Recent reviews of recommendations for charging a 6 volt battery with a 12 volt charger emphasize connecting the 6 volt batteries(2) in series then using a 12 volt charger to charge both. Must both batteries be of approximately the same size or could one of the batteries be a smaller lantern battery?
charging 6 volt batteries in series
- Thread starter CHerdic
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QGolden
Well-known member
Why?
Sparky
Well-known member
Batteries have internal resistance that determines the charging current. If you connect two batteries in series, and the internal resistances are too different, per Ohms law the charging voltage would not be split exactly in half (i.e. six volts) and one battery would overcharge while the other would undercharge.
You would be better off getting a charger that can switch between six and twelve volts.
Kurt in NJ
Well-known member
The batterys should be similar in type and size for best results, a drycell lantern battery will not work
johnbuckley
Well-known member
Just to be perverse I reckon as long as they are both the same type ( in this instance lead acid) it is OK to charge them in series- the internal resistance beautifully alters to accomodate the difference between the two as the batteries become charged. ( Which is why when you start to charge a lead acid battery with a basic type charger it shows (say 4 amps) passing - as it approaches full charge the current accepted by the battery reduces to 1-2 amps ( ish)- This is because of the internal resistance increasing not because of a an intelligent charging system. Anyway, why not try it , maybe keeping a regular check on each battery during the charging process and report back to us fordbarners?
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Tom Wesenberg
Well-known member
You can also put a 12 volt headlamp in series with the 6 volt battery to charge the battery with a 12 volt only charger. The 12 volt headlamp will act as a resistor to drop the current to a safe level for the 6 volt battery. If you use a 12 volt taillamp bulb the charging current will be even less for charging the 6 volt battery.
Best to just buy a good 6/12 battery charger with an ammeter that shows the actual amps while charging. I paid $7 for a new Sears 6 amp 6/12 volt charger at the used tool store. You can also find good deals at garage sales and swap meets.
Best to just buy a good 6/12 battery charger with an ammeter that shows the actual amps while charging. I paid $7 for a new Sears 6 amp 6/12 volt charger at the used tool store. You can also find good deals at garage sales and swap meets.
QGolden
Well-known member
Thank you, I understand, In reading your reply about 4 times, and re-reading the OP's question, I see now that I was perceiving that he was questioning the charging in Parallel and he clearly said Series. I misinterpreted his question.
Thank you.
Napa Skip
Senior Member
Just to beat this thing to death, and as mentioned (or alluded to) above, Ohm’s law (the current flow in a circuit is directly proportional to the potential and inversely proportional to the resistance, usually expressed in the form of i =E/r, where i is the current flow in amps, E is the potential in volts and r is the resistance in ohms) governs what’s going to happen when you attempt to charge a 6 volt lead acid battery with a 12 v charger, namely (depending on the capacity of the charger, how long you leave the charger attached, and whether the charger has any sort of current-limiting device) you’ll overcharge the battery (probably – as a consequence of the higher charging rate – overheating the battery and possibly warping the plates and/or boiling off some of the water in the aqueous H2SO4 electrolyte) and/or damage the charger.
Back to Ohm’s law; using the 4-amp starting rate for a 6-volt battery charger mentioned in an earlier post, the internal resistance of the battery would be (transposing Ohm’s law into one of it’s three forms, namely i = E/r, E = ir or r = E/i):
r = E/i = 6 volts/4 amps = 1.5 ohms.
Presuming the internal resistance doesn’t instantaneously change, then hooking up a 12-volt charger will give:
i = E/r = 12 volts/1.5 ohms = 8 amps.
Setting aside the fact that most [dry cell] lantern batteries (at least of the zinc-carbon type that uses an ammonium chloride paste as the electrolyte) are not meant to be recharged, the current likely obtained from attaching a 6-volt dry cell with 1.5 ohms internal resistance in series with a lead-acid battery might still be too high for the dry cell. As stated by others, the best situation (besides the obvious one of getting the correct 6-volt or 6/12-volt charger) would be to find a similar 6-volt lead-acid battery and charge them both at the same time, connecting the negative terminal of one battery to the positive terminal of the other and the charger to the remaining +ive and –ive terminals of the two batteries.
This is, after all, what a 12-volt lead-acid battery is; a set of 6 wet cells, each of which provides 2 vDC output (or two sets of 3 wet cells, i.e., two 6-volt batteries) connected +ive to –ive, such that all we usually see is the starting and ending +ive and –ive terminals.
Returning to ohm’s law, the two 6-volt batteries, each with a nominal internal resistance of 1.5 ohms, connected to a 12-volt charger, will give (remembering that resistances in series are additive; that is, R total = R1 + R2, whereas in parallel, R total would equal the product over the sum of the resistances, i.e., R1R2/[R1+R2]):
i = E/r = 12 v/(1.5 + 1.5) ohms = 12 v/3 ohms = 4 amps.
Don’t have a second 6-volt battery? You can achieve the same results (almost) by using a power (“power-dropping”) resistor with sufficient heat (power) dissipating capacity. In our example above, presuming the starting charging rate of 4 amps is what we’re concerned with, you’ll need a power resistor capable of dissipating:
P = Ei (or substituting E = ir) = i^2r = (4 amps)^2 x 1.5 ohms = 24 watts.
Note however (this is the “almost” part) that while the internal resistance of the battery will change (increase, in this case) as the fully-charged condition is neared (which is why the charging rate drops off as the charge nears its completion), the resistance of the power dropping resistor remains basically constant. Thus, near the end of charge (say when the charging rate nears ½ the starting rate, which is a rule of thumb for charging lead-acid batteries) when the internal resistance of the battery approaches 3 ohms, the voltage dropped across each component (the battery and the power resistor) will be:
E total = E battery + E power resistor = i x 3 ohms + i x 1.5 ohms = i x 4.5 ohms = 12 volts => (implies) i = 12 volts/4.5 ohms = 2.67 amps, resulting in:
E battery = i x 3 ohms = 2.67 amps x 3 ohms = 8 volts
and
E power resistor = i x 1.5 ohms = 2.67 amps x 1.5 ohms = 4 volts.
So now we’re getting back to overcharging the battery. This situation could be solved by getting a variable power dropping resistor (or – heaven forbid) adding additional power dropping resistors to match the increasing internal resistance of the battery, but at some point the light ought to go on (pun intended) and it becomes obvious that the simple and safe (and cost-effective) solution is to buy a 6-volt (or 6/12-volt) battery charger.
Or change to a 12-volt system, if you’re a real masochist…
Zzzzaaaap!!!
(Scotty, beam us up! Please!...)
Back to Ohm’s law; using the 4-amp starting rate for a 6-volt battery charger mentioned in an earlier post, the internal resistance of the battery would be (transposing Ohm’s law into one of it’s three forms, namely i = E/r, E = ir or r = E/i):
r = E/i = 6 volts/4 amps = 1.5 ohms.
Presuming the internal resistance doesn’t instantaneously change, then hooking up a 12-volt charger will give:
i = E/r = 12 volts/1.5 ohms = 8 amps.
Setting aside the fact that most [dry cell] lantern batteries (at least of the zinc-carbon type that uses an ammonium chloride paste as the electrolyte) are not meant to be recharged, the current likely obtained from attaching a 6-volt dry cell with 1.5 ohms internal resistance in series with a lead-acid battery might still be too high for the dry cell. As stated by others, the best situation (besides the obvious one of getting the correct 6-volt or 6/12-volt charger) would be to find a similar 6-volt lead-acid battery and charge them both at the same time, connecting the negative terminal of one battery to the positive terminal of the other and the charger to the remaining +ive and –ive terminals of the two batteries.
This is, after all, what a 12-volt lead-acid battery is; a set of 6 wet cells, each of which provides 2 vDC output (or two sets of 3 wet cells, i.e., two 6-volt batteries) connected +ive to –ive, such that all we usually see is the starting and ending +ive and –ive terminals.
Returning to ohm’s law, the two 6-volt batteries, each with a nominal internal resistance of 1.5 ohms, connected to a 12-volt charger, will give (remembering that resistances in series are additive; that is, R total = R1 + R2, whereas in parallel, R total would equal the product over the sum of the resistances, i.e., R1R2/[R1+R2]):
i = E/r = 12 v/(1.5 + 1.5) ohms = 12 v/3 ohms = 4 amps.
Don’t have a second 6-volt battery? You can achieve the same results (almost) by using a power (“power-dropping”) resistor with sufficient heat (power) dissipating capacity. In our example above, presuming the starting charging rate of 4 amps is what we’re concerned with, you’ll need a power resistor capable of dissipating:
P = Ei (or substituting E = ir) = i^2r = (4 amps)^2 x 1.5 ohms = 24 watts.
Note however (this is the “almost” part) that while the internal resistance of the battery will change (increase, in this case) as the fully-charged condition is neared (which is why the charging rate drops off as the charge nears its completion), the resistance of the power dropping resistor remains basically constant. Thus, near the end of charge (say when the charging rate nears ½ the starting rate, which is a rule of thumb for charging lead-acid batteries) when the internal resistance of the battery approaches 3 ohms, the voltage dropped across each component (the battery and the power resistor) will be:
E total = E battery + E power resistor = i x 3 ohms + i x 1.5 ohms = i x 4.5 ohms = 12 volts => (implies) i = 12 volts/4.5 ohms = 2.67 amps, resulting in:
E battery = i x 3 ohms = 2.67 amps x 3 ohms = 8 volts
and
E power resistor = i x 1.5 ohms = 2.67 amps x 1.5 ohms = 4 volts.
So now we’re getting back to overcharging the battery. This situation could be solved by getting a variable power dropping resistor (or – heaven forbid) adding additional power dropping resistors to match the increasing internal resistance of the battery, but at some point the light ought to go on (pun intended) and it becomes obvious that the simple and safe (and cost-effective) solution is to buy a 6-volt (or 6/12-volt) battery charger.
Or change to a 12-volt system, if you’re a real masochist…
Zzzzaaaap!!!
(Scotty, beam us up! Please!...)
MikeK
Well-known member
Napa Skip, I hate to throw a wrench in the works, but your basic calculation of internal resistance is incorrect because of an error in defining the problem. Your following calculations are also incorrect based on the initial error.
To determine the apparent internal reverse (charging) resistance a battery charger 'sees' you need to determine the voltage difference between the RMS output of the charger and the terminal voltage of the battery. Example: A battery charger (or alternator, or generator) is measured to output 7.2V at the battery terminals and is delivering an indicated 4 amp charge. When switched OFF, the terminal voltage of the battery reads 6.3V. The difference is 0.9V. Using Ohm's law with 0.9V and 4 amps, you get an apparent internal resistance of 0.225 Ohms.
If a '12V' charger/ generator.alternator that delivers 14.5V to the terminals is applied to that same 6V battery, the voltage difference is 8.2V. Using the calculated internal reverse charge resistance of 0.225 Ohms, the indicated charge would be 36.4 Amps!
To complicate matters, internal forward (discharge) resistance of a battery is independent of the apparent charge resistance and also much, much lower. If that same 6V battery shows 300 amps @ 5.2V on a load tester you would have a calculated internal discharge resistance of 0.017 Ohms.
To determine the apparent internal reverse (charging) resistance a battery charger 'sees' you need to determine the voltage difference between the RMS output of the charger and the terminal voltage of the battery. Example: A battery charger (or alternator, or generator) is measured to output 7.2V at the battery terminals and is delivering an indicated 4 amp charge. When switched OFF, the terminal voltage of the battery reads 6.3V. The difference is 0.9V. Using Ohm's law with 0.9V and 4 amps, you get an apparent internal resistance of 0.225 Ohms.
If a '12V' charger/ generator.alternator that delivers 14.5V to the terminals is applied to that same 6V battery, the voltage difference is 8.2V. Using the calculated internal reverse charge resistance of 0.225 Ohms, the indicated charge would be 36.4 Amps!
To complicate matters, internal forward (discharge) resistance of a battery is independent of the apparent charge resistance and also much, much lower. If that same 6V battery shows 300 amps @ 5.2V on a load tester you would have a calculated internal discharge resistance of 0.017 Ohms.