I'm having an issue with my NCP718 voltage regulator overheating to the point of thermal shutdown. I have attached a photo of my schematic. I am inputing ~15-20v and outputting 4V from the reg. The reg is rated to 300mA and I am drawing about 50mA at idle according to my switch mode power supply.
Even at such a low current draw the regulator gets too hot to touch and eventually shuts down. Any tips would be greatly appreciated, I am a little stuck right now!
The reg is rated to 300mA
That's one of its many limits! Not the only one. You're thermally limited.
Let's see what that means.
Linear Voltage regulators work like this:
They use an internal transistor as an "adjustable resistor", which they always adjust so that, given the current drawn at the output, the voltage dropped over the resistor is such that the desired output voltage is kept.
In other words: they're designed so that the difference in energy per charge (electron) in the voltage difference between in- and output is converted to heat.
That gives us the very simple formula of the power converted to heat:
$$P_\text{linear reg., heat} = \left(V_\text{out}-V_\text{in}\right)\cdot I_\text{out}\text.$$
In your case: 11 to 16 V drop × 50 mA current = 550 mW to 800 mW idle power.
Go into your data sheet, look for "thermal specifications" or similar, look for "thermal resistance". Look for your package variant. If you've got a heat sink attached to some part of your chip, look for the resistance junction-to-that-part, add the thermal resistance of the heatsink, and multiply with the power converted to heat.
You sadly didn't specify the package variant, so I'm assuming SOT-23-5. That's got a junction-to-ambient thermal resistance of 235°C/W, so 550 mW will heat it up by more than 100°C. Yowza!
So, clearly, this was a bit of a design mistake:
If you have an absolute top limit to the current that needs to be drawn, then you can drop some of the voltage with a series resistor before the regulator. Size the resistor to still give you at least the minimum headroom at the LDO at the maximum supply current. The LDO input capacitor must still be directly to the LDO for stability.
This is not an efficient solution, the best would be to use a switch mode DC-DC converter as the other answers state. However, it is a practical workaround to get an existing board working with minimal changes.
With 20V input voltage and 4V output voltage you have a 16V drop on the regulator. With 50mA load current this is already a power dissipation of 800mW. The small TSOT-23 package will really have problems cooling 800mW away and even the larger WDFN6 needs big groundplane beneath it.
You can either add a heatsink on top of the package or you should reduce the input voltage. If you reduce the input voltage to be only something like 9V for example the regulator has to drop only 5V, which reduces power dissipation to 250mW.
If you have to stick with this high input voltage it would be very reasonable to use a DC/DC converter, which can convert your voltage with high efficiency and that way it will stay cool.
You had to have a look of the total power consumption of your regulator:
(20V - 4 V ) * 50mA = 0,8W And this could be hot!!!
That regulator will need to drop at least 15 V - 4 V = 11 V at 50 mA that's already 0.55 W.
There are two variants of the NCP718, one in a WDFN6 package and one in a TSOT−23−5 package. Which one are you using?
Table 3 in the datasheet tells us what the junction to air thermal resistances are, for the WDFN6 is is 65 °C/W but for the TSOT−23−5 it is 235 °C/W.
Normal practice is to provide a large copper area on the PCB, this copper area then works as a heatsink.
Dropping more than 11 V or more across a small regulator isn't such a good idea unless the current is really small (less than a few mA). You can also add a zenerdiode or even a resistor to drop part of the voltage that appears at the input of the regulator. I would try to make the voltage drop across the regulator around 2 to 3 V at 50 mA.
