter.

--------------------------------------- Posted through

On a sunny day (Tue, 24 Jan 2012 21:38:58 -0600) it happened "vkj" wrote in :

Current transformers HAVE a core. :-)

Yes, but transformers operate at flux levels where the flux generated by the primary current is very nearly cancelled out by the secondary current. That's what they mean when they say ampere-turns cancel out.

The back emf you mention does work to keep a potential transformer's open circuit magnetizing current low. But that is because potential transformers have many turns on their primary, each providing a few volts of back emf.

A current transformer has one primary turn. Its primary current is mostly dependant on the downstream load and as a result, the core flux can easily reach saturation because the back emf generated by d(flux)/dt through one turn is too small to buck the source voltage. Its self inductance is very low. On the secondary, with many turns, the product of d(flux)/dt and N turns can be very high. In some cases, higher than the insulation's rated max voltage.

works fine with linear loads which draw sinusoidal current, doesn't work for phase control, bridge rectifiers feeding capacitors, or other active loads.

on the

That is not all, maybe not even the most important. The flux in the core when operated properly in the CT mode is nearly zero, and if the = secondary is open that is NOT true, and the core could easily saturate. BTW the relevant IEEE standards make a distinction between metering and protection/relaying CT and their performance properties in overload.

?-)

thus

someone

Wrong and wrong. The primary impedance applies _ONLY_ when the secondary is terminated with a proper load (resistor). Otherwise it becomes an ordinary voltage transformer, and the primary voltage is no longer in the mV range, but in the volts range, with corresponding kV across the secondary.

Clear enough yet?

?-)

An unloaded air-core CT is called a Rogowski coil. The output voltage is proportional to the derivative of the current. If all you do is look at the terminal voltage, there is no effect on the wire at all, nothing equivalent to a primary impedance.

If I may borrow a quotation, "Wrong and wrong."

Not true. There is flux in the core no matter what. It is what gives rise to a secondary voltage across a resistor.

thus

milli-Ohm.=20

someone

the

secondary

the

Not a "normal" current transformer in case you hadn't noticed.

?-)

We were discussing air-core CTs.