Step down switching regulator heating up with light loads

Question

I built a step down switching regulator based on a Diodes AP5100 using the datasheet example circuit with this layout. The voltage is being stepped down from 15V to 12V and the AP5100 is heating up to 60°C+ under light or no load at all(0-40ma). At about 80ma load the temperature is more reasonable at about 40°C and runs relatively cool(low 30s) above 150ma. The circuit that is being powered by this operates at low current(<20ma) 85% of the time and the other 15% at about 250-500ma, so it would be nice to have it run cooler under the light load. Should I try increasing the inductor size to improve the light load efficiency?

EDIT:

The circuit that I am using is shown above and it's for 3.3v. The only change that I have made is to use 3.6k with 49.9k to get 12V. And also, I have substituted the B230A with a STPS2L60A. The caps are all ceramics at the proper voltages including the 10uF(35V) and 22uF(25V). The inductor is 3.3uH with these specs.

Answer 1

I fixed the problem by using a higher value inductor(10uH) as suggested by the datasheet and the regulator does not heat up at all under light and no load conditions. I have tested it at half the maximum expected load (250ma) and I am getting 20mv pp ripple. I'll do more testing at the max load, but the results are very good so far.

Answer 2

Your output capacitor is rated at 6.3 V only. If you have changed the voltage divider for the circuit to output 12 V instead of 3.3 V, chances are you have destroyed the output cap. It may now act as a short (or heavy load), so you might actually be far from a light-load condition.

Don't worry about efficiency at no-load or light-load condition. No-load means P_out = 0. Efficiency is defined as η = P_out / P_in. At no load, even if your standby power is very small (e.g. P_in = 1 µW), with P_out = 0, the efficiency will still be zero. This, of course, won't look good in the efficiency diagram, so the manufacturers tend to not plot the curve all the way down to zero output power or current.

For such problems, it helps to think about the converter's losses instead of its efficiency.

Answer 3

This answer is incorrect

This graph for efficiency vs. load current is taken from the datasheet:

It looks like this IC is not supposed to be used for driving low currents in typycal configuration.

Note that the slope of the efficiency curve is very steep for \$I_L\leq\sim300mA\$. This means that lowering the current in this region of operation increases the power dissipated by the IC, hence you see the increase in temperature.

I'd say that you shouldn't use this typical configuration for driving a load which is 85% of the time is very light. You need to either change the configuration (which I believe to be a hard task), or use another component. In my opinion, the latter is the most promising and easy solution.

Mitigating the heating problem in the given configuration:

As suggested in the datasheet, you can try to use higher value inductor. This should reduce the current drawn from the internal MOSFET and will lead to higher efficiency at light loads, but will also lower the efficiency at higher loads and may lead to higher output voltage ripple due to insufficient current capability (which will require a bigger \$C_{out}\$ to mitigate). I can't say to what extend can you shift the efficiency curve - you'll need to find it out experimentally.