converting a DC power supply to lower voltage and current - help!

Use an LM317 adjustable voltage regulator (50 cents). Don't use a dropping resistor to supply voltage. Your voltage will vary according to your load.

BRW

One of the things you _may_ be confused about is that when there are three variables involved, knowing two of them solves the other one exactly. Ohm's law is like that, too: E=I*R. Three variables. If you modify your power supply to put out 1.2 volts, that specifies one variable, E. Whatever you plan to connect to it, the "load," specifies another, R. The third is then a consequence of the other two and not something you normally design a power supply to limit. In other words, if you modify the output voltage and hook up your load to it, the current that flows just happens as a consequence and you don't need to do much about that except to make sure your power supply is up to delivering whatever is required.

Now, a lot of "loads" are dynamic, in the sense that they may try to pull varying amounts of current over time. In these cases, you need to make sure your modified 1.2 volt power supply is able to adapt and keep the voltage steady over varying load currents. This is what a power supply's usual job actually is and why people go to any trouble at all designing one. So if your load is a varying one, like a radio may be, then you really do need a "power supply design" of some kind to hold that 1.2 volts over a varying R.

On the other hand, if you know your load's R very well and it doesn't change enough to matter, then you could just compute a "dropping resistor" value and let it go at that. In that case, the result would look like:

,---------, | | (+) | R

Here are the specifics of what I'm trying to do:

I've got a hair trimmer with a rechargable battery in it. I'd like to replace the battery with a constant power supply that plugs into the wall. The battery is 1.2V and 600mA. The closest power supply i have is 3.3V and 3A.

It seems like the load (hair trimmer) should be fairly constant, though maybe it changes based on whether it's cutting or not.

Anyway, thanks again for your input. If this helps clarify the best solution, I would love to hear it!

Thanks much - Eric

Well said, Jon. Eric - Are you sure your battery says 600 mA (current), or does it say

600 mAh (capacity)? I've never seen a battery with a current rating stamped on it, but almost all rechargable batteries have a capacity stamped on them (mAh). A battery may have a 600 mAh capacity, but it could possibly supply much more than 600 mA of current. You should measure your actual current draw.

BRW

BRW -

I think you're right. It's probably stamped 600mAh capacity. Unfortunately, the battery doesn't charge any longer, though i could probably put a 1.5V AA battery in to measure the current draw, right? Or can I go ahead and drop the voltage with the variable resistor and see if the supply can properly regulate the current?

Eric

It probably won't hold the voltage, anyway. The load, if you are using the trimmer, will change appreciably and so will the voltage being supplied, as a consequence of that.

Some suggestions (many others exist):

(1) You might be able to find a supplier of the exact battery on the web. I've found some cheap sources, with some effort, for some fairly odd-ball NiCad contraptions. You might be lucky, too. This assumes you still have some device for recharging the trimmer.

(2) Just try what you are talking about. Measure the current by using an alkaline D cell (no reason to test this with a AA -- you want the best voltage stability you can get cheap and easy.) Just check it quickly, first, to make sure the extra few tenths of a volt are okay (probably just fine.) Then go for longer and measure. Then measure it again under some reasonable load (cut some hair.) My wild guess is that the load current will about double, at most. Probably less than that, though. In any case, if you plan to just use a dropping resistor in spite of all this, you will either have too much voltage without a load or too little voltage under load. Whether that works out for you will depend. But you can, of course, experiment and see.

(3) Make a proper 1.2 volt power supply out of your current 3.3 volt one or else make up a new one.

(4) Make your own custom battery pack and wire it up. Use a D-cell NiCad and get a charger for it and add some wire to connect to the trimmer. Or use a D-cell alkaline, if that works fine. This isn't any more cumbersome than some modified AC-to-3.3V-to-1.2V supply and you can get standard, cheap batteries and use standard, cheap rechargers (if you get NiCads.)

Jon

You could use some diodes. Silicon gives 0.7V drop for each and 3 of them is

2.1V which 3.2V - 2.1V = 1.2V which is exactly what you need. (well ultimately its a good match)

The issue here is power dissipation but if your at 600mA then thats 0.7V*1A ~ 700mW or so max. You should be able to find some diodes that can handle this quite easily or if your worried you could run several in parallel to split the power dissipation(although there are a few problems with this I think it would be ok... just not to many because then you limit the current through each diode and it might not reach your 0.7V).

What I would do if I needed a quick and dirty solution would be to but 3 1W Si diodes in series with the supply. Then I would test it on a constant load like a resistor(you'll need something like 20-50ohmz for a decent test) then I would test it on the device and see how it performs. Ultimately I was worried about ruining the device I wouldn't do this but use a regulator... chances are though it will be ok.

Jon

Thanks for all of your help. This is my plan...

As I'd like to avoid rechargable batteries, i'll try the following: I've got a corded shaver (probably a different motor) that i'll look at to get a vague idea of the current draw (as i think you were right about the 600mA being 600mAhours of storage). I'll then try either a resistor or the tripple diode solution - maybe both to see which works best.

again - thanks for your help - i'll post back once i've tried these options.

best - eric

On Jul 10, 3:01 pm, snipped-for-privacy@yahoo.com wrote: > BRW - >

So do what I did. Get 2 1000mA NiCad cells with solder tabs and just replace the bad cells. I did that 3 years ago and it's going strong. The higher capacity cells require less frequent charging as well.

GG

Replace the battery with a 1200 mAh rated NiCd, then wire up this circuit to the trimmer:

----- Vcc ---+---Vin|LM317|Vout---+ | ----- | [.1uF] | [47R] | | | | +----------+----> + to trimmer Gnd ---+-------------------------> Gnd to trimmer

That will provide a constant current to the trimmer of ~ 26 mA. The NiCd battery will regulate the voltage to ~ 1.2 volts. You will need to use a 5 volt or higher supply - the closer to 5 volts the better. Leave it plugged in all the time. The setup will charge the battery very slowly - a bit less than

2 1/2 days for a full charge. Because it charges slowly, it can be left plugged in all the time.

If you make the connection from the circuit to the trimmer detachable, you'll have the best of both worlds - a cord free trimmer when you want it that way, and a corded trimmer when power is available.

If you can't get a battery that will fit into the trimmer, you can use the same circuit with an external battery. You do not want to use this circuit without a battery.

Ed