Transformer Question

I have worked a [very] little with transformers and wish to gain a better understanding of them. Here is my question. Suppose I have a

1:n step-up transformer. Assume it is wound using a toroid and leakage inductance is negligible. Let's also neglect winding resistance. I am interested in understanding "magnetizing inductance". Let's say I have an ideal impedance meter and measure the primary impedance with the secondary open circuited. I will measure an inductive impedance with equivalent inductance value L1. Now, I assume that if I use my meter to measure the secondary impedance with the primary open-circuited, I should measure and inductive impedance with a corresponding inductance of L2=n*L1.

So, if I were to model this transformer and refer all of the inductance to the primary, I would end up with Lp equal to L1 in parallel with the secondary inductance referred to the primary. Right? This bothers me in a sense, because if the secondary is open-circuited, no current should flow in the secondary, and I should not see the secondary impedance reflected to the primary. However, if I were to model Lp as equal to L1, it seems that I am ignoring the inductance of the secondary.

In any case, if I refer the secondary inductance to the primary, I end up with a referred inductance of n*L1/(n^2)=L1/n. Thus, the transformer inductance referred to the primary would be Lp=L1||(L1/n), or Lp=(L1^2)/[(1+n)*L1]. For large n, this approaches Lp=L1/n [or Lp=L2/(n^2)].

I think this answer is right, and I am comfortable with it unless I think too hard about it. Can anyone help the thought process?

-JJ

Reply to
wizard12342002
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You've gone through a great deal of reasoning that is hard to follow. But, there is a basic flaw in the first equation. The inductance is proportional to the number of turns squared; therefore, Lsec = n^2*Lpri. If the secondary is open circuited, it is not "seen" by the primary and it's existance and number of turns is immaterial and visa versa. However, winding capacitance plus any other loading of the secondary will reflect into the primary by

1/(n^2).

Thefore a 20 ohm load on a 1:2 transformer will reflect into the primary as

1/2^2 which is 5 ohms. Think of it his way: assuming no loss, the power in must equal the power out. So if the transformer delivers 10 volts into 20 ohms, the power must be E^2/R = 10^2/20 = 5 watts. The input must be 5 volts for 10 volts out (1:2), but the power must be 5 watts (no loss). Therefore, the reflected resistance from R = E^2/P = 5^2/5 = 5 ohms.

In normal transformer operation, the magnetizing current which is the current into the primary impedance because of the Lpri is a small percentage of the total primary current. Usually it is only a couple of percent of the main current. In otherwords, the primary reactance is high at the operating frequency and the greatest part of the primary current is loading reflected from the secondary.

Reply to
Bob Eldred

** Ok up to here.

** Now you have re-defined which winding is the primary.

The winding that has the power supplied to it IS the primary.

** Insanity - there is only one primary at any given time.

.......... Phil

Reply to
Phil Allison

I read in sci.electronics.design that snipped-for-privacy@yahoo.com wrote (in ) about 'Transformer Question', on Wed, 5 Oct 2005:

No, n-squared. Inductance is proportional to the square of the number of turns.

No.

It should bother you. That means you are half way to seeing your mistake.

No. The current through Lp sets up the magnetic field. Looking back into the secondary winding you don't see a secondary inductance but Lp*n^2 in parallel with the source impedance of whatever is supplying the magnetizing current.

Don't do it; it's wrong.

>
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Regards, John Woodgate, OOO - Own Opinions Only.
If everything has been designed, a god designed evolution by natural selection.
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Reply to
John Woodgate

oops, inductance scales with n^2. So that buggers up all your calculations.

The terms "primary" and "secondary" are arbitrary, and mostly historic. In general power flows from the primary to the secondary. For a 1:1 transformer, the voltages and currents are identical, so it really doesnt matter much which you call primary, and which secondary.

Besides, in some circuits the transformer has two or more primaries....

thats because of the mistake in the first sentence. A simplified model without leakage is an ideal transformer in parallel with an inductor. The inductor is, by convention, placed on the primary side of the ideal transformer, so its value is that of the transformers "primary" inductance, measured across the primary terminals with the secondary open-circuited.

If OTOH you wanted to draw the inductor on the secondary side of the transformer, you would use the "secondary" inductance, measured across the secondary terminals with the primary open-circuited.

Which will be the primary inductance times the square of the secondary-to-primary turns ratio.

I like to think in terms of AL (cant do a subscript capital "L").

L = AL*N^2

AL incorporates all of the core dimensions, material & gap information, and is often given in nH/turn^2. but for this analysis, we dont even need to know its value, just that it is constant for a given core, material and gap (OK, permeability is all over the show with time, temperature and sock colour, but lets just pretend).

the primary inductance is thus:

Lprim = AL*Np^2

likewise the secondary inductance is

Lsec = AL*Ns^2

the primary and secondary are wound on the same core, so there AL values are the same, so either of these equations can be re-arranged to solve for AL:

AL = Lprim/Np^2

AL = Lsec/Ns^2

either of these can be substituted into the other inductance equation, giving:

Lprim = (Lsec/Ns^2)*Np^2 or

Lprim = Lsec*(Np/Ns)^2

and

Lsec = (Lprim/Np^2)*Ns^2 or

Lsec = Lprim*(Ns/Np)^2

turns ratio is defined (IEEE Std 100-1996) as:

*turns ratio (electric power transformer)* the number of turns of a given secondary divided by the number of primary turns. Thus a ratio less than one is a step-down transformation, a ratio greater than one is a step-up transformation, and a ratio equal to one is a unity ratio

which makes a lot of sense. especially the last bit, which is perhaps unnecessarily redundant. The standard then goes on to define:

*turns ratio of a current transformer* The ratio of the secondary winding turns to the primary winding turns

(these are commonly referred to in power systems as CTs)

and, even more helpfully:

*turns ratio of a voltage transformer* The ratio of the primary winding turns to the secondary winding turns

(these are commonly referred to in power systems as VTs, or PTs - potential transformers)

its interesting that the VT turns ratio definition is the *reciprocal* of the CT and power transformer definition.

In practice I can never remember, and always fall back on the AL equations, and use Np, Ns explicitly.

alas, this is all wrong. Still, well worth asking, as apart from the initial mistake using n instead of n^2, the maths itself is correct.

Cheers Terry

Reply to
Terry Given

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