24 to 12v converter 442

What? Twelve watts?! I can think of plenty of automotive accessories which consume more power than that.

Eh? Of course it will! Adding say 12 Ohms in parallel with 100K Ohms will reduce the overall resistance to less than the lowest resistance. You appear to be confusing series and parallel, though with respect your logic still doesn't add up. Have a look at Ohm's law on the web and you'll see that to calculate parallel resistances you don't add them together as you'd do if they're in series.

It's calculated thus: 1-R = 1-R1 + 1-R2 + 1-R3 etc.

As soon as you add an appliance to the equation (effectively a parallel resistance with the lower PD resistor) you end up completely upsetting the potential divider, rendering the lower resistor redundant due to its high resistance and the appliance's much lower resistance.

A 12 Watt device will have a 'resistance' of 12 Ohms (12V into 12 Ohms causes a current of 1 Amp to flow, 12V*1A=12 watts). So you effectively have the 24V supply going through a 100K Ohm resistor and through the combined parallel resistance of the other 100K resistor and the 12 Ohms of the appliance to ground. The combined paralled resistance of the appliance and the 100K resistor is slightly less than and close enough to 12 Ohms to ignore the 100K resistor for our calculations.

So, we now effectively have a potential divider with 100K at the top and 12 Ohms at the bottom. The voltage across the lower part of the divider and hence the appliance? 2.88 millivolts if my calculations are correct!

I would love to know where you get that from! Why just up to one amp? What happens when you get past the one amp barrier? If you could give me some idea how you arrive at your calculations it would be helpful.

You cannot draw an amp from a 2 * 100KOhm potential divider supplied by 24V- it's physically impossible. You can't alter Ohm's law! The maximum current can only be 24-100000= 240 microamps and that's shorting out the bottom resistor!

What on earth has that got to do with anything? *Any* audio device, for example, draws a varying current, from £5 computer speakers to top of the range hifi devices. In fact, I can't think of many of any electronic gadgets that draw a fixed, constant current all the time.

OK, I've had a few drinks, so hopefully I won't screw this up...;-)

A 1 Watt 12V device will have an approximate 'resistance' (not really the correct term, but let's not get complicated!) of 144 Ohms and draw 83 mA. So now the potential divider has 100K Ohms at the top and slightly less than 144 Ohms at the bottom. That will give a whopping (approx) 35 millivolts across the lower part of the divider and hence the appliance with the remainder (23.965V) across the upper resistor.

Now for a 100mW device. It will draw approx 8.3mA and have a 'resistance' of around 1445 Ohms. Now we have 100K Ohms on the top and 1424 Ohms ( 1445 Ohms paralleled with 100000 Ohms) on the bottom. That will give a voltage of just under 337 millivolts!

Now a 1mW device. It will draw approx 0.83 mA and have a 'resistance' of 14,450 Ohms . Now we have 100K Ohms on the top and 12626 Ohms on the bottom. That will give only 2.68 plus 11V across bottom resistor and the appliance, and the rest across the top resistor.

Anyway, hopefully you get the jist of it. You cannot get a constant voltage from a potential divider when there is a load which varies or is an unknown quanbreasty. It's simply not possible. Nor can you calculate the potential divider without knowing *precisely* the load to be connected.

Therefore building a voltage dropper based on nothing more than two 100K Ohm resistors is not only impractical, it wouldn't work with virtually any device you care connect to it, as I have demonstrated above.

The only thing I can think of that could run off a 100K Ohm potential divider would be something that draws fairly constant current in the microamp range like an LCD digitial watch. Probably around 99.99% of devices will fail to work.

Dave