I'm using an MT3608L with an input voltage of 12 V to create a voltage of 18 V using the below schematic. As this is a cutout of a larger schematic, the input voltage is applied to the top left wire, the bottom is GND (obvious, but want to avoid confusion).

Schematic MT3608L

I'm using the 18 V to drive a bunch of LEDs for background lighting. The current going to my LEDs in total is about 300 mA max, so far below the spec of 2.5 A.

When using a normal linear voltage regulator like the AMS1117, you have to watch out for power dissipation in the regulator. This depends on the output current, but also the voltage drop Vout - Vin. But there, it's obvious that all the current flows through the regulator, hence power is being dissipated.

But how do I calculate the power dissipation of the MT3608L? Or does that make no sense, because the current is not running through the chip?


1 Answer 1


The AMS1117 is a linear regulator, it can only reduce voltage, and the input current is roughly the same as the output current. The excess power P = (Vin - Vout) · I is dissipated as heat.

A DC/DC converter is a different beast; it converts input power Pin = Vin · Iin to output power Pout = Vout · Iout, usually by using switching techniques and energy storage in a coil.

At 100% efficiency, Pin = Pout. No DC/DC converter is a 100% efficient, of course. The efficiency at different loads/output currents can be found in the datasheet (usually in the 80-95% range), and this will give you an idea of the losses/power dissipated as heat in the DC/DC converter's circuit.

Part of the losses/dissipation will be in the IC, the rest in other components in the circuit.

There is no general formula for calculating the power dissipation in an arbitrary DC/DC converter IC, so you will have to consult the datasheet.


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