I have heard that switching power supplies that convert, say, 110 VAC to 12 VDC, are very efficient. I understand that the power transistor that is regulating voltage is either on or off and spends very little time in its linear range, thus it doesn't burn much of the energy. Also I understand the principles of converting AC to DC via a full wave bridge. After the AC voltage is rectified, however, given that the voltage drops from up to 110VDC to 12VDC that energy (above 12) has to go somewhere. If the electrons coming from the AC are merely regulated down to 12 then I don't understand what happens to the lost energy. It has to go somewhere. In a very efficient system I would think that a small amount of current at high energy going into the AC input should produce low energy with higher current according to the equations. So I should be able to get by with a smaller gauge wire going to the input (at high voltage) of my switching power supply while having to use a heavier gauge wire for the output (at lower voltage). My problem is that I can't see how reducing the amount of energy on the same electrons can't result in some sort of heat generation. Where does the energy go? Can someone please explain?
Electronic – “Efficiency” of Switching Power Supplies and Theoretical Energy Loss
acefficiencypowerpower supplyswitching
Best Answer
This is a much simplified answer: -
There is this thing called a transformer - it has two coils that are closely magnetically connected. The two coils are called primary and secondary. If the primary has 1000 turns and the secondary has 100 turns it will step-down an AC voltage of 120V to 12V. There will be a slight power loss but usually transformers are better than 90% efficient. If we assumed near 100 % efficiency it follows that if we step down the voltage by ten times we also can step-up the current by ten times so, simplistically: -
\$V_{in} \cdot I_{in} = V_{out} \cdot I_{out}\$
The transformer is at the heart of an off-line switching power supply but instead of the transformer being fed with raw AC at 60Hz or 50Hz it is fed at a much higher frequency and this means a much smaller transformer can be used.
So the incoming 110V AC is rectified and smoothed (not to 110V DC) but to a voltage more like 154 volts DC. This feeds a power oscillator (that might run at 100kHz) and this drives the transformer primary.
On the secondary (the low voltage side) there is another rectifier and more smoothing capacitors. There may also be a feedback system that keeps the output closely regulated. Here's a typical block diagram: -
(source: philpem.me.uk)
Correct!