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Summary: I am using a LM317 to drive a liquid pump that draws 1A. Unfortunately my regulator is dissipating way too much power and getting too hot. Currently my input voltage to the regulator is 12V and I'm trying to create 7V so that's 5Watts! So I bought a heat sink BUT I would also like to drop the input voltage of the regulator a little lower, no need for it to be 12V.

Goal: To drop the 12V input voltage to around 9V for the regulator by using a zener diode, that way the power dissipation is lowered to about 2W.

Question: 1) Could I get a sanity check on my circuit I created? 2) Do I need to worry about the current going through the zener diode because its 1A? 3) Could I just use a 3.3ohm resistor instead of a zener diode at all that way 3.3ohmd*1A is 3.3V drop which should do the same thing.

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    \$\begingroup\$ No matter what trickery you employ to reduce the power dissipated in the LM317, as long as you're using a linear circuit (no switch-mode stuff), that power will still get dissipated somewhere. So a 3.3V zener with 1A flowing through it will dissipate 3.3W and it will get hot. Same goes for a resistor. \$\endgroup\$
    – brhans
    Commented May 5, 2016 at 16:40
  • \$\begingroup\$ Yea but doesn't it help a little to spread out that dissipation? Instead of all 5Watts on the LM317. If I have a 5W zener than it should be fine with 3.3W dissipated there. \$\endgroup\$
    – Suh Dude
    Commented May 5, 2016 at 16:45
  • \$\begingroup\$ You could also daisy-chain two or more LM317s to lower the thermal load on each of them. \$\endgroup\$
    – JimmyB
    Commented May 5, 2016 at 17:23
  • \$\begingroup\$ Another option is to run the pump from a PWM (pulse-width modulated) speed controller. That would switch the full 12 V on and off at high frequency. By adjusting the pulse width you can vary the motor speed. \$\endgroup\$
    – Transistor
    Commented May 5, 2016 at 17:31
  • \$\begingroup\$ Could I use 2 zener diodes in parallel to split in half the Pd. \$\endgroup\$
    – Suh Dude
    Commented May 5, 2016 at 18:17

3 Answers 3

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In principle this will work- the zener would have to dissipate more than 3W, so it will have to be something like a 5W zener- not cheap. The heat has to go somewhere in a linear regulator, adding the zener will move part of it to the diode.

The resistor will work if the output current is a constant 1A current, however motors, especially those used in applications such as yours, tend to draw much higher current during starting so it may end up stalling your pump and burning up the resistor.

At this level, I strongly suggest you use a switching regulator (buck regulator), such as those based on the LM2596. It will run cool. Inexpensive modules are available from China which may or may not use genuine LM2596s, but seem to work well enough.

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  • \$\begingroup\$ I probably will end up changing regulators, its just I already bought it and this is my first circuit I've designed/built so seeing all these problems is really teaching me by trying to solve them instead of just replacing one part with another. I planned on using a 5W zener just need to find out how much they are. \$\endgroup\$
    – Suh Dude
    Commented May 5, 2016 at 16:15
  • \$\begingroup\$ Another possibility is to add some series diodes on the input. You can then adjust the voltage in increments of about 0.6V. The total amount of power to be dissipated remains the same. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 5, 2016 at 16:25
  • \$\begingroup\$ So you are saying instead of dropping all 3V over the diode and dissipating all 3Watts there, I could throw some forward biased diodes lets say 2 of them and drop 1.4 across those that way not all 3Watts are being dissipated in the zener? I would have to find a diode that can handle 1A flowing through it. This sounds like a good idea though. \$\endgroup\$
    – Suh Dude
    Commented May 5, 2016 at 16:30
  • \$\begingroup\$ I looked at the LM2596 datasheet and the Thermal Resistance Junction to Ambient is just as high as my LM317. It says Junction to ambient is 65C/W so if I have 5V drop and 1A then this will be 5Watts so 65*5=325C so it seems like the heat problem will be here too? Can you explain? \$\endgroup\$
    – Suh Dude
    Commented May 6, 2016 at 13:38
  • \$\begingroup\$ The math is different for a switching regulator. See the datasheet calculation- Pd~= Vin * Iq+(Vo/Vin)*Iout*Vce(sat). It will be around 700mW typically for 12V in 7V out @1A. So probably no heat sink at all is required. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 6, 2016 at 13:57
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Assuming that you are using the TO-220 LM317 without a heatsink, it has a junction to ambient temp rise of 19 degrees C / watt. Given you need to dissipate 5 watts, this would give a temperature rise of about 95 degrees C. With an ambient temperature of 21 degrees C that is 116 degrees (HOT). Since the part has a maximum operating temp of 125C, it MIGHT not burn up immediately but it is too close to the limit.

If you got a heatsink that is 5 degrees C / watt and add the junction to back thermal resistance of 3 degrees C / watt that would give 8 degrees C / watt and give a temperature rise of 40 degrees C + the 21 degrees C ambient gives a temperature of 61 degrees C (about 142 degrees F) which is much better for the part but still too hot to hold your hand on.

If you use the Zener diode as you suggest, as others have noted, you need a 4 or 5 watt Zener diode, for most of these parts the junction to lead thermal resistance is about 15 degrees C per watt (with .25" leads) which would have the diode at a 49.5 degree C temp rise which puts it's temperature at about 71.5 degrees assuming its mounted to a significant board that can sink heat well through the leads. The Zener should be fine at this temperature. For the LM317 in this instance, it will need to dissipate 1.7 watts which is a 32.3 degree C temp rise and should be about 53.3 degrees C.

I suggest that the better option is to get a proper heatsink or use a different method of voltage regulation (i.e. switching).

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  • \$\begingroup\$ I actually have a heat sink so I believe the heat sink AND dropping the voltage via zener diode should work well. I found a 5W zener for 50 cents on mouser. Not expensive at all. \$\endgroup\$
    – Suh Dude
    Commented May 6, 2016 at 3:12
  • \$\begingroup\$ Also the data sheet I am looking at for the lm317 has a junction to ambient temperature rise of like 65C/W idk what data sheet you are looking at? Maybe I should attach the one I'm looking at. \$\endgroup\$
    – Suh Dude
    Commented May 6, 2016 at 3:14
  • \$\begingroup\$ I'm looking at the Texas Instruments datasheet for the LM317. link The "KCS" package is the TO-220 in 7.4 Thermal Information. You may still have problem burning up the Zener diode unless it has some amount of heatsinking through the leads. The 5W 3.3V Zener diode datasheet has calculations for temperature rise link \$\endgroup\$
    – euler357
    Commented May 7, 2016 at 1:32
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Your reasoning is flawed because you are just moving part of the power dissipation to the zener which is much less tolerant of that kind of current. It will most likely melt in seconds, unless its a large 4W zener.

You are better off trying to use heatsinks to improve your regulator performance. Aluminium is great for heatsinking, and special heat paste for heat transfer. When you use heatsinks its all about surface area and airflow for best cooling.

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