# Voltage regulator with heatsink gets overheated

First of, I'm sorry if I'm using the wrong terms or if I'm pointing out something that is obvious to most, this is not my profession, it's only a hobby.

I have an input voltage of 12V 5A which is needed for a water pump. I'm trying to use the same source too drive a Raspberry PI Zero W together with a bunch of sensors. To lower the voltage to 5V I'm using a voltage regulator, L78S05CV, which I had lying around. It should be able to give me 2A but currently I'm only needing about 0.27A.

The voltage regulator has a heatsink with a thermal resistance of 29 °C/W and the regulator has a thermal resistance from junction to case of 5 °C/W (Rthj-a of 50 °C/W). The regulator dissipate around 1.89W ((12 - 5) * 0.27), which would mean around 64 °C ((29 + 5) * 1.89). Assuming an air temperature of 25 °C it should end up at around 90 °C. The regulator should have a operating junction temperature of up to 150 °C. Still I do belive that the regulator gets overheated after some time, I can see a drop in voltage to about 2.5V.

I am trying to understand the math here, to see if I need another heatsink with a better thermal resistance or if I should just look at other alternatives.

I have read several other responses to similar threads, but too be honest, the answers is a bit out of my league.

• Resistor? Do you mean regulator? The whole heat dissipation thing relies on the heatsink having heat removed so that the local ambient temperature can be assumed to be 25 degC. If this does not happen then local ambient rises a bit then the chip then rises a bit (in response as you would expect) then local ambient rises a bit more and you have a problem. Commented Aug 26, 2018 at 18:43
• Unless you're doing something funny, your output is 5 volts, so power is (12-5)/.27, which gives a final temp of about 114. Commented Aug 26, 2018 at 18:58
• This is really a role for a switching regulator, not a linear one. Commented Aug 26, 2018 at 19:00
• @Whit3rd The specified thermal resistance of the heatsink should be the resistance between component case and ambient without a fan unless indicated otherwise. The specification usually requires sufficient airflow from convection, though, so it does not apply if convection is impeded by a tiny case or components nearby. If the regulator operates inside a case (even a big one allowing internal convection), the ambient temperature is the temperature inside the case, which in fact might be a lot higher than ambient outside the case. Commented Aug 26, 2018 at 22:10
• @Ganhammar Did you add thermal compound or a sil pad to get the heat from the case to the heatsink? If not, you might have a high thermal resistance there. If you use a sil pad, look up its thermal resistance. You most likely cannot neglect it. Commented Aug 26, 2018 at 22:13

The linear regulator is not the best solution when there's a big difference between the input (12V) and output voltage (5V). 7V are dissipated in heat.

If possible try to use a switching regulator.

If you want to keep the linear regulator you can attach an old pc fan in the heatsink(from a hard disk or graphic card for example) and use thermal paste between the regulator and the heatsink.

The improvement is dramatic.

Put a power resistor between the 12V supply and input to the 7805.

This will transfer the excess wattage from the 7805 to the resistor.

I used a Yageo 5.1Ω SQP500JB-5R1 resistor for a 12V powered 7805 with a 300 mA load in a product I manufactured for 18 years. No field service issues ever.

• This is a great suggestion, but I already bought a switching regulator and a bigger heatsink. Will probably use the switching regulator in the end. Commented Sep 1, 2018 at 4:51
• The switching regulator is the way to go if efficiency is of any concern. Linear is the way to go if reliability is a concern. With the Yageo resistor the heatsink is not needed. Commented Sep 1, 2018 at 5:02