I need to power a 5V device that needs more or less 1.2A in order to work correctly. I'm using a 13.8V power source with two diodes in series (1N4007) to limit the voltage to roughly 12V. I tried using the voltage regulator AZ1084T, since it has 5.0A of maximum output current, but even with a large heat sink attached it was heating up a lot, like 117°C (and the maximum admissible temperature on the datasheet is 125°C).

So, just to test, I replaced the AZ1084T by the old and good LM7805CV, which has 1.5A of maximum output current. It worked just the same, heating up just the same (and that's a good thing, because this one is much cheaper). But how can I avoid the heat? I'm pretty sure it's the power dissipation from 12V to 5V that's causing it, but this voltage regulator was supposed to work with that input voltage.

I tried this scheme, using TIP127 to "split" the current, but my device won't even turn on (the measured output current was showing peaks of 2A~3A, which kept turning off the power source which limit current is 1.7A). Scheme

Anyway, any suggestions on how to split the current while maintaining the output voltage or on how to decrease the power dissipation are welcome.

  • 2
    \$\begingroup\$ They heat up the same amount because they both have to dissipate the same amount of power (which is the voltage drop between input and output times the current). \$\endgroup\$
    – Joren Vaes
    Sep 12, 2017 at 14:00
  • 1
    \$\begingroup\$ If you have a 5-V out regulator, @1.2A, with 12-V input, you'll dissipate 7*1.2 = 8.4W. That's 40% efficiency. You'll need a good sized heatsink. Otherwise you might try a switching regulator. The efficiency would be much higher (more than 80%), therefore the power dissipation would be so small, not to require any heatsink. \$\endgroup\$
    – next-hack
    Sep 12, 2017 at 14:07

3 Answers 3


Just because a regulator is able to output 5 A of current, does not mean it can do so over it's entire input range - especially not without a good amount of heatsinking (potentially with fan). The reason both of the devices heat up the same is because they have to get rid of the same amount of power, given by

$$P = (V_{in}-V_{out}) \cdot I$$ which in your case is about \$(12\ V-5\ V)\cdot1.5\ A = 10.5\ W \$, which is a lot of heat! (there is a reason computers have big heat sinks with fans on top).

You have a few options. You can look at going to switch-mode powersupplies, which are far more efficient (because they don't lower the voltage by turning the excess into heat). Alternatively, you can look at using multiple regulators in series - Use a regulator to go to 9V, then to 7V, then to 5V. This way, the load is spread out over multiple regulators and each has less power to dissipate (of course the total dissipated power is the same!). Finally, you can look at using external components like you already were doing. You mentioned having issues - I would suggest trying that topology with lower loads and a good amount of protection on the powersupply end to see what is going on - could be that you are getting some kind of instability.


Your power dissipation is caused by the amount of voltage you drop to get to 5V times your load current plus the whiff of current the regulator uses to do it's thing. The latter is usually negligible.

As long as you are using such a large voltage source you will have to deal with that power if you are using a linear regulator.

You pretty much have three choices as shown below in order of preference.

  1. Use a lower voltage source. If that 13.8V is coming from a transformer, buy a different transformer that is closer to the lowest voltage the regulator needs. Also, you can get better regulators called low dropout regulators that need even less input voltage.

  2. Build a switch mode regulator circuit. There are many devices and circuits out there these days that will convert voltages for you at high efficiencies. There is some cost in complexity and noise though.

  3. Split the power: Add in additional power diodes or power resistors before the regulator to soak up some of that heat. You may still need to heat-sink all that but you may find the heat-sink requirement is quite a bit less to keep the active component within tolerance. Of course this method does NOTHING to improve efficiency and so comes last in the list.

Note: The current splitting technique you show is really intended to allow you to use the regulator at higher currents than the regulator itself is designed to provide. That approach does not really help you in this situation.


I assume you need noisefree 5V signal, since you use a linear regulator. I would drop voltage with a decent quality buck converter (>85% efficient) to 6.3V, and feed that to a LM350 (3A) or 7805 (1.5A) - both drop a nominal 1.25V regardless.

As already mentioned above, it is pointless to dissipate and lose so much power. W_dissipate = ( Vin - Vout ) * I_load is pretty clear. I always use step down converters -> linear regulators if there's more than 2-3 volts difference, and my load is moderate.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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