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Multiple Voltage Sources?

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% identical), similar to what I've read about batteries in parallel, 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 everything I've already asked, but mixing voltages just confuses me completely.

I think this is all I have to ask about this particular subject for the time being, though I'll surely think of something I forgot after I submit it. lol. Anyhoo, thanks in advance!

Reply to
FyberOptic
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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.

Reply to
John Popelish

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 adapters in series?

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?

Reply to
FyberOptic

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 internal resistance, without overloading the supply.

Yes. That is the difficulty using unregulated supplies.

Reply to
John Popelish

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