different batteries in parallel

Discussion in 'General Electronics Chat' started by piracyer, Jun 21, 2012.

Jun 9, 2012
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Hi, I kind of struggle with this question for a long time.
What happens when I put two batteries with different voltage rating in parallel?
Such as the circuit I have attached.
I saw one thread that ask about this in the forum, but the answer given is quite simple, saying that the higher voltage battery will charge the lower voltage battery and the lower voltage battery might explode. I am looking for more detailed answers - how the electrons flows in this case? what is the voltage across RL? how is the current distributed at node 1 and node 2? For the simplicity let's just assume the neither battery will explode.

If the voltage across RL is 6 volts, according by many website, why is that?

Also, according to the "All About Circuits Volume1 DC: Chapter 9: Thermocouples section" (http://www.allaboutcircuits.com/vol_1/chpt_9/5.html) the last several paragraphs, applications use thermocouples in the measurement of average temperature between several locations by connecting several thermocouples in parallel with each other. The reason is "The millivolt signal produced by each thermocouple will average out at the parallel junction point."

So why in the case of thermocouples the voltage average out? Why not in the case of batteries?

and please don't just reply saying connecting batteries with different voltages in parallel is dangerous.

Thanks.

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2. w2aew Member

Jan 3, 2012
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OK, since you've gotten a lot of advice in the past that told you that connecting batteries of different voltages in parallel can potentially be dangerous - then you already know this part, so I won't repeat it.

A battery is an electrochemical device that has a low internal impedance (generally) and will produce a given voltage. Some batteries are designed to be recharged, some are not. Rechargable batteries can accept current in the opposite direction of the discharge current. Generally, non-rechargable batteries can not - they can be damaged - sometimes violently.

Since batteries have a low internal resistance, connecting batteries of different voltages in parallel *can* cause a LOT of current to flow from the higher voltage battery into the lower voltage battery. The large current could potentially damage one or both batteries. Assuming they survive the large current, and assuming that the lower voltage battery is a rechargable type - then you will ultimately wind up with the larger voltage battery being discharged (dead), and the lower voltage battery getting potentially damaged via overcharging or excess charging current.

You didn't have a picture attached, so the above comments are based just on the description of putting two dissimilar batteries in parallel.

Bottom line - it's almost always a bad idea, unless you control the current, and ultimately intend to charge one battery from the other.

Jun 9, 2012
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Oops, my mistake, , forget to upload the picture.
Can you talk more about the current and voltage in the circuit and a comparison with the thermocouples?
Thanks!

4. strantor AAC Fanatic!

Oct 3, 2010
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1,989
Lets say in your drawing that both batteries are 9V, and there is no load - delete that resistor. Equal potential, so no current flows between them . Now, change the battery on the left to 6V like you have it drawn (load resistor still deleted). 3V difference. Unequal potential, so current will flow. 3V difference. Let's say each battery has .1ohms internal resistance. So you have 3v being applied across a .1ohm resistor. I = E/R, I = 3/.1 = 30W. 30W is a lot of power for something smaller than a fish stick to dissipate with no heat sink, so it heats right up to explosion temp. Not to mention, the 9V battery is also dissipating as much power, so would probably explode as well.
In reality though, household batteries are so pathetic they will probably go dead before they explode. When I was a kid I used to connect 9V batteries, + to -, - to + and hold them in my hands in cold weather as a hand warmer. I didn't realize how stupid that was, but I never had them blow up in my hand. I'm not recommending you try it, it is stupid.

5. cork_ie Member

Oct 8, 2011
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Each battery can be considered a Randles circuit - see attached schematic. The overall battery resistance consists of ohmic resistance, as well as inductive and capacitive reactance, this varies for each type and size of battery. There are various experimental methods of determining this for any particular battery.
This resistance is generally very low and directly connecting batteries of unequal voltage in parallel will most likely result in destructive currents.

The effect of the external resistance in your circuit is entirely dependant on the internal resistance of both batteries in your circuit.

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6. DMahalko Senior Member

Oct 5, 2008
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Series/parallel voltage mismatch problems can manifest naturally in the real world, where large industrial battery systems use these arrangements.

Eventually one or more of the cells goes bad, and that leads to damaging voltage counterflows and voltage drops and charging problems that can damage all the remaining good batteries.

I've tried to document these problems on Wikipedia.

Uninterruptible power supply -- Series-parallel battery interactions

http://en.wikipedia.org/wiki/Uninterruptible_power_supply#Series-parallel_battery_interactions

,

7. crutschow Expert

Mar 14, 2008
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3,353
Thermocouples connected in parallel will give the average voltage of the two since their resistances are equal for practical purposes. And since the voltage is very low there is no power dissipation problem or other concern about connecting them in parallel.

Batteries in parallel will also give the average voltage output if their internal resistances are equal. But since the resistances are typically so low, the current and power dissipation can be very high due to the current flow from the higher voltage battery to the lower.

8. Bipinchandra New Member

Jun 22, 2012
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Hi!,

In your posted circuit batteries and resistor all are in parallel.
You might be trying to know that "All parallel branches bears same voltage,then What if I force branches to have different voltages by connecting battery of different voltages?".
Well...

Theoretically, [note:Considering Internal Resistance of batteries] Internal Resistance of Battery is very low as compared to 10Ω.
Initially,You can consider no current will pass through 10Ω resistor.
Now,Two batteries are in negative series and if we know internal resistance of battery then we can calculate current also.
That current will be very very High.
By connecting Battery you can consider you have provided short that bypass 10Ω resistance.
If you have wire that can provide safe path for Higher current then you can try this experiment.
But it's not practical.

9. #12 Expert

Nov 30, 2010
16,655
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Referring to post#4

Putting a small, square, 9 volt battery in your pocket with some coins will cause enough heat to make you hurry to stop it!

It won't blow up, but you will fear it is going to burn you.

Jun 9, 2012
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Thank you guys for the reply!!
However, I can't resist the temptation to do a little experiment myself.
Of course, to ensure my safety, I only connect the circuits for a very short period of time for measurement.

The two power sources I use consist of three AA mercury-free alkaline batteries with same brand. One battery provides 1.5 volt. and the other two batteries are put in series to provide 3 volt.

So here are the result. (See picture 1 and 2)

1. By simply connecting the two power sources in parallel, there is around 0.5A current (the arrow indicate the direction of electrons flow). However, the current starts out at 0.55A and it start to diminish. It keeps diminish about 0.01A per 3 or 4 seconds until I disconnect circuit. So I imagine the current will keep diminishing if I left the circuit closed. I guess this might be explained by cork_ie's post about the inductive and capacitance reactance in a battery that
2. In this experiment, a resistor of 70ohm is placed in the circuit. A current of 0.03A is flowing through the resistor making a voltage drop of 2.16volt. This indicates the voltage from the two sources pretty much average out each other, and the lower power source does apply some power instead just being a resistor in another form, neither is the voltage across the resistor being simply the lower voltage of the two power sources. However, the current flow from the 3v source to the 1.5v source is still about 0.5A.

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11. bretm Member

Feb 6, 2012
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That's not something I would conclude from the data. It could just be power supply sag because the 3V source can't satisfy the applied load. The only way the 1.5V would supply power is if the normal chemical gradient inside the battery were in place, and I don't think this happens if the terminals are 2.16V apart. Put a 4.7 ohm resistor in place of the 1.5V battery and see what happens.

12. WBahn Moderator

Mar 31, 2012
18,064
4,904
Given just the idealized circuit you have shown, there would be an undefinable (think infinite) current flowing from the 9V battery positive terminal to the 6V battery positive terminal. You can't build an such and ideal circuit, nor is it a reasonable approximation to any circuit you can build. So if you want an answer that has any physical meaning, you are going to have to provide a better model of each battery. At the very least, you need the effective series resistance of each battery. This would let you estimate the initial currents and voltages on the terminals. But, as others have pointed out, the currents will be very high and the charge state of each battery is going to be changing very quickly, thus the internal resistance is going to be changing quickly, too. So your model will deviate quickly from a practical circuit unless it takes this into account.

As for the thermocouples, that is another case where you can't ignore the resistance in the model. Take three batteries that are reasonably close in voltage (say 6V, 6.1V, and 6.3V) and where each has in internal series resistance of, say, 1Ω to keep things simple. Now hook those in parallel. What will the resultant terminal voltage be?

Jun 9, 2012
32
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Yes, you are right, it could be the parallel resistance being so low, that the voltage has been dropped somewhere else (such as in the internal resistance of the 3v batteries).
Thanks for pointing that out.

Jun 9, 2012
32
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If talk about the flow of electrons (from negative to positive), I somehow can't get my head around this.
Doesn't the all the electrons flow to the most positive place in the circuit, that's the positive side the the 9V battery?
And can the electrons actually flow through the batteries? or they have to get to the other end of the battery through RL?

15. WBahn Moderator

Mar 31, 2012
18,064
4,904
In a circuit, electrons don't all flow to any one place. Everything stays neutrally charged (unless you are talking about some very specific circuits, but even there it stays true at a sufficiently large scale). If an electron leaves one end of a component (any component), then an electron has to enter another end of it. Think of a capacitor. When it is charged, there are not excess (or missing) electrons from the device as a whole. Instead, the external circuit pulled electrons from one terminal and pushed the same number of electrons into the other terminal. The same with batteries. As electrons leave the negative terminal and flow into the circuit, other electrons are flowing from the circuit into the positive terminal. The chemistry in the battery then move those entering electrons toward the negative terminal while increasing their potential energy along the way.

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