Finding out transformer's winding's reactance ...

Hello everyone,

I have a transformer that I want to find it's winding's reactance, how could I do so? The reason that I want to find the reactance is that I want to calculate the current in the primary winding of the transformer. I will use the following eqn. :

I(phasor) = V(phasor) / [ R(winding) + j w L(winding) ]

V(phasor) is 110 Vac (from mains) (considered to have zero phase angle, right?)
R(winding) could be measured using an Ohm-meter, right?
w = 2 * pi * (60Hz)
Now, for the L(winding) .. how could I find it ? Or am I doing the whole current calculation thing in a wrong way??

Thank you all in advance,

kypo said:

Hello everyone,

I have a transformer that I want to find it's winding's reactance, how could I do so? The reason that I want to find the reactance is that I want to calculate the current in the primary winding of the transformer. I will use the following eqn. :

I(phasor) = V(phasor) / [ R(winding) + j w L(winding) ]

V(phasor) is 110 Vac (from mains) (considered to have zero phase angle, right?)
R(winding) could be measured using an ohm-meter, right?
w = 2 * pi * (60Hz)
Now, for the L(winding) .. how could I find it ? Or am I doing the whole current calculation thing in a wrong way??

Thank you all in advance,

Click to expand...

If you can measure |I|, |V|, and R, you can calculate L. You don't need to be able to measure the phase between V and I.

|Z|=|V|/|I|
and
Z=R+jwL, |Z|=√(R^2+(wL)^2)

Solving for L,
L=√(|Z|^2-R^2)/w, where w=2*pi*f

Ron H said:

If you can measure |I|, |V|, and R, you can calculate L.

Click to expand...

Well, that would be true if I know |I| as you've said .. However, it is |I| that I want to calculate, so it's an unknown. I know |V| and R. But I have 2 unknowns: L and |I|.

Let me put it in a very simple way: How would you know the current through the primary winding ?

I'm sorry. I misread your original post.
Do you have access to any test equipment such a signal generator, oscilloscope, power supply, etc.?
Of course, an inductance meter would solve your problem.

BTW, I don't think you can do it if you have a load on the secondary. I'm assuming you don't.

kypo said:

Let me put it in a very simple way: How would you know the current through the primary winding ?

Click to expand...

Why not just measure it? - then you know it will be correct, instead of trying to calculate it - which will give a theoretical value which is bound to be wrong!.

You've also never mentioned what load is on the transformer, which is obviously vital - even if it's got no load!.

Or is this just a class or homework question?.

Ron H said:

Of course, an inductance meter would solve your problem.

Click to expand...

This is the first time I know that there is a thing called an "INDUCTANCE" meter !! :roll: So, is it cheap? Or it would cost me a fortune? Truly, I've never heard about it before! Maybe I could hack a DMM and turn it into an inductance meter?? Any ideas about that would be so helpful.

Ron H said:

BTW, I don't think you can do it if you have a load on the secondary. I'm assuming you don't.

Click to expand...

Could you clarify that please? At the moment, no I don't have a secondary load. However, I'll add one later. What I am building here is a power supply, and I am trying to figure out the maximum current that the primary winding could handle. That's why I am calculating the current and inductance.

Nigel Goodwin said:

You've also never mentioned what load is on the transformer, which is obviously vital - even if it's got no load!.

Click to expand...

I am afraid I didn't get that quite well Nigel! How it well affect the primary even if it's got no load? Later (but not at the moment), my load will be a rectifier bridge, some filtering capactiors and a voltage regulator (LM150 with 1.2 < Vout < 35 and max Iout 4.5A).

Well, I might have a good way to measure the inductance. If I place a high-power resistance Rt (to limit the current) in series with the resistance and inductance of the winding, then measure the current, I would be able to easily solve for L using the eqn. :

|I| = |V| / |Z| ---->>> |I| = |V| / [ [Rwinding + Rt] + j w Lwinding]

Now, everything is known (|I|, |V|, Rwinding and Rt) except the imaginary part of the impedance, that is, the inductance that I want to find. After that, I'll remove Rt. Then, the only impedance that will exist is that of the transformer's winding. Rwinding is measured using an ohmmeter. Lwinding
have just been calculated. So ...

Imax = V(mains) / [ Rwinding + j w Lwinding]

Any comments or suggestions about this way?? Oh another thing .. How could the secondary load affect the previous eqn ? Also if there was no load .. how the eqn is affected ?

Nigel Goodwin said:

Or is this just a class or homework question?.

Click to expand...

No, it's not a homework question. I am just a hobbyist, I am not an electronics engineer or something :lol:.

thnx.

Ahhh, now that you tell us what you are doing, I see that we don't really need to work it out too accurately, just a ballpark figure. (instead of it being a purely theoretical question).

The transformer primary will be passing maximum current at full load. You've given us the figure of 4.5A (secondary). I'm guessing this a 30V secondary winding (or close to it?). So this equates to (30 * 4.5) VA = 135 VA. Divide this VA rating by your mains voltage (110V), and we get a primary current of 1.27A.

The figure of 1.27 is, of course at a power factor of 1.0, which it will not be in reality, the power factor will be closer to 0.8 So we can say the primary current is (1.27 / 0.8) A = 1.53 A.

Allow a 50% safety margin, so you should rate the primary of your transformer at about (1.53 * 150%) = 2.3A

Phasor said:

...Ahhh, now that you tell us what you are doing...

Click to expand...

You see, I always prefere knowing the general method rather than just knowing the solution of a specific circuit in hand. Therefore, I am a big fan of equations :lol: and general solutions that work on all cases. However, I admit that it's a lot easier to solve a specific problem at hand than finding the general equation or rule that govern it ... anyways ..

Phasor said:

The transformer primary will be passing maximum current at full load

Click to expand...

I didn't get you here. What do you mean by full load? What I know from Ohm's law is that as the load decreases, the current will increase. So, you'll reach a maximum current at the minimum load. So, what did you mean by saying "primary will be passing max current at full load"?

Phasor said:

You've given us the figure of 4.5A (secondary)

Click to expand...

No, the 4.5A is the maximum curret the the LM150 would output. The secondary current would be larger than 4.5A. I've attached the circuit that I am working on. Have a look at it ...

Phasor said:

... So this equates to (30 * 4.5) VA = 135 VA. Divide this VA rating by your mains voltage (110V), and we get a primary current of 1.27A.

Click to expand...

What you did here is equating the complex power in the primary to the complex power in the secondary, right?? What about the dissipated heat? There is A LOT of heat dissipated by a transformer. So we just simply ignore it ?!

And about the power factor that you have mentioned, is it between primary voltage and primary current, or secondary voltage and secondar y current ? Also, why the power factor is 1 ? it means that the current and voltage are in phase ! but this can't be true because we have inductors in the circuit, so the current must be lagging !! .. help me out Phasor .. I am getting confused :cry:

Kypo, Ohm's law is I=V/R. As R goes down, I goes up. A low R is generally called a high load (because the current is high) unless you specify high load resistance. Also, in a transformer, primary and secondary currents are proportional (ignoring magnetizing current, which you are trying to measure, and I believe are unnecessarily worried about). As secondary current goes up, primary current goes up, and vice-versa. With no load (secondary open), you will have negligible primary current.
You had me confused when you suggested adding a series resistor so you could measure the primary current, because you told me you could not do that. I thought you meant you didn't have an ammeter. You apparently meant that you thought the current would damage the transformer.

Ron H said:

ohm's law is I=V/R. As R goes down, I goes up

Click to expand...

Well, that's what I said :lol: !

Ron H said:

Also, in a transformer, primary and secondary currents are proportional

Click to expand...

What is the formula for that ?

All I know is just a primary current and a secondary current. What about the magnetizing current And how did you conclude that I am trying to measure the magnetizing current? I even don't know what it is ! :wink:

Ron H said:

With no load (secondary open), you will have negligible primary current.

Click to expand...

This is what I can't truly grasp! Howcome? I mean if you have the secondary open, then you'll just have a normal L circuit, right?? Then what would make the primary current negligible? Have a look at the attachment and tell me if am wrong.

Ron H said:

You apparently meant that you thought the current would damage the transformer.

Click to expand...

Yeah, exactly. But I was afraid that the ammeter would be damaged, not the transformer. Why should I protect the transformer anyway? I mean transformers are usually just plugged into the mains without any resistors to protect them. So, why should I protect the transformer from the mains voltage? Some circuits place a fuse just to protect the load at the secondary from drawing excess current, but not to protect the transformer. Am I right in all this?? Or am I doing some huuuge mistakes around here :lol: !!?

thank you so much Ron ..

kypo said:

This is what I can't truly grasp! Howcome? I mean if you have the secondary open, then you'll just have a normal L circuit, right?? Then what would make the primary current negligible? Have a look at the attachment and tell me if am wrong.

Click to expand...

This was why I asked earlier about the intended load, for most purposes you can consider a mains transformer a 'nearly perfect' machine, they actually have a very high efficiency.

Basically the power in the primary is the same as the power in the secondary (obviously there will be a slight loss, but it is quite slight). So if your secondary output is 11V at 10A, your primary input will be 110V at 1A - both 110W of power. If your secondary current reduces to 5A, the primary current will reduce to 0.5A - and if the secondary current reduces to zero (no load) the primary current will reduce to a fairly low value, near enough zero for most purposes.

So you seem to be chasing complicated calculations for no reason, as your original premise is incorrect.

You would normally fit a fuse in the mains to the transformer for safety reasons, or at least use a transformer with a built-in thermal fuse, if the transformer goes shorted turns (or an external short is on the secondary) the transformer will cook itself, so it's important for safety.

Apparently you haven't considered that the primary inductance of a 60Hz power transformer is typically on the order of several Henries.
Magnetizing current is another name for what you are trying to measure. Do a Google search on "transformer magnetizing current".

kypo said:

And about the power factor that you have mentioned, is it between primary voltage and primary current, or secondary voltage and secondar y current ?

Click to expand...

The power factor is relating to the primary side. At no-load, the primary will display a low power factor (maybe 0.4-0.5), and this increases, approaching 1, as you increase the secondary load.

The secondary power factor depends entirely upon the load - if you have a purely resistive load on the secondary, the secondary power factor will be 1.0