# How does buck converter increases current

I am confused regarding this, in buck converter Pin=Pout, so if I reduce voltage then current has to increase but my resister load is fixed at, say, 5 ohms - but resistors follows Ohm's law and according to that current should decrease if the voltage is reduced.

So if you could tell me where I am wrong.

## 2 Answers

The available power is based on Pin = Pout,

BUT, if you reduce the voltage into a fixed value resistor, then the current reduces appropriately as per Ohm's law.

Your confusion is about the power staying constant, it is available if you change the resistor...

If you reduce the voltage supplied to a fixed load resistance then the load current must decrease according to Ohm's Law. But if you use a buck converter to decrease voltage the supply current will reduce more than the load current.

For example, if you power a 5Ω load from a 10V supply then the current will be 10V/5Ω = 2A. If you then drop the load voltage to 5V using eg. a series resistor or linear regulator, both the load current and the supply current will be 5V/5Ω = 1A. So the supply is delivering 10V*1A = 10W, the load is consuming 5V*1A = 5W, the other 5W is dissipated as heat in the voltage dropping device, and the conversion efficiency is 50%.

However if you use a buck convertor the output power will only be slightly less than the input power. If the buck converter is 90% efficient then 5W at the output equates to 5W/0.9 = 5.56W at the supply. 5.56W/10V = 0.556A, ie. 55.6% of the 1A output current to the load. So the output current of the buck converter has increased compared to the input current, and the supply current has reduced more than the load current.