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This is probably a basic electronics question, but I'm a self-taught hobbyist and I can't seem to find a direct answer to this.

I have a host device that I want to connect to that supplies 100mA at 5V.
The module I'm connecting runs on 3.3V and peaks at 200mA. I will have a voltage regulator and level shifters in between.

My question is, does running at 3.3V make any difference with the amp draw on the host? Or is it 200mA no matter what the voltage, meaning I will need an external supply.

Is there a general principle I can follow in the future?

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  • \$\begingroup\$ That depends on the regulator. But you can't get more power out (660mW) than you put in (500mW) regardless, so you will likely need another supply. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Jun 18, 2015 at 2:46

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Your power supply can generate $P = I(A) x E(V) = 0.1A x 5V = 0.5 W.

Now there are two ways to lower the voltage. A linear regulator or a switching regulator.

In the case of the linear regulator, the input current (at 5V) is the same as the output current (at 3.3V) - except for a couple of mA needed for the linear regulator to work. So, at the output you have 0.1A at 3.3V, or 0.33W. The difference (0.5W - 0.33W = 0.17W) makes the regulator heat up.

The switching regulator maintains the same power (not current) - except for a little power needed internally. So we calculator 0.5W at the input, then at the output we have 0.5W too (minus about 10% loss -> about 0.45W). So, the maximum current at the output is 0.45W/3.3V = 0.136A.

If the 200mA peaks are really narrow, a capacitor might save your life... (preferably at the input where the current is lower)

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