View Full Version : Reducing 12v DC to 5v DC
Inconel_601
Dec 05, 2005, 09:01 AM
Whats the most efficient, best way to reduce 12v DC to 5v DC? I'm using a 7805 regulator right now and it seems to heat up after prolonged use. The circuit is powered by a car battery in standby mode (5v PIC circuit) and ocassionally drives some switching relays (150 ma consumption). Any better more efficient regulators out there? Any ideas appreciated. Thanks.
vintage1
Dec 05, 2005, 10:20 AM
yep. Buy a smartBEC or equivalent switching regulator.
Mr.RC-CAM
Dec 05, 2005, 11:36 AM
Some practical info on basic efficiency of VRegs: http://www.rc-cam.com/dc-dc.htm
Dimensional Engineering, a sponsor here, has a few affordable switching Vregs. And the UBEC series are popular too.
Afroman
Dec 05, 2005, 07:03 PM
How much current do you need to handle? If a 7805 worked even semi-well, a DE-SW050 will do the job perfectly. It is a direct drop in replacement for the 7805, and is the cheapest 5V switching reg we have.
http://www.dimensionengineering.com/DE-SW0XX.htm
(Btw I own half the company so anything I say will be biased)
Johan B
Dec 06, 2005, 10:47 AM
I'm not convinced you need a higher efficiency converter.
By definition, the 5V regulator is about 40% efficient, given a 12V source (5/12 X 100). The only problem you mention is the heat.
3 ways to lessen the heat:
Put on a heatsink. Even a clip-on one should make a lot of difference.
Put a fan on it.
Put a resistor between the 12V supply from the battery and the input pin of the 7805.
Sizing a resistor:
I assume that your battery can drop to 10 V minimum when almost totally drained, that the regulator needs 7.5 volts to maintain line regulation (got that from an online spec sheet) and that you need to be able to maintain a current of 500 mA (3 relays plus 50 mA for the circuitry).
Given that, you can put on a 5 ohm resistor, which will drop the input voltage by 2.5 volts when your circuit is pulling 500 mA. Power converted to heat in the resistor at this max load will be 1.25 watt (2.5 X 0.5), so a 2 watt resistor should do. cheapest is probably two 10 ohm, 1 watt resistors in series.
At maximum heat wastage (12.5v supply, 500 mA), this would result in the 7805 only generating two thirds as much heat. rest of the heat is generated in the resistors. If you want to relax the requirements a bit, you can up the resistor ohms.
On the other hand, if the 7805 doesn't get too hot to touch, it should be fine.
This information is worth every cent you paid for it. It comes from a guy who spent a couple of days in fourth year electronics engineering nursing a transistor-shaped blister on his circuit-heat-testing finger. :rolleyes:
Johan Beyers
Eflight 101
ZAGNUT
Dec 06, 2005, 01:32 PM
if your circuit spends most of it's time just supplying a few mA to a PIC and only ocassionaly draws 150 mA then i wouldn't bother with a switching regulator. just add a heat sink if you think it's getting too hot.
dave
siberia37
Dec 06, 2005, 05:56 PM
Ok so who has suggestions for reducing from 12V->5V at 2.5 Amps? Any easy circuits out there, or better yet a place to buy a device that does it? This isn't for an RC application so it doesn't have to be small or light.
robe_uk
Dec 06, 2005, 07:47 PM
Ok so who has suggestions for reducing from 12V->5V at 2.5 Amps? Any easy circuits out there, or better yet a place to buy a device that does it? This isn't for an RC application so it doesn't have to be small or light.
Hi
You could try mic range in this thread, they had good spec's. or even the first reg mentioned as you are below the 14v cut off.
http://www.rcgroups.com/forums/showthread.php?t=431990
Acetronics
Dec 07, 2005, 04:42 AM
Whats the most efficient, best way to reduce 12v DC to 5v DC? I'm using a 7805 regulator right now and it seems to heat up after prolonged use. The circuit is powered by a car battery in standby mode (5v PIC circuit) and ocassionally drives some switching relays (150 ma consumption). Any better more efficient regulators out there? Any ideas appreciated. Thanks.
Hi,Siberia
:rolleyes: THE Best way would be to use 12v relays or loads ... powered directly from the car battery. ( the mod. is no so hard, even on an existing circuit !!! )
A small 78L05 would then be sufficient for a PIC and a couple of Led's ...
Without any change in the circuit, a switching supply, based on an inexpensive LM 2575, is the good solution.
See National LM2575 datasheet for details AND Calculations ...
Alain
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