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I'm working through a circuit analysis text on my own for the purposes of getting started with Arduino. I just came across a box that explained why putting two different voltage sources in parallel is a bad idea but the explanation didn't quite make sense and now it's bugging me.

I'm aware there is another, similar question here: "What happens if I connect two different DC voltage sources in parallel?" but the answers aren't quite what I'm looking for.

What I want to know is: how do I calculate the voltages, currents and resistances in a circuit that clearly defies idealizations? For instance, how much current will my poor different-voltage batteries encounter, respectively?

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Kris Sceagan
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3 Answers3

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As a conceptual model, a voltage source maintains its specified voltage even if doing so requires the supply of infinite current. Obviously that is not a component you can buy.

Connecting two set to different voltage sources with non-resistive wiring breaks their definition, so is an unanswerable problem.

In the real world, many devices can be approximated by the Thevenin equivalent model, which consists of an ideal voltage source in series with a resistor representing the internal impedance and response of the non-ideal source to load (there is a dual, the Norton model, which has a shunt resistor across a current source).

If you interconnect Thevenin or Norton source models, you can use basic circuit analysis to calculate the results.

Components such as batteries may be imperfectly approximated by these models. With care they can give you some idea if you are likely to run down, (over)charge, or explode interconnected parts but for obvious reasons it's an area where you want to allow huge safety factors.

Chris Stratton
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  • That's a step in the right direction. And yes I will err on the side of caution. Thank you. I'd upvote but I just registered for this site. – Kris Sceagan Aug 13 '12 at 14:13
  • Instead you could accept. – Wouter van Ooijen Aug 13 '12 at 14:34
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    And ...... done! – Kris Sceagan Aug 13 '12 at 15:12
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    An alternative model which is even more realistic is a piecewise-linear model, where supplies have some number of current ranges; within each range, they behave like a Thevenin source, and the sources on either side of each range boundary would produce the same voltage at the current represented by that boundary. A supply might, for example, behave as a 1M resistor to +5.1V when output current is negative, a 0.1ohm resistor to +5.1V when output current is 0-1A, and a 1ohm resistor to +10V when over 1A. Solution solving is not as nice as with a purely-linear model, but... – supercat Apr 02 '14 at 19:02
  • ...the piecewise-linear model will often be much closer to reality. – supercat Apr 02 '14 at 19:02
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As a practical matter, the internal resistance of batteries can be very low.

This is a little deceptive because when you study Norton or Thevenin circuits, your classroom-examples are going out of their way to use reasonable theoretical resistances, so that the calculated results are within reasonable bounds that don't make the student ask if he's doing something wrong. The real world isn't like that.

If a 12-volt car battery sags to 8 volts when cranking 1000 amps, what is its internal resistance? Pretty darn low.

Say you go to the golf cart dealer and he lets you take four 6-volt junkers that won't make it 18 holes anymore, and you accidentally grab an 8-volt one. You decide to parallel them anyway, giving a 12 and 14 volt pack. Those batteries are pretty violently going to try to murder each other. (Or to be more precise, the 14 volt battery will try to "charge" i.e. overcharge the 12 volt battery at high current, limited only by their internal resistance.) You can do the math but the numbers will be ugly.

Of course, diodes change everything. If you must parallel batteries, a diode per series string will solve a whole bunch of problems. It will cause asymmetrical loading of the batteries, but that is certainly better than the alternative.

With lead batteries you would need to worry about discharging the taller pack to levels which fatigue the pack (a unique flaw of lead-acid chemistry is that dipping it below 70% full will prematurely age it, the deeper, the faster.) So for instance your 14 volt pack would deplete nearly to 0% capacity before the 12V pack would carry much load, and that would quickly age it to death. At that point you'd want a little bit of intelligence to at least cut off the lead-acid pack before a damaging level of discharge. Or better, have a fairly sophisticated buck converter which monitors battery levels and draws down each battery in proportion.

This paralleling could be worked to your advantage with dissimilar packs. E.g. A reliable, indestructible 10-cell NiFe pack has a slightly higher normal voltage than a 6-cell lead-acid pack, but much higher internal resistance. So the diodes would block the lead battery for routine service loads the NiFe can handle easily. But when an engine is being cranked over, the NiFe would sag out, and the lead-acid pack would shoulder much of the load - just the kind of momentary burst load that lead-acid is perfect for.

Harper - Reinstate Monica
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For 2 voltage sources in parallel, the calculation is quite simple:

Subtract the lower voltage from the higher voltage. Add the internal resistances. Divide the voltage difference by the resistances for the current that will flow from the higher voltage source to the lower (this is why it's not a good idea - e.g. in the case of a non-rechargable battery it will do nasty things)
Multiply the current by the higher voltage sources internal resistance, then subtract the result from the higher voltage to get the voltage "between" the batteries (you can do this with the lower resistor/voltage source and add)

For example:
If we have a 10V, 10Ω source and a 5V, 40Ω source in parallel.
10V - 5V = 5V
10Ω + 40Ω = 50Ω
5V / 50Ω = 100mA
10Ω * 100mA = 1V
10V - 1V = 9V
So the parallel voltage will be 9V, and the current flow from the higher source to the lower source will be 100mA.

For more complex networks, grab any half decent book on circuit theory. These links on network analysis from All About Circuits should get you started.

Oli Glaser
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  • "Subtract the lower voltage from the higher voltage." I know that this is done, I just don't know why. "For more complex networks, grab any half decent book on circuit theory." I've got one. That stuff comes up next. – Kris Sceagan Aug 13 '12 at 15:13
  • @KrisSceagan - I tried to show why in the answer - you said you wanted to know how to calculate the voltages and currents so I gave an example of how to do so. If this doesn't help though, try the link, it goes into a lot of detail and is pretty well written. – Oli Glaser Aug 13 '12 at 15:31
  • The voltage across the resistance is the difference between the voltages at its two terminals. – Chris Stratton Aug 13 '12 at 15:37
  • I'll figure it out somehow. – Kris Sceagan Aug 13 '12 at 16:24
  • What I'm aiming for is a deeper theoretical understanding of why this happens, not just a method for calculation. I'm sorry I didn't explain that earlier. – Kris Sceagan Aug 13 '12 at 17:13
  • Supposing the answer is correct, i like it for simplicity. It shows HOW, but not WHY. – Rev Feb 14 '13 at 15:06
  • If one connects a hose between two pressurized tanks, fluid will flow from the tank that's at higher pressure to the tank that's at lower pressure, and the rate of flow will be kinda sorta proportional to the pressure difference (the behavior of real fluids is complicated due to factors like turbulence and such, which generally don't apply with electrons). – supercat Feb 14 '13 at 23:51