# Step down switching regulator heating up with light loads

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.

• Which circuit did you use - I didn't see an example of a 12V output circuit. Please be clear and supply component values. – Andy aka Aug 22 '13 at 22:43
• I have just edited my question. – user26200 Aug 22 '13 at 23:05
• @Andyaka - FWIW, the datasheet states VOUT Output Voltage 0.81 to 15 V  under the recommended operating conditions, so I think it's OK. – Connor Wolf Aug 23 '13 at 6:05
• C1 and C2 don't look large enough to have the required capacitance. Can you provide the schematic for the board, as you have assembled it, rather then the schematic from the documentation. – Connor Wolf Aug 23 '13 at 6:09
• @Connor Wolf - The circuit schematic is exactly the same as the datasheet with the changes mentioned in the EDIT. – user26200 Aug 23 '13 at 7:17

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.

• Congrats, great. This is why answering your own questions is not weird but encouraged. It would be awesome if you could still improve your answer including some scope screenshots showing the "before" and "after" effect and indicating how the higher value inductor is handled better by your controller IC. It's still unclear why the IC did what it did. – zebonaut Aug 24 '13 at 8:19

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 Pout = 0. Efficiency is defined as η = Pout / Pin. At no load, even if your standby power is very small (e.g. Pin = 1 µW), with Pout = 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.

• I actually used the correct voltages and I have just indicated that in the edit to clarify. – user26200 Aug 23 '13 at 17:53

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.

• I guess putting in a larger inductor as suggested in the datasheet won't be enough to help. – user26200 Aug 23 '13 at 8:27
• @user26200, I don't really know. It can certainly help to some extend, but I can't say whether you'll be able to reach 80%-90% efficiency with <20mA (I guess not). See the edit to the answer – Vasiliy Aug 23 '13 at 9:10
• @user26200, do you really need this low current operation mode? You may consider switching the regulator OFF when you don't need high currents. – Vasiliy Aug 23 '13 at 9:13
• The datasheet does NOT show that power dissipation is increasing below 300mA -- only that it is not decreasing as fast as the output power is decreasing. – david Aug 23 '13 at 9:22
• @david, the load power is $\propto I_L$. If the power dissipated by the IC itself would be constant, then $\eta =\eta _{max}−C\frac{I_LV_{out}}{P_{int}}$, where $P_{int}$ is constant, $C\approx 1$. Since the slope of the efficiency is clearly steeper than a linear, $P_{int}$ is rising. – Vasiliy Aug 23 '13 at 9:51