Monday, February 26, 2007


Well, appears I was right. My 5 Volt power regulator is failing. I dont think its a dud or anything, I just think its being overloaded. I wonder if I attach a heatsink if it would last longer. I dont know if its simply getting too hot, or if theres too much current going through it. So what makes a good heatsink anyway? Getting a bigger/better power regulator might solve it for now, but I need to have a better solution at somepoint.

I did consider making a stepdown via resistors, and then feeding that into the power regulator, but I suspect the resistors will just heat up. Anyone any ideas?


Levente said...

Hi Mike!

You can verify if the power regulator is the root of those problems... just measure the current drawn by the circuit (connect a multimeter in series between the output of the power regulator and the input of the circuit). LEDs generally draw a lot of current; multiple LEDs can mean too much current for a small(er) linear power regulator. If the current drawn by the circuit is close to the rating of the power regulator chip (you can look that up in its datasheet), you can start suspecting this. ...Second, most of these power regulators feature overheat-protection; they will regulate the output current "down" so that they stop heating up even more. A heatsink can be of great help if this is the case.

To eliminate most of the heat problems and all that stuff, most people use switching mode power supply chips (stepdown voltage regulators in this case) that provide relatively good power efficiency, utilitizing a coil + capacitors as external passive parts. With that, they also avoid having to get rid of (well, most of) the heat, as the voltage drop isn't dissipated, unlike it's done with using linear power regulators. (These chips are a little more expensive and require a little more caution than their linear counterparts, though).

A series resistor (as you suggested) may help dissipating the voltage drop; if you run the circuit from 12v (battery) and the circuit requires 5v inside, the regulator must drop 7v. If you measure the current drawn by the circuit, you can make a guess on the average resistance of the regulator in operation (R = U/I), and then select a power resistor of a suitable value, so that the voltage drop (thus the power dissipation) will be shared between the resistor and the regulator. Say, let it drop 3-4 volts, so that the regulator still has some margins in voltage. (Note that this would somewhat also limit the operational current range of the regulation... too high currents won't be possible to draw anymore, but that would probably also be unintended).

And good luck with experimenting! :)

Mike said...

Cheers Levente, yep...I've still been playing with it a little. I'm not certain that it can handle to voltage and the current but its just getting too hot..

But I'll possibly try either a proper PSU or some resistors to see what happens first.