Electronic – Voltage regulator waste too much energy

heatvoltage-regulator

I'm trying to obtain 5V DC from a 37V DC power supply. The load is a computer fan that at 5V drains 31mA (around 0.15W).
I'm using a LM350 voltage regulator (LM350 Datasheet).
I get a very precise 5V output but the chip becomes really hot. I measured that the the power supply provides 49mA at 37V. I think that the LM370 use 48mA – 31mA = 17mA, so it dissipates 37V * 17mA = 0.63W. (Please correct me if these calculations are wrong)
It bothers me that the voltage regulator requires more energy than the load. It's normal or I'm doing this in a stupid way? There is a better way to supply a computer fan from a 37V DC tension?

Thanks 🙂

P.s. On the datasheet it says that the maximum input-output voltage differential is 35V. I'm using it with a 32V voltage differential. Is it safe? I need a better voltage regulator?

Best Answer

You are correct in that the linear regulator appears to be using 17mA at 37V so that is a power of:

Vin * Iregulator
37V * 0.017A = 0.629W

But you need to also consider that the linear regulator that you are using is also dissipating power of:

(Vin - Vout) * Iload
(37V - 5V) * 0.031A = 0.992W

Thus you have a total power loss in the linear regulator of:

0.629W + 0.992W = 1.621W !!!

You really need to be using a switching buck regulator for this job. If you do not want to build such a thing you can find a plethora of ready made modules on web sites the like of Amazon or eBay. They do not cost a lot either.

enter image description here

enter image description here