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?)

  • \$\begingroup\$ You've got it pretty much spot on. If you're interested in the efficiency of buck regulators, you'll have to determine how much power you're going to pull from it and then look at the datasheet of the buck regulators you're browsing. The more current you consume from a switching regulator the more efficient it is. There should be a plot that shows this efficiency on all buck datasheets. Bucks would be the way to go here as they're very simple and operate efficiently. \$\endgroup\$
    – horta
    Commented Feb 10, 2015 at 5:56
  • \$\begingroup\$ How much current for each rail? If you are consuming only a few mA at the 1.8V rail, there probably isn't much need for a switch-mode power supply. If it's hundreds or thousands of mA, a switch-mode power supply is a really good idea. \$\endgroup\$ Commented Feb 10, 2015 at 6:19
  • \$\begingroup\$ @horta Thank you for answering both of my questions. 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 (it's charged by a solar panel). I'm having a hard time looking through possible switching regulators though. Is there any particular resource you would recommend that I start with? \$\endgroup\$
    – Christina
    Commented Feb 10, 2015 at 6:43
  • \$\begingroup\$ @DwayneReid My load current for the 3.3 V rail is 546.2 mA and for the 1.8 V rail it's 1.154 mA. Thus, the 3.3 V rail could draw a little over half an amp, so it seems like I would definitely need a switch-mode power supply. \$\endgroup\$
    – Christina
    Commented Feb 10, 2015 at 6:48

1 Answer 1


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: -

enter image description here enter image description here

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.

  • \$\begingroup\$ Is there a way to determine how much power it takes to power the regulator itself? (The active components of the regulator circuit such as op-amps). Is this already factored into the efficiency? \$\endgroup\$
    – Christina
    Commented Feb 10, 2015 at 9:17
  • \$\begingroup\$ @Christina yes it's factored into the efficiency figure but the data sheet for the device will give you the standby/no-load power consumption (or current consumption). \$\endgroup\$
    – Andy aka
    Commented Feb 10, 2015 at 11:40

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