I am trying to convert 24-30V input to 24V 3A regulated output. I tried using ti's 5A adjustable regulator LM338 but I faced heatsink problem which seems like impossible to keep cool while trying to supply that much. Do I have to buy a DC/DC converter or is there any other 'diy' way of doing it?

This regulator is going to be used for powering a water pump which requires 24V 3 Amps.

Heat dissipation calculation:

lm338 has a characteristic of 'Junction to Case, RθJC(top) 15.7°C/W'

The load will be dissipating 24v*3A=72 Watts of power, and the regulator will be dissipating (30v-24v)*3A = 18 Watts of power. So total is 90 Watts.

90*15.7= 1413°C seems too much isn't it?

  • 1
    \$\begingroup\$ Was LM338 TO3 or TO220? Input-to-output voltage differential is a minimum of 3V. So you'd need 27V minimum for the regulator to work. \$\endgroup\$ Mar 21, 2017 at 17:49
  • \$\begingroup\$ It was TO220 but power dissipation was so high that it shuts down automatically \$\endgroup\$
    – Alian4life
    Mar 21, 2017 at 18:09
  • 1
    \$\begingroup\$ 18W * 22.9 = 412.2 deg Celsius and maximum = 260. Junction to ambient. Add ambient temperature. Decrease voltage to 27V. 3V means 1/2 of load, so it will work. Load dissipation is not part of calculation. Personally, I'd go with a DC/DC convertor. \$\endgroup\$ Mar 21, 2017 at 18:15
  • 1
    \$\begingroup\$ +1 for StainlessSteelRat 's answer. Just to correct your calculation: Junction-to-Ambient thermal resistance, \$R_{th(j-a)}\$, is more meaningful because it gives you directly the felt temperature of the heating element. Assuming the device has TO-220 package, total temperature rise will be \$\Delta T = (V_i-V_o) \cdot I_{LOAD} \cdot R_{th(j-a)} = 6V \cdot 3A \cdot 22.9°C/W = 412°C\$ and the felt temperature is \$T_f = \Delta T + T_{amb} = 436°C\$. Of course this is too much. Go for a DC/DC converter or, as SSR stated, decrease the input voltage. \$\endgroup\$ Mar 21, 2017 at 18:30
  • \$\begingroup\$ Thanks for correction. So it seems like I would definitely need a converter. Can't I parallel 2*24V 1.5A regulators and some heatsink on them for that half amount of heat to take care of? \$\endgroup\$
    – Alian4life
    Mar 21, 2017 at 18:52

3 Answers 3


You are partially correct with your calculations. The load power has nothing to do with the power dissipation of the LM338.

30V - 24V = 6V

6V \$\times\$ 3A = 18W

$$P_D = \frac {T_{J\ Max} - T_A}{\theta_{JA}} $$ $$T_{J\ Max} = P_D \times \theta_{JA} + T_A $$ $$18W \times 22.9^{\circ}C/W + 20^{\circ}C = 412.2^{\circ}C $$

Assuming ambient temperature is \$20^{\circ}\$. The maximum for the TO220 package is \$260^{\circ}C\$ (at least the one I reference). So thermal shutdown.

If you decrease input voltage to 27V. You meet the minimum Input-to-output voltage differential. The LM338 will not work below 27V to provide 24V.

To minimize losses and to keep parts cooler, it is best to operate linear regulators at the minimum input voltage.

27V - 24V = 3V

3V \$\times\$ 3A = 9W

$$9W \times 22.9^{\circ}C/W + 20^{\circ}C = 226.1^{\circ}$$

Below \$260^{\circ}C\$, so this should work with a heatsink. It will be hot and waste power, but it will work.

LM338 datasheet


The real question is: why the hell do you want to feed regulated DC to a motor?....

If it needs to be run at a constant RPM, use a PWM speed controller with a speed sensor.

If it will overheat and burn at 30V (but not at 24V) use a PWM to keep the average voltage on the motor at 24V.

Most likely the motor will run just fine on 24 - 30VDC, and you don't need any regulation at all.

LM338 has 2-3V dropout voltage anyway, so with 24V in, you'll only get 22V out maximum.


You are using entirely the wrong thermal resistance. You will not be connecting the top of the case to a heat sink. Instead, look at the number for the bottom of the case (the metal tab). You'll notice a number of 0.7 deg/W. Add another 1 deg/W for a good heat sink, and you'll get a temperature rise of of about (1.7 x 18), or 30 degrees above ambient.

This brings up a question which you have not addressed: exactly what sort of heat sink were you using when you found it "impossible to keep cool"? If you used a pcb heat sink such as enter image description here

well, there's your problem. You need something a lot beefier. That particular heatsink has a thermal resistance of about 24 deg/W under natural convection.

Instead, you need something like this bad boy which has 1 deg/W. Of course, it's 4 1/4 wide by 5 1/2 long, and will run you more than 20 bucks new, but that's what it takes. You can, of course, go with a smaller heatsink and get a bigger rise, or you can mount a fan to it to greatly improve airflow and cooling.

But really, 20 watts on a TO220 is no big deal.


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.