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Multiple Voltage Sources?

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

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  1. FyberOptic

    FyberOptic Guest

    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

    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

    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!
  2. FyberOptic wrote:
    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.

    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.

    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. FyberOptic

    FyberOptic Guest

    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?
  4. 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.
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