Seeing all the transformers around me looks like a huge piece of iron. I was wondering why transformers have such a huge mass. Consider if we need 11V/ 440V voltage transformation. Why don't we have 11 no of turns on primary and 440 turns on the secondary. Theoretically this would give us the required voltage ratios and much less weight. Why doesn't it work? What am I missing?
Why don't we have 11 no of turns on primary and 440 turns on the secondary. Theoretically this would give us the required voltage ratios and much less weight. Why doesn't it work? What am I missing?
Why not have one turn on the primary and 40 turns on the secondary you might as well ask. With pretty much any normal size of core you might get somewhere between 1 uH and 100 uH inductance for one turn. Let's say you get 10 uH and let's say your AC voltage was at 50 Hz. The single turn primary winding has an inductance of 10 uH and at 50 Hz, this is an impedance of 3.14 milli ohm i.e., with no load attached to the secondary, the primary alone is drawing a current from your 11 volt AC supply of 3500 amps. Now nobody wants that.
With 11 turns, the inductance isn't just 11 times higher it's 11 squared times higher so now the inductance becomes 1.21 mH and the reactance is 0.38 ohms and will take an unloaded primary current of 29 amps - i.e. a lot better but still not that great but, do you see the point here - irrespective of the lump of iron being a transformer, it still has a residual primary inductance that can take too much reactive current from your 11 volt AC source and it makes sense to seek to minimize that. If you went for 44 turns the inductance would rise by 16 to 19.4 mH and the residual current would fall by 16 to about 1.8 amps - this might be more reasonable.
You might want to go higher of course but then you have a lot more copper and there becomes a trade-off between residual magnetization current (which leads to core saturation losses) and copper losses. More turns means less core loss but maybe unacceptable copper losses (\$I^2R\$).
This is why SMPS devices use switching frequencies of typically 100 kHz - 10 uH will have an impedance of 6.28 ohms (compared to 3.14 milli ohms at 50 Hz).
- Using the example of 1 turn having 10 uH inductance drawing 3500 amps means a magneto motive force (MMF) of 3500 ampere turns.
- With 11 turns, the inductance is 1.21 mH and the current is 29 amps or an MMF of 319 ampere turns
- With 44 turns the MMF is about 80 ampere turns.
To accomodate a lower number of turns and prevent saturation either the core cross sectional area has to increase (generates more inductance per single turn) or the core length needs to increase (decreases the H field) so when you say: -
this would give us the required voltage ratios and much less weight
You are misled.
This is really done to limit the no load magnetising current. You might be aware of the fact that the induced emf in a transformer is proportional to the flux linkage. The flux linkage is itself proportional to the magnetising current and the square of the number of turns. Hence, by increasing the number of turns, we reduce the current needed to establish the same flux in the core and hence, produce the same emf.
If you're not acquainted with the concept of flux and magnetising current, you'll find it in any standard Electrical Machines textbook.
Imagine the case when there is no load. That means there is no current in the secondary, in which case it might as well not be there. Now you are left with some turns of wire around an iron core. Then the transformer is just an inductor connecting the leads of the power input. If that inductor is too small, a large current will flow.
For instance, if you have a 120 VAC -> 12 VAC 60 Hz transformer, and the primary has 0.01 henry of inductance, the no load current would be
120 V / (2 * pi * 60 Hz * 0.01 henry = 32 amp. In principle, if the inductor were perfect, the current would be out of phase with the voltage, and the net power would be zero, but your wires always have some resistance. So instead you want to increase the inductance up to a higher value, say 1 henry, so now you only have 300 mA of magnetizing current, which is much more reasonable.
The way you increase the inductance is to increase the number of turns. So your core size and material and the input voltage determines the number of turns you need.
Was measuring a transformer just today--- 3 windings, each of 10 turns around a railroad spike. To limit primary current (and protect the old signal generator against shorted loads), a 2,700 ohm resistor was in series with the primary winding.
With 10 voltspp from the generator at 10,000Hz, the primary only had 3 millivolts PP across that winding. Why? a very low inductance, because of the flux return path was mostly through air. Z(primary, at 10KHz) is 1/3,000th of the 2,700 ohm protection resistor, or 1 ohms. Inductance? 16uH, including the railroad spike permeability. About 1uH, modeled as air core.
Other recent questions ask about the largest inductors available. The ABB power transformer, rated 1Billion Watts, had nearly 2,000 Henries. Why?
So the impedance at very high voltages would limit the current. The impedance of 2,000 Henries at 377 radians/second (60Hz) is +j754,000. At 1 Megavolt, the magnetizing current is 1.3 amps, or 1.3Million watts just to get the core into idling condition, ready to serve as a transformer.
It depends on the power thoughput of the transformer.
Power is the product of volts, and amps.
There's a limit to how many volts per turn you can generate around a transformer core. The larger the area of the core, the larger the volts per turn.
There's a limit to many amps you can push through a copper before it gets too hot. The larger the area of the wire used in the winding, the more current it can handle.
Taken together, these mean that the power throughput of a transformer is more or less proportional to its volume, certainly a larger transformer of the same style can shift more power than a small one.