I use L7805CV voltage regulator (TO-220) to get 5 V from a Li-Po 7.4 V 2200 mAh battery. However, after a while, voltage regulator gets really hot.

Current measurement showed 200 mA.

P=I*V=0.2*(7.4-5)=0.428 W.

From datasheet:

Thermal resistance:

junction-case = 5 °C/W, junction-ambient = 50 °C/W, total 55 °C/W.

In this case, temperature rise will be 0.428*55=23.55 C. Seems low to me, but I can't touch voltare regulator for more than one sec. Do I need a heatsink or it isn't necessary. Datasheet refers there is thermal overload protection. In this case, can I calculate how long does it take for voltage regulator to shut off (if this will happen)?

  • 1
    \$\begingroup\$ Besides the thermal overload protection, note that semiconductor lifetime is very much related to temperature. \$\endgroup\$ – Tut Jun 16 '14 at 15:27
  • \$\begingroup\$ Spehro's answer seems to address the likely issues, but see this question for possible extra ideas. \$\endgroup\$ – Russell McMahon Jun 16 '14 at 16:03
  • \$\begingroup\$ @Tut Nice advice. \$\endgroup\$ – dempap Jun 16 '14 at 18:38
  • \$\begingroup\$ The output of your power dissipation equation should be 480mW, not 428mW. \$\endgroup\$ – marcelm Oct 9 '16 at 12:56

That should be okay, in a TO-220 case with no heatsink, assuming moderate environmental temperature.

You don't need to add junction to case to junction to ambient, all you need is junction to ambient, which should heat it to less than 48°C (assuming 25°C ambient). Remember, it's the temperature rise you are calculating.

Too hot to touch is around 60°C, at least for my calibrated finger. If you're getting it too hot to touch there are several possibilities:

a) Your 200mA measurement is not accurate (perhaps the meter drops the voltage)

b) Your 7.2V measurement is not accurate

c) You have really sensitive fingers

d) It's really hot where you are testing it (30-40°C)

To answer your question about the maximum temperature and thermal shutdown- the data sheet specifies an abs. max. junction temperature of 150°C, but to be conservative I like to see no more than 100°C. That would imply a maximum ambient temperature of 75°C, which is reasonable.

The actual thermal shutdown is kind of a last ditch thing, and does not come on until you've well exceeded the absolute maximum junction temperature, typically around Tj = 170°C. It's very unhealthy for the chip to experience this, but it does prevent immediate destruction. That would correspond to an ambient temperature of 145°C which seems mighty hot, even for southern Europe.

  • \$\begingroup\$ "If you're getting it too hot to touch there are several possibilities: ..." - Also, the junction-to-ambient resistance is usually assuming free-standing in air, allowing for convection. If you violate that assumption (e.g., packinging in foam), the thermal resistance may be considerable higher. \$\endgroup\$ – marcelm Oct 9 '16 at 13:00

Your best bet is to get a heat sink installed for that regulator. Last thing you want is for it to get too hot and cutout, effectively shutting down your electronics. When the power level in the regulator is in the area that you describe it could turn unto a thermal oscillator wherein it gets too hot, cuts out and cools down some and then comes back on to restart the cycle. This can get very annoying when trying to use your electronics for anything productive.

An alternative to the heatsink idea is to purchase a switching regulator module that would operate more efficiently and cooler. It is possible to find modules that have pinning similar to the 7805 that you are using now.


A heat sink (also commonly spelled heatsink[1]) is a passive heat exchanger that transfers the heat generated by an electronic or a mechanical device to a fluid medium, often air or a liquid coolant, where it is dissipated away from the device, thereby allowing regulation of the device's temperature at optimal levels. In computers, heat sinks are used to cool central processing units or graphics processors. Heat sinks are used with high-power semiconductor devices such as power transistors and optoelectronics such as lasers and light emitting diodes (LEDs), where the heat dissipation ability of the component itself is insufficient to moderate its temperature.

  • 1
    \$\begingroup\$ Ok, but that doesn't really address the question. \$\endgroup\$ – Wesley Lee Oct 9 '16 at 12:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.