# Rough lab measurement of transformer inrush current?

## Main Question or Discussion Point

Hi, Id like to measure inrush current for various salvaged transformers (unknown windings, but probably 120/24) to the nearest 10 mA. I only have a simple DMM and soundcard scope.

It seems that a clamp meter has this capability; if a clamp meter is the best approach, can I construct my own clamp meter attachment to work with my normal DMM somehow? (Im guessing not since the transient will be too fast to notice)

Another idea, I could use fuses at various values, and find the one that breaks. Is there such thing as a low current variable re-settable fuse?

Are there practical equations for transformer inrush current that could ballpark my measurements?

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tech99
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Hi, Id like to measure inrush current for various salvaged transformers (unknown windings, but probably 120/24) to the nearest 10 mA. I only have a simple DMM and soundcard scope.

It seems that a clamp meter has this capability; if a clamp meter is the best approach, can I construct my own clamp meter attachment to work with my normal DMM somehow? (Im guessing not since the transient will be too fast to notice)

Another idea, I could use fuses at various values, and find the one that breaks. Is there such thing as a low current variable re-settable fuse?

Are there practical equations for transformer inrush current that could ballpark my measurements?
The DMM is a sampling device so I don't think it could be trusted for a pulse measurement. A fuse will rupture depending on both current and duration, so is unreliable. I think the profession method would be a current transformer with a storage oscilloscope. The clamp is a current transformer. Not sure if your sound card 'scope is fast enough but it might be OK.

meBigGuy
Gold Member
Only the soundcard scope has a possibility of measuring the current pulse. You can measure the voltage across a small current sense resistor between the transformer and ground. Expect high currents if you apply DC steps to the transformer.

jim hardy
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Transformer inrush is very interesting, you should google it.

How much you get depends on where in the line cycle you close the switch.

About one time in ten you'll close at the zero crossing and get a huge inrush, 10 to 100X normal excitation current, like this:

great pictures of lamp inrush too, at
http://www.beta-a2.com/EE-photos.html

Observe he got 40 amps inrush into a ~1.5 amp transformer., 27 X rated current..
Also observe current peaked well after sinewave voltage peak.

What is going on is this:
SInce e=ndΦ/dt
Φ=(1/n) ∫edt

Aha ! voltage is positive for the whole first half-cycle
so flux must have positive slope for that half cycle, that's what integration does
but flux started from zero not negative peak like it does in steady state operation
and the core saturated .
That repeats for subsequent several cycles, and you'll find it well described in numerous places.
Observe once you reach steady state, at zero volts flux is at negative peak.
but if both start from zero you'll push flux up past the B-H knee so current goes sky high.
http://www.vias.org/eltransformers/lee_electronic_transformers_02_07.html

So don't be surprised, be prepared.

That's why Solid State Relays (SSR's) come in zero switching and peak switching flavors, the former for resistive and the latter for inductive loads.
You might want to try one of each, it'd give you more consistent inrush than random switchclosing.

If you apply only half voltage to your transformer it'll not saturate from that cause.

Have fun

old jim

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jim hardy
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Is there such thing as a low current variable re-settable fuse?

i never saw an adjustable one
but an assortment wouldn't be prohibitively expensive.
http://www.littelfuse.com/ptc

berkeman
Mentor
How can a transformer alone have inrush current?

jim hardy
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How can a transformer alone have inrush current?

most of the time it won't have enough to notice.
Think for a moment about the sinewave voltage - flux relationship.

At normal operation they're out of phase by ninety degrees so when one is at zero the other is at peak.

When you unplug transformer from the wall voltage of course goes to zero , and so does flux because good transformer iron doesn't have much remanence..

When you plug it back in, what's the voltage ?
Pure chance, where's the sinewave when contact is made ?
Fortunately a sine wave spends most of its time not so far from peak and very little around zero.
So chances are voltage won't be at zero.
But if it is,
you are starting the transformer from worst possible initial condition - voltage and flux in phase not 90 deg out..

On the rare occasion voltage is zero, you have an entire half cycle of positive(or negative) voltage . Flux must increase for that entire time to counter the positive applied voltage.
Volt-seconds would be a measure of flux because Φ = n-1∫vdt
and you'll find that parameter in transformer literature.

Somewhere in that first half cycle flux passes the knee of BH curve as explained in post 5.
The core saturate and it takes ridiculous primary current to make counter EMF.
That's shown in the oscillograph - notice current started rising after the peak , when voltage was decreasing.

here's a credible reference
http://www.eeweb.com/design-articles/view/beware-of-zero-crossover-switching-of-transformers

Transformers could be built with enough iron to not saturate on a worst case closure, but we always go for low bidder.

This causes trouble when powering from inverters , . the occasional big inrush gives nuisance trips or drops bus voltage upsetting instruments connected to it..

The easy fix is to buy a bigger transformer and operate it at half voltage.
Or SSR's that close on sinewave peak .
http://www.crydom.com/en/products/catalog/p_s24.pdf

meBigGuy
Gold Member
If someone had asked me this morning how to avoid transformer inrush currents, I would have told him to turn on at the zero crossing.

jim hardy
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If someone had asked me this morning how to avoid transformer inrush currents, I would have told him to turn on at the zero crossing.
Yes it's well known you can eliminate capacitive inrush, as drawn by computers, by feeding them through a ferroresonant (aka Sola) transformer.
But you have to use the Sola at less than its nameplate input voltage or it'll saturate every now and then from this effect.
I had good luck using them at 70%.

thanks for the kind words,

old jim

The Electrician
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Transformer inrush is very interesting, you should google it.

How much you get depends on where in the line cycle you close the switch.

About one time in ten you'll close at the zero crossing and get a huge inrush, 10 to 100X normal excitation current

old jim
There's a scope capture at higher sweep speed showing the effect, plus explanation here:

jim hardy
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There's a scope capture at higher sweep speed showing the effect, plus explanation here:

I wonder who that fellow might be ?

This may be an insignificant tangent, but suppose instead its a DC source on the primary. What if I used a a normal AA battery instead of a sine wave generator? Wouldn't there still be an inrush just from the transient; when contact on the primary means going from 0 to 1.5V in a split second?

Could this simplify the analysis with minimal parts by eliminating crossover randomness (which is indeed a complex design consideration)? How could I make a theoretical prediction of the max inrush current in that case (with an unknown transformer, is it even possible)?

The Electrician
Gold Member
Hi, Id like to measure inrush current for various salvaged transformers (unknown windings, but probably 120/24) to the nearest 10 mA. I only have a simple DMM and soundcard scope.
Why do you need to measure the inrush current to the nearest 10 mA? The peak current will be in many amps, and it will be quite variable. If you have read the thread so far you see that the peak current is highly dependent on just when the transformer is connected to the grid voltage.

I hope you know better than to use a sound card scope to make measurements on the grid connection to a transformer.

There is no formula, but you can get a good idea of what the maximum inrush current for a particular transformer would be like this:

Since the inrush current is caused by deep saturation of the core, to a good approximation the only thing limiting the current is the DC resistance of the winding. The peak of the inrush current would then seem to be equal to the peak grid voltage divided by the winding resistance, but this isn't quite right; read on.

I connected a 50 VA transformer (winding resistance of 10 ohms) to the grid through a switch so I could repeatedly turn the excitation on and off until I managed to get a good event, where the connection was made just on the positive going zero crossing of the grid voltage. I also had a 50 amp shunt in series with the winding, and I captured the event on an oscilloscope. This particular scope has isolated inputs so it can be safely used to measure grid voltages and currents. The current is the purple trace at 1 mV per amp; the grid voltage is the green trace. The purple trace is noisy because with a 50 amp shunt, I had to turn the gain of that channel to maximum:

There are two green cursors shown. The left (dotted) one shows the peak grid voltage. Look over on the right side of the image for the "Cursor 1" label: the voltage is 170 V, the peak grid voltage. The right cursor, "Cursor 2", shows a voltage of 120 V when the current reaches its peak value of about 12 amps (purple trace).

The peak current occurs some time after the peak grid voltage. If the peak inrush current had occurred at the time of the peak grid voltage, then the value of that peak current would have been "peak grid voltage/winding resistance = 170/10 = 17 amps".

But since the peak of the inrush current is delayed until sufficient volt-seconds have accrued, which occurs when the grid voltage has passed its positive peak and is decreasing, we might say peak inrush current equals "a certain delayed grid voltage/winding resistance = 120/10 = 12 amps", which is what we see.

The value of the grid voltage when the maximum peak inrush current occurs is coincidentally about equal to the RMS value of the grid voltage. Perhaps it wouldn't be too unreasonable to say that the maximum peak inrush current of a transformer will be about equal to the RMS grid voltage divided by the DC resistance of the transformer winding (the primary usually).

Most of the time when the transformer is connected to the grid, the inrush current won't be this large. But occasionally such a large peak inrush current will occur.

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The Electrician
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This may be an insignificant tangent, but suppose instead its a DC source on the primary. What if I used a a normal AA battery instead of a sine wave generator? Wouldn't there still be an inrush just from the transient; when contact on the primary means going from 0 to 1.5V in a split second?
If you connect a DC source to a transformer, there won't be a transient inrush current that dies out after a while. The current will increase rapidly to a value determined by the applied DC voltage and the winding resistance. Usually this will be equivalent to a very low resistance, almost a short, across the battery.

Could this simplify the analysis with minimal parts by eliminating crossover randomness (which is indeed a complex design consideration)? How could I make a theoretical prediction of the max inrush current in that case (with an unknown transformer, is it even possible)?
In the preceeding post I gave you an approximate way to calculate the maximum peak inrush current when connecting a transformer winding to the grid.

jim hardy
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Wouldn't there still be an inrush just from the transient; when contact on the primary means going from 0 to 1.5V in a split second?
That's wrong as can be.

Back to basic inductance:.
Does not inductance oppose change in current ? Of course it does.
What is ohm's law for an inductor? E = LdI/dt.
When you apply step function voltage to an inductor you get not a step function current like for resistance, or impulse function current for capacitance,
you get ramp function current, constant derivative.
Look at those three functions here.
http://lpsa.swarthmore.edu/BackGround/ImpulseFunc/ImpFunc.html

The slope of the current ramp is of course E/L .

In an ideal inductor, sans resistance , the current would increase forever.
There'd be no inrush.

Flux too would increase forever, because dΦ/dt = E/L and all three terms are constant

In a real inductor two things change that picture.
1. Current cannot increase forever. Resistance of the wire limits current to E/R. Current approaches that value with time constant L/R.
2.. Flux cannot increase forever. Iron will only pass a little more than 1.4 Tesla. So, when flux reaches that value , core is said to be "saturated", counter-emf disappears and current goes to E/R. That is what The Electrician is describing.

Sometimes it helps to think things out in DC and steps and ramps instead of sinewaves, because a sinewave and its derivative look so similar. But steps and ramps have derivatives that look quite different.

Now - if you apply a DC step voltage to a real transformer it'll allow a constant rate of current increase until it saturates.
How long does that take?
That depends on the size of the core and the number of turns. Knowing those , it'd be easy to figure out beforehand for a step function voltage because voltage is constant: how long does it take for flux to reach ~1.4 Tesla ? One could also apply some volts and measure elapsed seconds.
Electrician's measurements showed his core saturated after about 5.7 milliseconds with a 120Vrms sinewave applied..
What's integral from 0 to 0.0057 of 177sin(377t) ? That's the volt-second parameter λ for his transformer.
see page 30 here
http://www.ieee.li/pdf/introduction_to_power_electronics/chapter_12.pdf

You will see this effect in power transformers when you apply an analog ohm-meter. There's a delay between making the connection and start of needle movement.

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Thanks, you are right of course. I wish I had an oscilloscope capable of the interesting experiments above. With my soundcard scope, I suppose an extremely low voltage sine wave generator would not work since the core would never saturate.

And yes, there's no reason I need such a small resolution (I honestly didn't realize it was so chaotic). I think my cheapest option might be the fuses.