Transformer Question

Thomas_S_Hunterson

Just a quick question about AC transformers.

Will a transformer always carry maximum load on its primary regardless of the load placed on it's secondary?

For example, an adaptor for a mobile phone, will it always drain the same amount of power from the mains, regardless of whether or not it is plugged into the phone


Thanks in advance for the replies

Michael Collins

Nope, the load on the primary will depend on the load of the secondary. Specifically if the load on the secondary is Zsec, then the load on the primary will be:

Zprim = (n1/n2)^2 . Zsec

Where n1 is the number of turns on the primary, and n2 is the number of turns on the secondary. So for your average mobile phone adapter with an output of 6 V say, n1/n2 = 240 V / 6 V = 40, hence the load on the primary will be 40^2 times the load on the secondary i.e. 1600 times the load on the secondary...

Of course this is for an ideal transformer. In actuality you'll have some winding loss on both sides, series inductances, eddy & hystersis losses etc. but hopefully it answers your question.

dave2pvd

Michael Collins wrote:

In actuality you'll have some winding loss on both sides, series inductances, eddy & hystersis losses etc.

So be green and unplug unused transformers/powersupplies

Thomas_S_Hunterson

thanks guys, although i'm not too sure of the logic of it.

My own reasoning was that if the resistance of the primary was a certain figure, then regardless of the number of turns, ohms law would give a constant current through the coil, provided the voltage was constant, thus there would be a constant load.

dave2pvd

sean_keevey wrote:

thanks guys, although i'm not too sure of the logic of it.

My own reasoning was that if the resistance of the primary was a certain figure, then regardless of the number of turns, ohms law would give a constant current through the coil, provided the voltage was constant, thus there would be a constant load.

Sean,

Think of it this way: when you connect a load to the secondary, there is a 'reflected load' seen at the primary. Load being resistance. So there is current feeding the secondary load - this has to come from somewhere, right?

The primary does have a constant no-load resistance all right, as Michael said: eddy & hystersis losses etc. This is why the transformer is warm, even with the secondary in a no-load state, i.e. your phone is not plugged into the adaptor.

Dave

Michael Collins

But if that were the case the transformer would be constantly drawing a specific amount of power from the supply...where would this go if there was no load connected on the secondary?

You are correct in that the resistance of the secondary coil remains constant - but only at DC. When you apply an AC voltage (as you have to in order for an AC transformer to work!) the primary is no longer seen as a simple resistance, the primary coil acts an inductance (with the coil resistance in series, but for an ideal transformer we ignore that), as does the secondary coil, and these two inductors are "coupled" with the iron core. Coupled inductors have a specific relationship, which is hard to describe here, but basically as the AC current in the seconday increases, this acts to try and decrease the voltage in the primary (Faraday's & Lenz's law) but since the voltage across the primary is fixed, instead the current in the primary coil increases and hence the coil voltage increases by just enough to accomadate the voltage drop.

As for the relationship in my last post, that can be derived from the ideal transformer equations...

v1/v2 = n1/n2
i1/i2 = n2/n1

Now the input impedance of the primary is:

v1/i1 = ((n1/n2)v2)/((n2/n1)i2)

But v2/i2 is the impedance of the secondary, Zsec, so we have:

v1/i1 = Zsec.(n1/n2)/(n2/n1)

Some simple algebra gives us:

v1/i1 = Zsec.(n1/n2)^2

or Zprim = Zsec. (n1/n2)^2

That concludes transformer crash course 101 and I hope satisfies your logic.

Thomas_S_Hunterson

Gotcha, thanks a mil

mathias

Mr Collins ,
Surely there is a mistake in your post above , How on earth can the load in the primary be 1600 times the load in the secondary if the secondary is only 6 volts.

Clearly there is an error here , a domestic power socket is capable of 240V x 13 Amps = 3120 Watts. Assuming we are talking domestic supplies here and not some exotic step up mains fed transformer !!

Your winding equations all point to the practical fact that the power in the primary must equal the power in the secondary ( ignoring small losses etc. )

Then what this means is that due to the inherent equality of VI primary = VI secondary then as the voltage in the secondary is nearly always a fraction of the voltage in the primary it follows that the current ( load ) in the secondary is nearly always more than the current ( load ) in the primary.

For example a 12v device with a 2 amp load connected to secondary will have a power rating of 24 watt , corresponding current then in the primary will be W/V = I = 24/240 = 0.1 amps.

Or , roughly speaking , the load in amps on the primary will be less than the load in amps in the secondary in the same ratio that the number of turns on the primary is more than the number of turns in the secondary.

I think you have things backwards in your calculations for your first post in this thread.

Michael Collins

Mathias,

When I mentioned load in my above posts I was refering to an impedance, not a current. This is the usual meaning of the term "load" (although it can also mean power in other contexts). Otherwise what you say is correct, the current flowing in the seconday winding will be a lot more than the current in the primary for the exact reasons you pointed out. This is really just a restatement of what I said since, for a given voltage: the higher the load (impedance), the lower the current.