Looking for a voltage converter

[seeds]

I'm trying to find a 24vdc to 12vdc converter to drive a 12vdc sensor on a 24vdc craft. Does anyone know where I can find one? It only needs to supply about 1-3 amps at the most, but everything I find is either really cheaply made, or way too expensive. Thanks.

[Tim B]

You want an LM7812 voltage regulator and a plastic box. Every electronics distributor will stock it. Search google for datasheets and wiring diagrams.

Cheers,

Tim B.

[colinstone]

Concur LM7812 2A plus 0.1uf and 0.47 uf capacitors across supply and feed legs. STM website has full manual. I'm using several for all sorts of applications on a mainly 24v DC vessel - 12v solenoid valves, 2 speed PC type fans (12v and 24v) and a cctv camera.

[seeds]

Thanks to both of you. We're going to give this a try and I'll let you know how it works out.

-Justin

Just one more question. Which STM website? I did a google search and came up with a bunch...lol.

[Tim B]

The capacitors are only there for ripple-rejection, which (unless you are using them to power logic circuits) is generally unnecessary. Suppliers like RS and Farnell will stock them and have data-sheets. Their pretty simple, three connections: in, out and ground.

Cheers,

Tim B.

[marshmat]

FYI, here is the technical data sheet on the LM7812 (this particular version from Bay Linear). Looks like a suitable product for the application being discussed here, and at a much better price point than the more efficient, but more complicated buck-down converters.

[BillyDoc]

Um . . . I'm afraid I have to differ with the consensus here, it really isn't that simple. There is the matter of heat to dissipate, and the original post mentions up to 3 amps of current. The LM7812 is rated at 1.5 amps, tops.

Assuming you do only draw 1.5 amps, if you are putting 24 volts in and taking 12 volts out that means that the LM7812 is dissipating as heat the 12 volts at 1.5 amps it is handling. Eighteen watts, in other words - - - which would be very difficult to cool and wasteful besides.

On the other hand, switching regulators have come a long way. Check out the pdf below for an example of a 5 amp capable, 92% efficient, regulator that is also very easy to design and costs less than $20 by the time you buy all the components. That 92% efficiency means that only 8% of the input current is dissipated as heat, making the thermal part of the design much simpler.

But before going to this higher current approach it would be a very good idea to actually measure the current required by the sensor to be powered. Most sensors use very little, more like 1 to 3 milliAmps, and it could be that the LM 7812 is perfect for the job.

BillyDoc

[Tim B]

Sorry, billy, I think you're maths is a little flawed there.

An LM7812 won't dissipate 18W at 12V,1.5A. At the very most it will produce (Vin-Vout)*Current Watts. Dissipating the heat is not a huge problem. but you'll need a heatsink.

There are two solutions to getting the current-rating, either use several LM7812s in parallel, or use the very similar TS1084, which will cope with 5 Amps.

Cheers,

Tim B.

[BillyDoc]

Originally Posted by Tim B

Sorry, billy, I think you're maths is a little flawed there.

An LM7812 won't dissipate 18W at 12V,1.5A. At the very most it will produce (Vin-Vout)*Current Watts.

Tim B.

As I understand the problem, Tim, the Vin is 24, the Vout is 12 and the Current is 1.5. So, (24-12)*1.5 as you say - - - which, I think, is 18. I admit, however, to being very bad at math. If you would show me where I went wrong I would appreciate it!

Simply paralleling multiple 7812s brings up the same issue as paralleling multiple bipolar transistors to increase current capacity, as the output stage of these devices is, in fact, a bipolar. The transistors are never exactly matched so one of them takes most of the power, gets hot, and runs away into breakdown. You can get away with it using the LM78xx series because there is a 0.5 ohm resistor in series with the emitter of the output transistor. This resistor will compensate for the inevitable mismatches between units. It is NOT good practice, however. A much better plan is to use a separate power transistor to take the current that is controlled by a single 78xx, and the circuit to do this is provided by most data sheets for the device.

As for dissipating 18 watts of power concentrated in a TO220 package, sure you can do it . . . but it isn't easy by any means! Huge heatsinks and fans will do it.

Fundamentally, though, I think this entire discussion is moot because I can't believe that any sensor draws 1 to 3 amps. I think that seeds really meant 1 to 3 milli Amps . . . which a LM7812 can handle nicely in the TO220 package. (24-12)*0.003=0.036 Watts - - - right?

BillyDoc

[Tim B]

yes, you're right, 24-12 is 12!

Tim B.