On Monday (01/18/2010 at 07:47PM +0000), Tony Duell wrote:
Yes. There is a 15 ohm, 2W resistor from the input to the output of VR-1,
a 7805 regulator.
OK. Let's analyse that circuit. A quick look at a 7805 data sheet will
convince you it can only pull the output toiwards the input voltage, not
towards ground. So if the output is already higher than 5V (in this
case), the 7805 wioll do nothing.
15 ohm, 2W is what my original schematics say. It's entirely possible they
changed this value over time. If need be, I can take the lid off the
machine and see what is really in there... but I think we've converged
on the basic idea.
Consider the resistor on its own. The OP said there
was an input voltage
of 11.5V. SO there's 6.5V (11.5-5) to be dropped between the input and
output -- without the 7805, that means the resistor is droping 6.5V and
carrying 6.5V/15R = 0.4333A.
If the load drwas less than 0.4333A, the output voltage will be higher
than 5V. If it drwas more, then part of that current will be carried by
the 7805, and the output will remain at 5V (without the 7805, of course,
the output votlage would be less than 5V in this case).
Is it fair (as in rational, smart, non-stupid) to put multiple 7805 in
parallel? Can't say I've ever seen that done. Only the more typical
design of adding a pass transistor to carry more current.
Light bulbs are an approximatley constant-current
device. A 12V car bulb
that draws 0.433A (at 12V) is a 5.2W bulb. If the load is significantly
less than that, then of course the output voltage will be too high.
Danger Will Robinson.
I could understand mentioning Heath Robinson here...
Yes.. of course! That works too.