Transformers

Wizard

I've been trying to brush up on my knowledge of transformers.
I have a rather embarrassing question to ask, as I feel the answer should be obvious, but haven't managed to find it through google etc.

Here goes.

Question 1:
A transformer consists of two coils next to each other ... with power transferred from one to the other through electromagnetic induction.
I think I understand that bit.

What I don't understand is the behaviour of a transformer plugged into the mains on the primary end ... and with no load on the other.
This shows the mains being directly shorted out through the windings of the primary coil.
I understand that the (many) primary coils will offer some resistance ... but don't understand why the coil doesn't just get super hot and burn out ... like a light bulb.

How does it work then that a cellphone charger transformer can stay plugged into the mains and only draw such small current with this apparent direct short?

Question 2:
In one of the threads it was mentioned that the sound of a valve amplifier can be "warmed up" by upgrading the output transformer.
What characteristic of a substandard transformer would have a negative / "cooling" effect on the sound?

makepeace

good questions that i would also love to hear answers to.

just some stabs in the dark:

question 1:
i would imagine that if there is no load across the secondary, the primary current would be relatively low, ie only enough to induce currents to "supply" the losses caused by the core material and the switching of the direction of current flow.

question 2:
surely it would be the quality/suitability of the materials? ie the core for one.

Gearhead

answer 1: the transformer now turns into a choke. An ideal choke would conduct current, but not consume a lot of energy since the current would be 90 degrees out of phase with the voltage. Less than ideal chokes even worse, this is why electrical equipment is mostly switched off on the primary side.

answer 2: ask two transformer manufacturers and you will get two different answers. The consensus seems to be that for an output transformer the core material, the air gap (which stops the core from becoming saturated by the DC current running through the output tubes), the interleaving (which transfers magnetic energy directly instead of through the core) and copper/insulator quality need to be right.
The easiest way to make an OPT sound thin is by taking out too much bass by making the air gap too big.

djpauw

The short answer to q1 is that the answer lies in the deeper theory and math of the matter, I will only summarise here.

Knowing the basic purpose and operation of a transformer you would agree that I can say that when looking into the secondary side, I perceive a lower voltage than the mains it is connected to at the other side. Agreed? Similar applies to current etc and the factor being the winding ratio.

Same applies in either direction over the transformer.

Know you should be able to calculate that the perceived resistance seen from the mains side is also some multiple of the load resistance, which in the open loop case is theoretically infinite....

Any number multiply by infinity equals...? infinity
Therefore at the mains side the system perceives an infinite resistance and no current will flow. (in practice there is of course a very small current flowing...)

And a small note: cellphone chargers and other such adapters are switchmode power supplies, it's a whole different ballgame

Hope that helps

DJ

Wizard

djpauw wrote Now you should be able to calculate that the perceived resistance seen from the mains side is also some multiple of the load resistance, which in the open loop case is theoretically infinite....
Any number multiply by infinity equals...? infinity
Therefore at the mains side the system perceives an infinite resistance and no current will flow. (in practice there is of course a very small current flowing...)

I understand the maths; but was wondering about the physics.
How can a directly connected coil offer infinite resistance?

The best guess that I can think of is:
- AC current in primary generates an oscillating magnetic field
- this magnetic field would induce a voltage in the secondary if there were a load
- without a load, it induces a voltage into itself of equal and opposite magnitude; thus presenting infinite impedance

Does this make sense?

TomCat

A couple of things:

Transformers only work under AC (alternating current) conditions. in DC conditions the coil is almost a short circuit. When transformers are designed for power applications the input/output winding ration will determine the input/output voltage ratio as well. The size of the core and the wire diameters will determine the current capacity. The mains side has a lot more windings than the secondary side and is usually relavively thin wire so the impedance is quite high, hence not burning out.

Gearhead is absolutely right about the current being 90 degree out of phase with the voltage. Same goes for capacitors except that for transformers the current leads the voltage (+90 degrees) and for capacitors it lags the voltage (-90 degrees). This is what gives the transformer and capacitors it's impedance (along with the frequency of the AC signal. Transformers have a rising impedance with frequency and capacitors a falling impedance with frequency.

Another factor with transformers as an inductor is the Q of the windings/core combination etc. Air cored inductors typically have a low Q and iron cored inductors a much higher Q. This is important to consider when inductors are designed for passive crossovers in loudspeakers etc.

Karel-Mars

Nice question Wizard!

I thin Q1 is pretty much answered. Q2 is about what makes a transformer substandard. I'll try to answer it:

The function of a transformer, esp. the output trannie in a tube amp, is to pas a certain bandwidth of frequencies at a certain power from the primary to the secondary. Ideally it should do this without burning out or distortion.

So from a manufacturer's point of view, less copper/steel used is more profit. The other thing is the extent to which you magnetise the core for it to transfer power (measured in Tesla and called flux). The more flux (less windings per volt) you use, the more the core distorts the signal (nice for guitar amps), the less copper and steel you need.

The other thing is the bandwidth. You have to choose what lowest Hz you want to pass at full power. The lower you go, the bigger the transformer. So small transformer compromise on bass, more distortion (less punch). A better transformer (with a bigger iron core) will give you a tighter sound.

The thinner copper wire you use, the more the internal resistance of the primary goes up. Music is turned into heat. A bigger transformer (with thicker copper wire inside) will be more powerfull and give better bass. You have to hear the difference to believe it.

The economies of scale will always make manufacturers choose the smallest/cheapest transformer possible for the job.

Wizard

Thanks for the excellent answers !
?

guidothepimmp

Karel... very insightful and useful input. thanks

Wizard

My mind is not easily satisfied.
In my Q1 I was asking why an AC current on the primary of a transformer flows very little when the secondary doesn't have a load.
The answer appears to be one of AC impedance.

Brushing up on my electrical basics, I'm busy reading the foundation chapter of Art of Electronics by Horowitz & Hill.
It says:
"... a transformer of turns ratio n increases the impedance by n2. There is very little primary current if the secondary is unloaded"

This gives an observable characteristic.
I'm still interested in the reason why.

Is it a matter of self induction, where (under an unloaded secondary) the changing magnetic field produced by the changing current induces a changing current (almost) equal and opposite to the supplied current, thus preventing the current from flowing? (and where this "self-induced-opposing-force" is called the impedance?)

Gearhead

Start off by looking at it from an energy perspective. The only way a tranny looses energy is by heating up the copper or the iron and otherwise there's stray megnetism. So no energy used up in the secondary means no energy used in the primary. In this primary, energy is voltage times current for every little timeslice of the AC cycle.
Then, imagine what ways you could stop the current through the primary: induction. Remember that voltage is the accelerator, impedance is the brake and current is the speed of the thing. Rather than considering the magnetic effect that the secondary has on the primary as an increase in impedance, think of the impedance causing an opposing voltage. There is no drive anymore and no AC current will occur.
Remember also that this counts only in the AC domain; the DC current that runs through transformer and output stage has nothing to do with this effect.
The whole notion that a lack of current in the secondary makes for enormous voltage peaks in the primary is also clear when you think of how a tube amp without a speaker is likely to blow its output tubes and possibly transformer.

Gearhead

You could also see the primary-without-load-on-secondary as a John Wayne saloon door. If John moves slowly in one direction, the doors open as if they were not there (dc sees no inductance). If John now wants to go in and out 1000 times per second, this door will start to really become a nuisance, with all its inertia and wind drag.

Wizard

Gearhead wrote Remember that voltage is the accelerator, impedance is the brake and current is the speed of the thing.

This is a nice metaphor - thanks !

... think of the impedance causing an opposing voltage.

I suppose my question was what was causing this impedance, which was causing the opposing voltage ... and it can only be the source.
So it is, in essence, stopping itself.
Just like John Wayne is hindering his own progress in the high speed swinging door situation.

In the unloaded secondary scenario, the brake is connected to the accelerator.
The harder you try accelerate, the harder the brake is applied.
Ne?

While you're in such excellent explanation mode - what exactly is a "choke" and what is it used for?

As an unrelated, but similar experiment ... I was thinking of trying to push two magnets of equal poles together.
The harder you (try) push them together, the greater the opposing force becomes (as a result of pushing them together) and ... nothing happens.

Gearhead

A choke is nothing but half the transformer you posted in the original post. There is no secondary windings so the effect is the same as when the leads are unconnected: it just becomes a self inductance.
You use it to make sure DC comes through and AC not, but (like the John Wayne door) some low frequency AC does come through. There are two dominant places for chokes in audio, mostly in conjunction with caps:
- in power supply filters, where they keep the current constant while the voltage through of the rectifier wobbles with 100Hz;
- in frequency filters, where they let low frequencies through and block the high ones.
See also http://en.wikipedia.org/wiki/RLC_circuit

Wizard

So ...
- you use capacitors to let AC through, while blocking DC, and
- use chokes to let DC through, while blocking AC?

Gearhead

Spot on! ?
You do have to take into account that the practical imperfections of the things themselves make that you do get leaking, maximum current and voltages etcetera.