Electronic – What does a buck converter do with the energy that a linear regulator would otherwise waste as heat

buckswitch-mode-power-supply

I've been reading for hours about how switching regulators work but I'm still unable to understand what it does to keep the power equity between the input and the output (ignoring internal losses).

Let's say you step down 5v to 2v with a buck converter. What happens with the 3v dropped? Are somehow "converted" to current?

Another simple example: Let's say I have a 3.7v battery and I want to feed an IC which has an input VDD range of 1.8 to 3v so I use a buck converter to step down the voltage. If I step down the voltage to 1.8v would the battery last longer than if I step down the voltage to 3v?

Best Answer

Yes, crudely put, it converts the voltage into current. Ignoring the internal losses, if you halve the voltage you double the (available) current.

So if you convert a 5V 1A input into a 2V output the current capacity becomes 2.5A minus internal losses.

Say you have a switching efficiency of 85%, then you can work it out as:

$$ \frac{5V}{2V} = 2.5 \times 0.85 = 2.125 $$ $$ 1A \times 2.125 = 2.125A $$

You can think of it as the DC equivalent of a transformer for AC.