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.