# Multiple Voltage Sources?

Discussion in 'Electronic Basics' started by FyberOptic, Jul 10, 2006.

1. ### FyberOpticGuest

Hiya folks, I've been trying to understand more about circuits and
electricity in general, and I've been left confused about a few things.
My brain has been trying to think of it all in a strictly procedural
way as from a programmer's standpoint, but obviously electricity isn't
quite so linear/predictable. So for the moment, I'll stick to a
particular subject for my question askery, otherwise this would end up
being a huge post.

Basically, I'm curious what happens when two (or more) seperate voltage
sources are available in a circuit, and how they interact. Say you
have a chip which runs off of 5v, an IC of some sort, and you want to
trigger its inputs using TTL signals (5v) which originated from a
totally seperate source (a parallel port, for example). I know this is
possible, but I'm curious about the underlying results.

Now, it's kind of hard to put into words everything I'm trying to ask,
so I'll do my best. Does the 5v from the signaling TTL line (the
parallel port) travel through the IC to the IC's ground? I'm thinking
it would, in which case, how do amps come into play? I'd assume you'd
then be pushing the TTL line's amps into the IC's power circuit. And
if so, would this increase the overall amperature in the IC's circuit
everytime that TTL line is active? My question is valid in the reverse
as well I guess, if I were to be pushing data from the IC into the
parallel port.

Perhaps another thing to ask is simpler. From what I understand, if
you took two 5v ac adapters, and hooked both positives and negatives
together (putting them in parallel basically), you'd still end up with
a 5v source, but with more amps, correct? Though I also think this
would put more strain on the weaker supply (since no two would be 100%
right? In any case, isn't this concept similar to the above, with the
IC? Would you be putting strain on one of the two circuits by mixing
them?

A third curiosity could also involve ac adapters. If you took two of
the same voltage, could you power a circuit by using the positive line
from one, and the negative line from another? I guess technically
you're breaking the "loop" in this case, though, despite them both
being connected to the same AC line.

Also, what would happen if you took a 5v and 12v adapter, and ran a
circuit off of the 5v, but connected a component's ground to that of
the 12v circuit? Or vice-versa? I guess this kind of generalizes
completely.

time being, though I'll surely think of something I forgot after I
submit it. lol. Anyhoo, thanks in advance!

2. ### John PopelishGuest

FyberOptic wrote:
(snip)
The point to remember is that no circuit voltage exists in isolation,
except for floating charges. All voltages are the difference in
potential between two nodes. So, for a 5 volt signal from one source
to be connected to a device powered from a different source, both
sources have to be connected at one point, at least. That point would
be the negative sides of those two sources, in this case.

(snip)
Better said, you would have a source with a lower series resistance,
or more current capability for a given voltage droop. At no load, the
pair would be indistinguishable from either identical source, alone.
Paralleling sources is different than just connecting one side of
them, so that both voltages share a common reference point.

As to how they share load, it depends on whether or not they are one
way sources (like transformer rectifier supplies) or two way (able to
deliver power or absorb it) as with batteries. But in both cases,
details count.
For the sake of argument, I will assume these are transformer
rectifier supplies with some capacitor filtering across the output.
You could if you connected the remaining pair of supply terminals
together, effectively putting the two supplies in series, like
stacking a pair of cells.
Yes. They are isolated from any connection to the AC lines, so they
represent floating voltage sources.
(snip)

All sorts of parallel and series connections are possible, but again,
you have to consider whether each supply can only deliver power, or if
it can also absorb power through the load from the other supply. You
have to look at what the total circuit does to each supply, as well as
what the supplies do to the load.

3. ### FyberOpticGuest

Aha, that seems to make sense with everything I've read, if I
understand it properly. Basically I'd have to connect the ground of
the IC's circuit to a ground wire coming from the parallel port,
otherwise those TTL lines from the parallel port would provide no
"signal" to the IC, correct?
So hooking two batteries in series is different than hooking ac

I'm confused on the whole ac adapter thing in general. Referring to
your typical generic unregulated wall-wart, they have a voltage and
milliamps rating. Does this mean the adapter has to be drawing the
rated amount of milliamps before it will deliver that particular
voltage? Or does it mean you have to have a load with a particular
amount of resistance before you get that number of volts?

I'm also curious if my speculation is correct on the resistance. Can
you figure out the internal resistance of an AC adapter using the
standard formula? Like using a 9v 500ma adapter, could you just do
9v/500ma = 18ohms? I guess I should actually ask if this 18 ohms is in
fact the internal resistance, or is this the amount of resistance your
circuit has to create before the adapter will generate the 9v? And if
the latter is the case, why would the adapter be rated in milliamps and
not resistance?

I keep hearing that it's okay to use higher-rated milliamp ac adapters
on devices if you don't have the device-rated amount available. But
wouldn't this result in higher voltage going into the device?

4. ### John PopelishGuest

Well, no easily predictable signal, since it would involve accidental
leakage paths between the two supplies (like other signals between them).
In some cases. If you connect three cells in series, with one
backward, you get 1 cell of net voltage, with two of thew cells
canceling each other out, but the load current passing through all
three (including the one that has current going backwards through it.

The adapters include diodes, so current cannot go backwards through
them. That is the essential difference.
Something close to that. The fact that they are unregulated implies
that their output voltage sags as load current increases.
What particular resistance?
No. That is the way you calculate the lowest load resistance they can
drive, since that is the voltage across that load, and the current
through it.
The second is a close guess.
Because you do not normally want to use it at a load where its
resistance is the limiting factor.

You could model the supply as an ideal voltage source (with zero
internal resistance) in series with a resistor, and find the resistor
this way. Measure the unloaded, open circuit voltage, and make that
the source voltage. Then, short the supply with an amp meter (a very
low resistance device) and measure the short circuit current. Be
quick about it, because this may damage the supply very quickly.
Divide the first voltage by the short circuit current and you have an
approximation of the supply internal resistance.

A less damaging way to measure the internal resistance is to load the
supply lightly while measuring the load voltage and current and at
something around rated current. Then take the ratio of the difference
of the two voltage measurements divided by the difference of the two
current measurements, in the direction that gives you a positive
ratio, (V1-V2)/(I2-I1) and you come up with something close to the