Efficiently Regulate 12VDC Down to 3VDC

I am building a project that is powered by a 12VDC 1A wall adaptor. Within it is an digital voice recorder IC that will only run on 3VDC or less. It only consumes a few mA though.

What is the simplest and most efficient method for providing this lower voltage?

I had been using an LM7805, with two iN4007's added in series with its

5V output. The regulator is heatsinked but still gets predictably hot. I know this is not best practice.

Any suggestions would be appreciated. Prefer low cost, basic parts.

Thank you,

Kevin Foster

Reply to
Kevin Foster
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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.

Reply to
jurb6006

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.

Reply to
krw

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

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Reply to
Spehro Pefhany

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

Reply to
Johann Klammer

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

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

** The LM317 adjustable voltage regulator operates from 1.25V upwards. Same pak as the 7805 and needs two resistors to set the voltage at 3V.

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

You don't need anything fancier.

.... Phil

Reply to
Phil Allison

LM78L03

Whoa! how much is "a few"

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Reply to
Jasen Betts

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.

Reply to
Sylvia Else

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Reply to
sms

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

George H.

Reply to
George Herold

DK = Don't Know???

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Rick C
Reply to
rickman

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

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

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

Reply to
Johann Klammer

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.

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Tim Wescott 
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Reply to
Tim Wescott

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

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Reply to
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.

Reply to
jurb6006

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. .

Reply to
bloggs.fredbloggs.fred

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

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John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com
Reply to
John Larkin

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

Reply to
P E Schoen

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