Re: 24 to 12v converter


"Paul Weaver" <usenet@xxxxxxxxxxxx> wrote in message
news:1154768509.816664.102350@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Morse wrote:
If it's a linear regulator it has to be regulated, a purely resistive
dropper won't work because the current consumption of the appliance is
unknown. At the very least it'll be a zener + pass transistor based
affair.

Most devices I've seen are limited to arround 1A,

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

or even less, meaning
a resistance of a few ohms. A 100k resistor on each side isn't going to
be affected by a few ohms in parallel.


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.

http://www.aikenamps.com/AddingComponents.htm

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!

If it's a switching converter then it's almost certainly regulated by the
very nature of the vast majority of switching supplies.

If if if

For a simple low current appliance, CPC will
sell you a 100k resistor for 25p inc vat, Put two of those, and a
little solder, into your appliance to convert to 12V input.

How do you calculate that then? To calculate a resistance to drop a set
voltage one must know the load, and that load must be perfectly steady
and
fixed.

Not neccersary, it can vary between a few miliamps and an amp or so
with no real difference.


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!

If you are talking about using them as a potential divider to half the
voltage then that in theory will give exactly half of the supply voltage-
until you try to draw any current at all from it which will cause the
voltage to drop in proportion to the load.

Only if the load is high. As I said, it depends on the quality of the
device you're buying,

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 2.691V 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
quantity. 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.

and the device you're using.


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



.

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