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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!

enter image description here

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  • \$\begingroup\$ Which package regulator? SOT or WDFN? \$\endgroup\$ – Justme Apr 7 '20 at 8:37
  • \$\begingroup\$ There's about only one reason why you'd use linear regulators in the year 2020, and that is EMC. So unless you have some sensitive RF electronics, consider using a modern switch regulator instead. With proper layout, switch regulators don't give much noise at all nowadays. \$\endgroup\$ – Lundin Apr 7 '20 at 8:41
  • \$\begingroup\$ @Lundin uhhh that is not entirely true. We could discuss this, but it would be very much off-topic for this question! In this use case, it's very clear you're right: use a switch-mode regulator to drop 11-16 V. \$\endgroup\$ – Marcus Müller Apr 7 '20 at 8:45
  • \$\begingroup\$ @MarcusMüller BOM, board space and cost-wise, switch regulators are a quite sensible choice nowadays. Internal MOSFET is standard and vendors even seek to place the inductors on-chip. And of course in terms of efficiency, switch regulators are superior. Only down side I can see is EMC and more knowledge needed by the PCB designer. \$\endgroup\$ – Lundin Apr 7 '20 at 8:54
  • \$\begingroup\$ @Lundin noise, board space, integrateability, cost, vibration sensitivity, accuracy, requirement for stabilizing capacitors, and low-current efficiency are chief among my concerns why a linear regulator might be right for a specific application. This is engineering, don't approach it with broad generalizations; know your tools and when they're appropriate. Again, this application calls for a switch-mode regulator. \$\endgroup\$ – Marcus Müller Apr 7 '20 at 9:12
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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:

  1. When dropping reliably high voltage differences, the way to go is usually a switch-mode regulator. Linear regulators "burn" all the voltage difference between in- and output. Switch-mode regulators just store energy and deliver at a different voltage than the input – which means you have a bit of losses here and there, but not "burn all the energy per electron that's in the input voltage-output voltage".
  2. You didn't calculate how much power you'll be converting to heat in your linear voltage regulator, and thus didn't use one where you could have attached a sufficiently sized heat sink.
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  • \$\begingroup\$ Thank you for the very detailed reply! Yes, my package size is wrong, I failed to calculate proper heat generation. Thanks all! \$\endgroup\$ – Michael Apr 13 '20 at 9:09
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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.

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  • \$\begingroup\$ I'll allow myself to add: Let's calculate what an appropriate resistor would be: if you still want some headroom for regulation, considering the used linear regulator is actually an LDO, 5 V Vin would more than suffice. So, aiming for 5 V at the input, at the max input voltage of 16 V, the resistor should be dropping 11 V at the max current draw of OP's VCC – I don't know that, but let's assume it's 110 mA. That would give us a 10Ω resistor. Make sure it's rated for at least 800 mW! Many smaller resistors are rated for ¼ W only. \$\endgroup\$ – Marcus Müller Apr 7 '20 at 10:26
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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.

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  • \$\begingroup\$ If insisting on using linear regulators, rather than using mechanical heatsinks, it might rather be wiser to do the voltage divide in several steps, with multiple regulators in series. Though a DC/DC switch regulator is of course the best solution. \$\endgroup\$ – Lundin Apr 7 '20 at 8:39
  • \$\begingroup\$ You are right, distributing the dissipation over several converters might be a solution as well. But as you said, I would favor a DC/DC converter. \$\endgroup\$ – jusaca Apr 7 '20 at 8:42
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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!!!

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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.

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