Electronic – Choosing a linear regulator vs. switching regulator for low-power application

batterieslinear-regulatorswitch-mode-power-supplyswitching-regulatorvoltage-regulator

I have a 3.6 V battery that I want to regulate down to 3.3 V and 1.8 V rails. I have already asked about the configuration of the two rails here.

My question here however is whether I should use a linear regulator or switching regulator. I know that regulating from 3.6 V down to 3.3 V results in fairly high efficiency with a low dropout regulator (3.3 / 3.6 = 91.67 %), but going to 1.8 V results in half the power being wasted as heat (1.8 / 3.6 = 50%).

I want to compare this 50% loss in using a linear regulator vs. using a switching regulator. This involves considering whether or not the extra power needed to power the regulator itself is greater than the power conserved in using it.

How would I go about calculating this power lost vs. power conserved in order to compare it with the linear regulator?

I know that the power wasted in a linear regulator can be calculated simply by (Vout – Vin) * current drawn, but how do I calculate this for a switching regulator?

What kinds of switching regulators would be best for this application? (Buck regulator?)

Best Answer

I believe I may need a buck-boost however, because the components will need at least 1.8 V and 3.3 V, and my 3.6 V battery could drain below that

First hit on Linear Technology might be useful: -

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Try going to LT to get an idea and use this parameter search engine.

for the 1.8 V rail it's 1.154 mA

Use a linear regulator - it's just not worth the effort I believe to use a switcher from 3V3 to 1V8. However, if current was over 50mA I might start considering a switcher.

Generally, you can assume a buck regulator will give you about 90% power efficiency near full load and a buck-boost will be about 85%. Ball-park figures.