# Can I safely balance a series/parallel cell circuit to power a 12V soldering iron?

I want to power a 12V soldering iron with 2x 21700 '40T'batteries and 3x 18650 '25R' batteries with a torroidal rheostat between them.

The 3x25R batteries will be in parallel.

The rheostat will balance the resistances on a 6.4:3.6 ratio with 2x40T batteries in series.

The available apparent power of the 40T batteries (rated 30 amps) in series will come to root(2, 30^2 + 30^2) =42.4264068711929 volt-amperes.

With concern for exciting current between this value and that of the 3x25R parallel pack in the series, I will include a rheostat to accomodate for differing resistances.

The 3x 25R (rated 20 amps) parallel pack will share the loads amp stress.

1/((1/20) + (1/20) + (1/20)) = 6.6• amps average stress.

(It would be great to know how to calculate volt-amperes apparent power across the parallel pack. I imagine this calculation would help to prove efficiency of the circuit. For now I'll assume peak efficiency can be based on the a healthy draw from the parallel pack.)

The parallel pack will have a combined capacity of 7500mah (25R 2500mah each), while the 40T series batteries have 4000mah each.

6.66 amps are drawn from the parallel pack (7500/6.66=1125) and 3.25 amps are drawn from both '40T' batteries (4000/3.25=1231).

Stress over the 40T to 25R batteries will be 1125:1350 when the soldering iron demands 54.54 watts ( = 3.25x8.4 = 6.6x4.2).

All batteries in circuit will be 4.2V fully charged. They will exert the lowest frequencies over the rheostat and the load when fully charged.

(The following calculation is not sequential from the last).

When the rheostat will balance the operational amplitude of the 4.2 volt 2x'40T' batteries with the 3x '25R' pack at a ratio of resistance 6.4:3:6 because the shared stress of the 25R parallel pack will equal the volt-amperes of the 40T series apparent power stress. (6.4x6.6a= 42w/~va).

resistive ratio * avg current:(resistive ratio * avg current)/parallel cells = 11:14

I imagine that this is a negligible difference.

I assume that I dont have to add resistors; relay resistors at the inverting input; or include a ground; that the soldering iron will work at around 55 watts until the circuit pack voltage goes beneath 12V and no battery will be damaged or exhausted beneath 3.2V.

I would greatly appreciate a more experienced estimate and any explanation in terms.

After consideration of 'hijacking' somebody elses post, I should ask specifically for some-body to confirm for me that my circuit will be safe and efficient. This is in the interest of a rubric for myself and others. I have put dissimilar 18650 batteries in series before and damaged batteries.

The rheostat I plan to use is rated to 25w and is 20ohms. The soldering iron has a power switch. I don't intend to leave the soldering iron on for very long at a time, enough to damage the rheostat.

• Can’t imagine why you think that works as well as a commercial battery power iron with a temperature regulated tip Jan 11 at 18:58
• So you have 3 x 2500mA cells in parallel, in series with 2 x 4000mA cells also in series to make 12V, right? Exactly how is the rheostat connected to them? Please draw a schematic of the circuit. Jan 11 at 19:01
• Thanks Bruce ive added a picture Jan 11 at 21:56
• Thanks Tony but I'm a bit skint at the moment. I was concerned they would be weak and with non-replaceable batteries. Jan 11 at 21:58
• No worries if you havent got time/ not a fair turn-taking ratio ques Jan 11 at 22:00

Samuel, you seemed to be treating a pure DC circuit with a resistive load like it was an AC circuit with a reactive (eg. inductive) load, and like there was an amplifier in there somewhere with an "inverting" input. That noted, the drawing currently up is a straightforward matter, so I think you are on a better track now.

I would just put three 4.2 volt batteries in series and forget the balancing act you are trying to do. The rheostat won't help unless you are using it to adjust the soldering iron power (but that's a terrible way to do it).

Careful calculations cannot preclude variability between batteries, which is usually substantial, and which is the reason that one cell in a series battery pack can be damaged. The cell with the lowest capacity dies first then the other cells push current through it, which pushes its polarity into reversal, destroying its chemistry (and possibly even making it leak or explode or burn). The best simple protection against this happening is to be sure that all the batteries are as evenly matched as possible, and never discharging the pack very far. I see by your reasoning that you understand this, but possibly do not understand that battery capacity ratings are only approximations. You cannot guarantee much by trying to balance capacity ratings. Balancing is more for the sake of making the whole thing work as long possible, rather than only working until the weakest battery dies. Without a more sophisticated battery management system -- a much more complex circuit -- you are going to be stuck with simply being sure you recharge all the cells before any of them die.

This is made more critical with lithium batteries (compared to lead acid or alkaline) because they have very flat discharge curves. The one cell with the least capacity will drop voltage rapidly when it meets the end of its capacity while the others in the battery will still be pumping out electricity at almost full power. So you have one flat battery that is easy to be pushed to death by the remaining full-power batteries.

For these reasons I have to say "no, it is not safe to do that with lithium batteries". If you ignore that warning and go ahead anyway, carefully watch all three cells to make sure you stop using the battery if one of them is exhausted before the others. Since lithium cells have a cutoff voltage around 2.5 volts, you probably want to stop using the pack well before any of them go that low while under load. Definitely don't wait for one of them to go even lower.

• Thanks Daryl. You have clarified the problem that I'm aiming to resolve very well. Jan 12 at 14:19
• I have a scant understanding if the implications of what Daryl clarifies. As opposed to lithium batteries, lead acid batteries require a diode in charging circuits to be saturated. The low rated rheostat will quickly reach a self-resonance where it ceases to be useful as an inductor, the ceramic will dissipate accumulating voltage gain, and the square coils will four-fold increase the frequency, leaving a state of high-static high-q offset. Jan 12 at 14:25
• Samuel, you continue to use terms appropriate to AC circuits when discussing a DC circuit. Honestly, I can't even understand most of your statements, as a consequence. Going from the schematic currently posted, the whole is a simple series circuit. (For all practical purposes the 3 cells in parallel are a single cell electrically, just combining their Ah capacities.) Jan 12 at 17:36
• The resistors (soldering iron and rheostat) take current and produce heat. The rheostat does nothing else. It just wastes power. It can make the battery last a little longer, but the soldering iron will be a little colder. Jan 12 at 17:45

After testing stubbornly, Daryl is right. The rheostat heated up slightly quicker than the soldering irons tested. Open circuit the voltage measured 12.35V. Soldering Iron 1 with thinner lead wire left 1.85V measureable after load. Soldering Iron 2 with thicker lead wire left 4V after load. The single cells lose voltage at an equal rate, no heat at nickel tabs. But soldering iron not hot enough even after a half hour.

Jacobi's law

No worthwhile capacitance through ceramic that encased square coils of resistive wire and no markable inductive voltage gain.

In retrospect, I've found a 19V 5amp laptop charger would sufficiently heat up the soldering iron. 12-15v plug wouldn't do. Possibly because car battery that it is designed to draw from has high amperage.