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All,

Cant seem to find something simple. Is there a website where i can order a chip or small device which will change 14,4 dc to 8v dc. Output for 50mW is enough.

Or does anyone has something lying around or can make it? If it costs something thats ok ofcourse. So if anyone knows, thanks

// Erik (binkey.nl)

You can connect 4 or 5 diodes in serial to reduce it to 8 V.

Well like i said.. i'am not electro expert. Last time i did something like that is 15 years ago. Dont know if it matters, but the power is from a battery accu. So if you do something to desolve the power so battery life will drain its not so good. So an url to some chip which can do the trick, and then what needs to be done with it would be best.

// Erik (binkey.nl)

That can easy be done with a 7808, just go to Conrad and search for 7808. They come in all variations.


pin setting
left: IN
mid: ground (0)
right: OUT 8 volt

Ah, great.. i'am not a wizard with those things. I ordered a few + some other crap to at least get free postage. (otherwise its 8 euro's)

Now all i need to do is find my multimeter.. its suppose to be somewhere here.. then i can check if i connected it correctly. (i ordered 5 so i can screw up a few times)

// Erik (binkey.nl)

They're a brilian solution for all your quick power needs, and like said: They come in a great variety of voltages and currents. When using them, just obey these simple rules:
<ul><li>Never let the voltage on the output exceed the voltage in the input by more that 1 volt. This may seem unlikely, but it frequently occurs when you're using large output capacities and the regulator will die for sure. Just add a diode between output and input to prevent this situation.</li><li>Make sure your input voltage is at least 2 volts higher that the required output voltage. It needs this bandwith to regulate properly.</li><li>Use two tantallum capacitors of about 100nF each between input/ground and output/ground to reduce regulation noise. Keep them as close to the regulator as possible (short leads).</li><li>Use a capacitor on the input (behind the rectifier) large enough to keep the dips on the ripple above the 2 volts as mentioned in point 2, even under maximum load.</li><li>Use a capacitor on the output of about half of the capacity on the input.</li><li>Cool well.</li></ul>
Good luck!

you just blown away my "this is going to be easy" idea.. i thought i just use that what pieter showed, and i'am done. O well i can always experiment a little bit. I did order a heat disappater (or however you call it) to go with it. Its going to be mounted on the outside, so it should have enough cooling by itself then almost.

points 1,2 check. points 3 and 4 are those really needed? In the case that the power drops for some reason will it brake down? The device behind it is a small camera.. if it has less power/voltage it will just stop working i gues.

// Erik (binkey.nl)

Didn't mean to scare you. But it's still really simple, don't worry. Two tantallums, two capacitors and a diode is all it takes to be perfectly safe in any condition. Use a 4700uF capacitor at the input and a 470uF at the output and you cannot go wrong.

Try it out first I think the UA7808 will work good.

Ok, i will first play a bit

// Erik (binkey.nl)