# “Efficiency” of Switching Power Supplies and Theoretical Energy Loss

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?

• Energy is not voltage – Andy aka Mar 9 '15 at 18:31

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)

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).

Correct!

• Yes. I know about transformers. They are the primary element in a linear power supply (AC to AC converter). What I didn't realize is that switching power supplies still use them, though in a different form. Thanks. This answers my question. – Paul McKneely Mar 9 '15 at 21:38

There are many different topologies for switch-mode power supplies. I think you might be getting some of the concepts mixed up.

First off, in all cases that I know of, AC is converted to DC using a rectifier of some sort. The AC isn't connected to the switching element.

There's a family of SMPS called off-line converters. These connect the high-voltage rectified DC and the switching element to a transformer. This provides two major benefits. The first is that a high switching frequency lets you use a smaller transformer. (That's how they make those iPhone chargers small enough to fit inside a power plug.) The second benefit is isolation from the mains voltage, which is a safety requirement.

More basic SMPS are often called DC-DC converters, and I think these are really what you're asking about. You can think of these as energy relays. They use an inductor to store and limit the flow of electrical energy based on the duty cycle of the switch. Look at a buck-boost converter -- there's never a conducting path between the source voltage and the load:

A lot of people think that energy = voltage, and then get confused when inductors show up. Remember, just because there's a current and a voltage drop, it doesn't necessarily mean the energy is lost.

In switchmode supplies, the energy lost (heat generated) is in the freewheeling diode, and in the inductor/capacitors used, and as you mentioned, the switching element.

The combination of ESR from the capacitors, forward voltage drop from the (always a Schottky diode) freewheeling diode during the off cycle, and the inductor's DC resistance are what combine to make the efficiency losses.

The important thing is also the max/min duty cycle of the converter, and some are not as good at keeping low duty cycle conditions. They may be very inefficient at very high or very low duty cycles (usually determined by the input voltage or the difference in input/output voltages) and more energy than usual is burned through the diode.

You may want to refer to an animated diagram showing a buck converter with diode and inductor and capacitor setup and see where the current flows during the cycle, and see where the losses occur.