Efficiently Regulate 12VDC Down to 3VDC

A switcher, PWM and a coil and cap.

As cheap as possible ? Find an old PC motherboard. The coil will be right there.

Also, why do you have to pump 12 volts into it ? They got walwarts that put out 5 volts.

Define "a few mA".

Probably a power supply module but it really depends on what "a few mA" means.

I may be but if it's getting hot, you're talking about more than a few mA. 9V x 3mA = 27mW. Not much.

A switching power supply module? Measure the current, though. Without that, you're shooting in the dark.

If you're not in a hurry buy one of those switching supply modules and adjust the output to be 3V before hooking up your chip. You can get them with a voltmeter built in for less than the cost of a chip and coil.

--sp

I think recom had those 78xx footprint switchers popular with hobbyists Don't know if they're any good... never used them...

I tested the CUI 5-volt equivalent, P7805-S. Seemed to work fine. I think they have a 3.3 volt version.

Idle current is very low and needs about 3mA load to work properly.

You don't need anything fancier.

.... Phil

LM78L03

Whoa! how much is "a few"

I wouldn't expect a 7805 to get noticeably hot with only a few milli-amp load when delivering 5V from a 12V supply.

Sylvia.

That was my suggestion. There are butt loads of DC-DC converter modules on DK these days.

George H.

DK = Don't Know???

Use the Five Second Rule: if you can press your finger against a part for 5 seconds, it's not too hot.

once the nerve endings are charred I can go for a lot longer than 5 seconds...

There's TI versions, too. It's been about a decade since I had occasion to use one, but when I did it was trouble-free.

How much is actually "a few mA"?

Simplest would probably be an LM317. Depending on your "a few mA", a switcher might actually draw more power.

Posting a schematic on dropbox, or wherever, would allow us to help you more "efficiently" ;-) ...Jim Thompson

more "efficiently" ;-) "

Incidentally Dropbox says they no longer render in a web browser. First of all I don't see how they can prevent it and second of all it is not a selli ng point if I ever decide to pay for it. So far the free space has been ple nty but now it looks like I am gong to have to go with something else, mayb e even real hosting.

If by efficiency you mean least power dissipated in regulator then a simple power resistor in series with an adjustable will do the trick, like so:

Please view in a fixed-width font such as Courier.

. . . R limited adj reg . LM317 . ------- . | | . 12V Input >-[Rs]--+-- |IN OUT|----+------+-->3V . |. | ADJ | | | . | ------- [270] | . | | | 10u| . 0.1u --- --------+ --- . --- | --- . | [390] | . | | | . | | | . --- --- --- . com com . . . . . Imax= maximum load current + 5mA , 5mA is regulator bias. . . Vmin= minimum voltage required across regulator . . Pmax= maximum power dissipated by regulator . . . Then maximum allowable Rs to maintain proper operation . . of regulator at maximum load current, Imax is: . . Vinput- Voutput- Vmin . Rs= ---------------------- . Imax . . 2 . 1 (Vinput-Voutput) . Pmax= - x ---------------- . 4 Rs . . . and this occurs at load current . . . 1 Vinput-Voutput . I= - x ------------- . 2 Rs . . . . Plugging in some numbers: . . . Vinput = 12V Voutput=3V Imax=100mA Vmin=1.5V . . 12-3-1.5 . Rs = -------- = 75R . 0.1 . . . then . 2 2 . 1 (Vinput-Voutput) 1 ( 12 - 3 ) . Pmax= - x ---------------- = - x -------------- . 4 Rs 4 75 . . . . = 0.27W this is a 38oC junction temp rise to free air . . . 2 2 . Power dissipation in Rs is Imax x Rs= (0.1 ) x 75 =3/4W use . . 2 W 75R for Rs. . . , . . Compare to circuit without Rs, Pmax= Imax x (12 -3 )= 0.9W . . which is a 125oC temp rise and probable regulator shutdown. .

I can touch 50C forever, and 60C for about 1 second, and I can interpolate pretty accurately between.

The wall wart probably is unregulated and putting out as much as 18V lightly loaded. The LM7805 should handle 1 watt even without heat sink (65C per watt - which is pretty hot) so about 1/2 watt should be about 30C above ambient. So I = 0.5/(18-5) = 38 mA. A modest heat sink should allow well over 1 watt before it gets too hot to hold.

Paul