I'd like to get 3.3V from a 12Vdc input. The load current is <0.5A. (The 3.3V output is used to power a little motor). I was thinking of using the LM317 because it's cheap! However, I'm concerned that it would generate a lot of heat.
Would I benefit a lot by using a low dropout voltage regulator instead? These regulator are a lot more expensive.
The answer is: NO ..
You wouldn't be better off if you go for LDO regulators ..
The difference between an LDO regulator and the LM317 is that the 317 regulator needs the Vin to be at least 3V higher then the Vo, whereas in LDOs the difference is sometimes <0.5V or lower, but both are linear regulators and both will have to convert (Vin-Vo) * Iload into heat ..
If you really want to benefit, go for a simple switching regulator ..
Here is one example: https://www.onsemi.com/pub/Collateral/LM2575-D.PDF
The switching regulator seems to be an expensive solution with lots of external components.
I tested my circuit with the LM317. The regulator gets super hot even thought I have a heat sink on it. I guess it must be because I am trying to drop 12V down to 3V.
Is there a better method for doing this?
If you don't want to use switching regulators and would like to "stick" with the LM317 then try to drop some volts accross external power resistor ..
If the LM317 "sees" instead of 12V, say, 7V dc it won't be as hot as it was before ..
I'd like to get 3.3V from a 12Vdc input. The load current is <0.5A. (The 3.3V output is used to power a little motor). I was thinking of using the LM317 because it's cheap! However, I'm concerned that it would generate a lot of heat.
At 0.5A you will be be burning off about 4.5W. You can easily find a heatsink that can dissipate 4-5W in still air. That would still be cheaper than a swtching regulator.
One thing you may try that is very simple is to use voltage divider and than use lm 317. in other words divide 12V into two 6V with resistors, and than attach lm 317
Added after 12 seconds:
One thing you may try that is very simple is to use voltage divider and than use lm 317. in other words divide 12V into two 6V with resistors, and than attach lm 317
Thanks all for the suggestions!
I think that I will try out IanP's power resistor first, because that sounds simple. While I'm at it, I'll also add in a diode to drop an extra 0.7V.
I just did a search for power resistors. My local electronic store has some 3W metal film, metal oxide film, and wirewound resistors. Would those work or do I need actual "power resistors"?
All resistors have a power dissipation rating in Watts.
The description POWER Resistors is usually used for those of above the common value of 0.25W - dissipation values above this rating are normally noted on schematics.
In your case: reasonable voltage across LM317 is 2v7 (minimun spec 2v0) for it to work correctly.
So, required volt drop = 12v - (3v3 + 2v7) = 6v0
Power to disipate in resistor= V x I (dc) = 6v0 x 0.5A = 3W0
Resistance to drop 6v0 at 0.5A = V/I = 6v0 ÷ 0.5A = 12Ω0
So a resistor of 12Ω (5w rating) will drop 6v0 at 500mA and dissipate 3W
Your LM317 will dissipate 2v7 x 0.5A = 5W4 at an output of 3v3
Hi,
I think LM317 not good way for generating 3.3V from 12VDC.
You can use LM2575M-3.3 or LM2575M-ADJ. Datasheet at www.national.com/pf/LM/LM2575.html. This Switching Regurator Power Suply.