Open circuit short circuit (test) TX flux effect

tim9000
Hi,
It's been a while but from what I remember when you short circuit a transformer, electrically it's like you've series'd up the coils and shorted the magnetisation branch. And when you open circuit a side of the TX it is like the current flows through a side then through the magnetisation branch. I also remember hearing that when a TX short circuits, that is reflected to the primary.
I'm just wondering, if you imagined you had a single phase TX and are only interested in looking at (thinking about) the flux, is the flux effected at all as one side is short circuited and open circuited? Or is it best to just think about it like concervation of power?

Thanks

Gold Member
Dearly Missed
I'm just wondering, if you imagined you had a single phase TX and are only interested in looking at (thinking about) the flux, is the flux effected at all as one side is short circuited and open circuited? Or is it best to just think about it like concervation of power?

Try visualizing a simple transformer like this one.

When secondary is open, it might as well not be there and you have an inductor, which you just studied in another thread. So build on that.

http://www.edisontechcenter.org/Transformers.html , an interesting page !

Now - when current flows in secondary, it'll make an opposing MMF that pushes back against primary MMF.
So you get opposing MMF's.
So , flux would begin to drop, reducing counter EMF in primary
so more primary current would flow
restoring flux to a value that'll balance applied voltage at new current.

Note that with secondary shorted the currents can be quite large, so some of applied voltage makes IR drop in primary instead of induction, so flux will be a little lower. Power will be dissipated in the windings in unusual amounts, so conservation will not be a straightforward approach. You have huge MMF's opposing one another so leakage flux will go up.

So the answer to your question is - i think you should think in terms of what are the internal processes. That'll make the equations intuitive.

Be aware that there exist transformers that operate at very low voltages, with current in the primary set by something external.
In those transformers flux is kept quite low by assuring that secondary is very nearly shorted. That's up to the user to guarantee.
Since primary current is set externally , only enough secondary current will flow to cancel primary amp-turns..
Primary and secondary amp-turns are almost exactly equal so very nearly cancel, keeping flux low..
They're called Current Transformers and you'll get to them later.

That's why it is important to be aware of what's going on in the iron core.
Train your mind to push flux around imaginary cores. I was baffled by current transformers for a long time, until i finally learned my magnetic units and started pushing flux.....

old jim

tim9000
That's an excellent explanation Jim, exactly what I was trying to remember. I couldn't remember if the shorted secondary would oppose or assist the MMF of the primary, good to be clear it's the former.

But what did you mean by:
"Note that with secondary shorted the currents can be quite large, so some of applied voltage makes IR drop in primary instead of induction, so flux will be a little lower"?

Thanks again

Staff Emeritus
Jim is referring to the transformer circuit model, where there's a resistance in series with the primary winding (to account for ohms in the windings). Greater primary current means you get more heating in that resistance: wasting electrical energy.

tim9000
Jim is referring to the transformer circuit model, where there's a resistance in series with the primary winding (to account for ohms in the windings). Greater primary current means you get more heating in that resistance: wasting electrical energy.
I understand that, but what does that have to do with the induction?
Thanks

Staff Emeritus
Power lost as heat in the primary misses out on being transferred by transformer action to the secondary circuit.

tim9000
Ooh so induction was meant in a context other than 'inductance (L)', more like to be induced?
i

Gold Member
Dearly Missed
more like to be induced?
precisely.
Vprimary = Vinduced + I X Rprimary , so it's possible for induced voltage to be quite low. That'll wreck an ordinary power transformer in short order because all the power is dissipated in the transformer windings, I^2R. A genuine short circuit load has zero ohms so can't dissipate any power, I^2 X zero is zero..

This is acceptable in a current transformer where some impedance quite external to the transformer limits primary current. Current in a shorted secondary will be in accord with transformer turns ratio .

Thanks, Nascent !

tim9000
So when shorted secondary: Vinduced in the Primary is from V = flux*N*w, which is really small because the flux is small and Iprimary goes up a lot; which is squared and multiplied by a constant resistance, to disipate power in the primary. Is this what you meant? What do you do to limit such current in a CT?
Thanks

Gold Member
Dearly Missed
So when shorted secondary: Vinduced in the Primary is from V = flux*N*w, which is really small because the flux is small
/
Yes. The flux is small because of the shorted secondary, where substantial amp-turns cancel out the amp-turns from substantial primary current. So the sum of the two opposing mmf's is quite small, even though each is substantial.

and Iprimary goes up a lot;
Sure, because nearly all the source voltage must now be dropped across the primary winding's resistance which is small.

which is squared and multiplied by a constant resistance, to disipate power in the primary.
"constant resistance" is the resistance of the primary and secondary windings. It'll make enough heat to wreck the transformer unless something limits the current..

Train your brain to push flux around . And to recognize current isn't only charge in motion it's also MMF.

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Gold Member
Dearly Missed
deleted. This was an attempt to get system to accept my complete post but it refuses to include first part. Maybe to do with from where i quoted?

Anyhow, i just appended first part in separate post.

Gold Member
Dearly Missed
For some reason the system refuses to allow me to include first part of above post, so i put it here.
No idea why. Have had lots of trouble lately.

[QUOTE="tim9000, post: 4950742, member: 480143] What do you do to limit such current in a CT?[/QUOTE]

A CT has its primary connected in series with a load whose current we wish to measure, like a motor or transformer or lamp or something.
So the amp-turns in primary are determined by that load and the number of turns on the CT.
Many CT's are built like a donut with the current to be measured flowing through the donut hole, in the wire going to the load.. That's a one turn primary.

You'd route a wire to load through the center. A stepped down current will be available in the secondary that can drive a small meter.

tim9000
...And to recognize current isn't only charge in motion it's also MMF.

Interesting way to put it, I'm going to write that down to drum it in to my head.

For some reason the system refuses to allow me to include first part of above post, so i put it here.
No idea why. Have had lots of trouble lately.

[QUOTE="tim9000, post: 4950742, member: 480143] What do you do to limit such current in a CT?

A CT has its primary connected in series with a load whose current we wish to measure, like a motor or transformer or lamp or something.
So the amp-turns in primary are determined by that load and the number of turns on the CT.
Many CT's are built like a donut with the current to be measured flowing through the donut hole, in the wire going to the load.. That's a one turn primary.

You'd route a wire to load through the center. A stepped down current will be available in the secondary that can drive a small meter.[/QUOTE]

But what happens if you shorted the secondary on a CT? I imagine it would still be quite small, but the reason I ask is I'm wondering about how to calculate the current from the turns ratio of the CT. Would the voltage of the single turn primary be the difference between source to load? (Which is arround the CT as the wire passes through it from source to load) Just seems quite different to how I would think of a CT voltage-turns ratio.

Another top post (Old) Jim! ;)

Gold Member
Dearly Missed
CT's are operated differently than voltage transformers. They're still just transformers.
The difference is only in how we use them.

Remember from several poosts ago
amp-turns in secondary cancel amp-turns from primary

so those few primary amp-turns will cause modest mmf,
which will cause modest flux
which will induce modest voltage in secondary ,causing just enough voltage to push current in the amount required to cancel nearly all those primary amp-turns.
If secondary is shorted that takes but very very little voltage.
You're left with a transformer that's not working hard at all. Flux is low, current isn't out of control.

Now the counter emf induced in primary will be small because flux is small.

Current transformers are designed to operate with low flux and small voltages.
Therefore they must an have excellent core that'll pass flux with small mmf, remember mmf seen by the core is sum of amp-turns from both primary and secondary. The idea is to have those as nearly equal as possible.

What is dangerous to a CT is an OPEN secondary and here's why:
With secondary open you have no secondary amp-turns to keep flux low.
So, if there's much primary current at all, (remember something else controls that not the CT,)
the flux goes way up 'cause secondary doesn't cancel primary mmf..
IF the core saturates, which it can because it's so good, , flux will swing between positive and negative saturation.
That'll be sudden snaps not the smooth cosine wave flux we like to see..

Think about Voltage induced in the open secondary , where there's lots more turns:
during those 'snaps' it can become great enough to puncture the insulation.
That's why you NEVER allow a CT's secondary to become open while there's primary current.
It can hurt you and wreck equipment including itself.

In days of vacuum tube hi fi amplifiers the same applied to the speaker outputs, for tube amps have transformer output (see RCA Tube Manual)
so you always made sure your speaker is working before turning up the volume. Better sets had a 100 ohm resistor permanently soldered in across the amp's speaker terminals to protect the transformer from "OOOPS's".

So the current transformer seems wierd but all it is is an excellent transformer with a high permeability core.
The difference lies in how we operate it.

I'll be away a few days. I hope some other old timers will help you out, because you are on a roll here !

It is rewarding to see your progress.

Have fun ! Read up on "Fluxgate Magnetometer" they're a lot of fun.... heart of the electronic compass and treasure hunter's shipwreck finder. Janicke's "Magnetic Measurements Handbook" has plans.

old jim
.

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zoki85
You'd route a wire to load through the center. A stepped down current will be available in the secondary that can drive a small meter.

But what happens if you shorted the secondary on a CT? I imagine it would still be quite small, but the reason I ask is I'm wondering about how to calculate the current from the turns ratio of the CT.
Well, as you indicate, CTs are used in the short circuit regime (or close to it), . For instance, you can put ammeter across the secondary and measure current. Transformation ratio of CT is used for calculation with high preceison: I1 : I2 = N2 : N1 . In design with one wire through the donut hole N1=1. Jim makes a good point about danger of leaving secondary open.

tim9000
Well, as you indicate, CTs are used in the short circuit regime (or close to it), . For instance, you can put ammeter across the secondary and measure current. Transformation ratio of CT is used for calculation with high preceison: I1 : I2 = N2 : N1 . In design with one wire through the donut hole N1=1. Jim makes a good point about danger of leaving secondary open.

Ah, of course.

Think about Voltage induced in the open secondary , where there's lots more turns:
during those 'snaps' it can become great enough to puncture the insulation.
That's why you NEVER allow a CT's secondary to become open while there's primary current.
It can hurt you and wreck equipment including itself.

So the fact that the core has such a high permiability, this is saturating the core causing a large change in flux and is causing a secondary Voltage similar to a large di/dt from open circuiting an inductor? Just thinking about this and the turns ratio, so would the OC Vsecondary = N*secondary*d(fi)/dt + (Rsecondary*Isecondary = 0) =
Nsecondary*d(fi)/dt ? Which is really big

I'll be away a few days. I hope some other old timers will help you out, because you are on a roll here !

It is rewarding to see your progress.

Have a Merry Christmas