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Mains transformers

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Sceadwian

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I know this is a basic question but it's always bugged me. On a generic mains transformer what limits the current that's always passing through the primary? Is it just the bulk inductance of the primary winding? If this is the case isn't determining the inductance of the primary on a transformer simply a matter of determining the bulk resistance of the primary and subtracting that from the quiescent current?
 

duffy

Well-Known Member
Subtract resistance from current? No, the thing that limits the current is the inductive reactance of the primary. Unfortunately, the inductace changes with the loading on the secondary. This works like the shades on a shaded-pole motor. It has the effect that it makes the core look smaller ("shades" it) when it is heavily loaded.

At this point you are dealing with an inductance that has a smaller apparent core, and so a lower inductance, lower inductive reactance, and higher current.
 
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Sceadwian

Banned
I just want to find out what the equivalent inductance of the primary is when the secondary is unloaded, I know this changes under loading of the secondary. If the secondary is completely unloaded and I read the amount of current going through the primary when it's connected (there's always some) then I simply find out what the inductor value needs to be at 60hz to allow that much current to flow. When I meant subtract the resistance I meant subtract the effect the coil's resistance has on the current to get closer to the actual inductance.
 

duffy

Well-Known Member
Oh, yeah - that will work.
 

duffy

Well-Known Member
Divide the voltage by the current, divide by 2, divide by π, divide by 60hz.
 

duffy

Well-Known Member
Technically you should do that, subtract out the real component from the complex impedance -

- but the reactance is going to have such a lion's share of the impedance it will be like the difference between XL=1130Ω and Z=1147Ω for a 200Ω winding resistance.
 

fernando_g

New Member
The current that one always sees in a transformer's primary winding is called the MAGNETIZING CURRENT, and is the current required to complete the hysteresis loop in the core.

This current waveform is NOT sinusoidal, and varies with the slope of the curve, which will rapidly increase with the input voltage as the core reaches its saturation flux density.

See the following link:
Electronic Transformers - Magnetizing Current
 

i_build_stuff

New Member
Transformers work on the principle of mutual inductance.

If you just plug the transformer in, with no load at the secondary, then it will act like a large inductor. Because the same flux goes through both windings, though, any current that you allow to flow in the secondary will cancel the magnetic field created by the primary. Any such "equal but opposite" component of current in a transformer is invisible to the primary / secondary inductances, so the energy just goes right through (this is why a circuit being driven by a transformer doesn't look like it has a huge inductor in series with it).

As for measuring inductance, you could always apply a higher frequency signal than normal (like 1kHz or something) so that the resistance is not significant to get a rough idea.

A better way to get a set of figures for a transformer is to measure both the voltage and current, as well as the phase angle between them, at the primary and secondary with the secondary open and shorted, as well as taking DC resistance measurements at both sides. This can be a little more complicated, but it lets you fill in a more complex model for your transformer that includes an approximation for core loss due to hysteresis and eddy currents.
 

Sceadwian

Banned
Increasing the frequency will change the measured inductance though. I have no problem measuring the current so it should be easy to do right from the wall. My meter is RMS so the current being non-sinusoidal shouldn't matter. I just want an idea of the primary inductance out of curiosity. Got what I need thanks for the posts.
 
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Diver300

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Very small transformers are really lousy and primary resistance can be a lot.

For any transformer over about 50 VA the efficiency will be 80% or better at full load. That means that at full load the losses are less than 10 W at full load for a 50 VA transformer.

The primary resistive loss is I^2 * R so on 120 V ac input, the primary current is about 0.42 amps, so R can't be more than 57 Ω, and that would be with all the losses in the primary. It is more likely to be less than about 30 Ω.

The primary current with no load will be about 0.1 A, so the impedance is around 1200 j Ω. The "j" indicates reactive impedance, effectively at right angles to the resistive impedance. The 30 Ω would only increase it to sqrt(1200^2 + 30^2) = 1200.37 Ω

What that means is that the primary resistance makes no practical difference to the primary current at no load. Any effect is lost in measurement errors.
 

The Electrician

Active Member
Very small transformers are really lousy and primary resistance can be a lot.

For any transformer over about 50 VA the efficiency will be 80% or better at full load. That means that at full load the losses are less than 10 W at full load for a 50 VA transformer.

The primary resistive loss is I^2 * R so on 120 V ac input, the primary current is about 0.42 amps, so R can't be more than 57 Ω, and that would be with all the losses in the primary. It is more likely to be less than about 30 Ω.
You have done this calculation based on the assumption that the transformer is under full load, and you haven't accounted for the fact that a transformer that can supply 50 VA to a load will absorb more than 50 VA at the primary under load. The primary current under full load will be more like .5 amps.

The primary current with no load will be about 0.1 A, so the impedance is around 1200 j Ω. The "j" indicates reactive impedance, effectively at right angles to the resistive impedance. The 30 Ω would only increase it to sqrt(1200^2 + 30^2) = 1200.37 Ω
But now you are considering the unloaded transformer, so the 30 Ω is no longer the relevant resistance.

What that means is that the primary resistance makes no practical difference to the primary current at no load. Any effect is lost in measurement errors.
Here are some measurements on a 50 VA transformer (25.2 VAC output at 2 amps).

Primary DCR = 9.5 Ω
Secondary DCR = .74 Ω
Measured turns ratio = 4.08
Rated secondary current = 2 amps
Rated primary current = .49 amps

With 120 VAC applied to the primary and the secondary open:

Primary current = .109 amps
Watts loss = 3.6 watts (measured with precision wattmeter)

Since the current is so low (compared to full load primary current), this loss figure is essentially all core loss, and maybe .1 watt copper loss.

We can calculate the unloaded primary impedance as 120/.109 = 1101 Ω

Let's resolve this into reactance and resistance components. First calculate the resistive component. Knowing that the loss is 3.6 watts, we have I^2*Reff = 3.6, so Reff = 303 Ω. Then using the relationship Z^2 = X^2 + R^2, we have X (the reactance) = SQRT(1101^2 - 303^2) = 1058.4 Ω.

From this we can calculate the effective unloaded primary inductance. Leff = 1058.4/(2*PI*60) = 2.81 henries.

To measure the copper loss, we short the secondary and with a variac apply gradually increasing voltage to the primary until the primary current is equal to the full rated primary current (.49 amps). In the case of the transfomer under consideration, this happens with about 10 VAC applied. With such a low applied primary voltage, the core loss is entirely negligible, and the measured loss is the copper loss under full load. We measure the loss under this condition with a precision wattmeter; the value is 5.5 watts. Strictly, the transformer should have been allowed to reach normal full load operating temperature to account for the increase in the copper resistance with elevated temperature, but I don't feel like waiting that long.

So, we have the core loss as 3.6 watts, and the copper loss as 5.5 watts, for a total transformer loss under load of 9.1 watts.

This means that the transformer loss resistance under full rated load is 9.1/(.49^2) = 37.9 Ω, compared to 303 Ω with the secondary unloaded.

In either case, the loss resistance is a small part of the overall primary impedance. Under full (resistive) load, though, the primary impedance is almost purely resistive (about 240 Ω) compared to the almost purely reactive primary impedance when the secondary is unloaded.
 

Diver300

Well-Known Member
Most Helpful Member
My calculations were just estimates. The 30 Ω was just the upper limit of what the primary resistance could be, and that won't change with load.

From your measurements, I was being pessimistic about the resistance. I had also ignored the resistive component that comes from the iron loss.

The point, as your calculations and measurements show, is that the transformer is almost entirely inductive at no load. The inductance can be worked out within 5% by measuring the unloaded current, voltage and frequency.

For a more accurate model of the core losses, it might be better to consider the iron losses as a parallel resistance. In Electrician's example, a parallel resistance of 120^2/3.6 = 4000 ohms. That will be virtually constant as the load changes. As the parallel resistance is a lot more than the overall impedance of the unloaded transformer, the primary current is still mainly controlled by the primary inductance.
 
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