Loading two super caps in series at 5 Volts


i have 2 2.7v/100f super caps on desk , planning charge them (in series) using 5v/20a power supply. since charging time (to 62%) defined t = r * c, keep r small achieve faster charging.

let's want charge them using 1 amp @ 5 volts. ohm's law (v = i*r) tells need 5 ohm resistor achieve that. however, have lot of 1/4 watt resistors, if use single 5 ohm 0.25 watt resistor pretty sure resistor burn out. solution: use 25x 125 ohm resistors in parallel (25 * 0.25 watts = 6.25w, caps shouldn't burn out). overall resistance r = 1/(25*(1/125)) = 5 ohm. loading time @ 5v , 1a t = r * c = 5 ohm * 50 f = 250 seconds charge 2 caps in series ~62% (its 50f , not 100f because using 2 caps in series; take ~500 seconds charge them 85%).

is calculation ok or think run trouble? btw, asking myself if healthy caps charge them @ 1.5 or 2 amps... not sure, has experience that?

edit: found datasheet ( db, see page 8 ). says "max continuouscurrent 5 a", wow. able charge caps @ 5v using 5 amps??

if caps one
http://www.mouser.com/ds/2/257/maxwell_hcseries_ds_1013793-9-341195.pdf
which has 6.7a rating max keep temperature rise under 15 degree f (page 4, , note 8 page 6), 1a charging should okay.
the time may longer calculate - cap voltage goes 0v 5v inverse exponentially, charging 5v gets slower cap voltage gets higher, , voltage across resistor decreases.
value of resistors in parallel 1/rtotal = 1/r1 +1/r2 + etc.
1/5ohm = 0.2
1/125 ohm = 0.008
0.2/.008 = 25, math seems correct. each resistor dissipates v^2/r = 5*5/125 = 0.2w, 1/4w parts may warm charging starts but okay cap charges , current drops.


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