I have two 2.7V/100F super caps on my desk and I was planning to charge them (in series) using my 5V/20A power supply. Since the charging time (to 62%) is defined as t = R * C, I would like to keep R small to achieve faster charging.

Let's say I want to charge them using 1 Amp at 5 Volts. Ohm's Law (V = IR) tells us that I would need a 5 Ohm resistor to achieve that. However, I have a lot of 1/4 Watt resistors, so if I use a single 5 Ohm 0.25 Watt resistor I am pretty sure the resistor would burn out. My solution: I will use 25x 125 ohm resistors in parallel (25 * 0.25 Watts = 6.25W, so the caps shouldn't burn out). The overall resistance will be R = 1/(25(1/125)) = 5 Ohm. So the Loading time at 5V and 1A would be t = R * C = 5 Ohm * 50 F = 250 seconds to charge the two caps in series to ~62% (its 50F and not 100F because I am using two caps in series; it will take ~500 seconds to charge them to 85%).

Is my calculation ok or do you think I will run into trouble? Btw, I am asking myself if it would be healthy for the caps to even charge them at 1.5 or 2 Amps... I am not sure, anyone has experience with that?

EDIT: I just found the datasheet ( DB, see page 8 ). It says "Max ContinuousCurrent is 5 A", wow. So I would actually be able to charge the caps at 5V using 5 Amps??

If your caps are like this one http://www.mouser.com/ds/2/257/Maxwell_HCSeries_DS_1013793-9-341195.pdf
which has a 6.7A rating max to keep temperature rise under 15 degree F (page 4, and note 8 page 6), then with 1A charging you should be okay.
The time may be longer than you calculate - the cap voltage goes from 0V to 5V inverse exponentially, so charging to 5V gets slower as the cap voltage gets higher, and the voltage across the resistor decreases.
Value of resistors in parallel is 1/Rtotal = 1/R1 +1/R2 + etc.
1/5ohm = 0.2
1/125 ohm = 0.008
0.2/.008 = 25, so your math seems correct. Each resistor dissipates V^2/R = 5*5/125 = 0.2W, so 1/4W parts may get warm as charging starts but be okay as the cap charges up and current drops.

Your math looks right but I wouldn’t put the capacitors in series (unless they are rated for 5V) because you cannot guarantee the voltage will divide evenly. I think you need a 5V capacitor, or lower the voltage.

And it’s just silly to put 25 resistors in parallel… Order a 5 or 10W resistor…

I see. Maybe there is a way to achieve an even voltage between the caps? hmm, I will look into this.
yes you are right, but 10w resistors are not cheap, I will use 5watt. Now that I know I can use max 5 Amps I ordered 10x 5 ohm (5 watt) resistors. I will use 4 of them in parallel to get 1.25 ohm (this will cost me just 70 cents). 5 volt / 1.25 ohm = 4 Amps charging! 5 v * 4 a = 20 watt (just enough for 4x 5 watt resistor)

But if you discharge each cap before you start to series charge them I wouldn't be to worried. Especially on super caps. You need a pretty big difference in capacity to have a big voltage difference. Although I might not charge them up to 5V over a long time (in which they may degrade).

Maybe it was not a good idea to charge them in series, hmmm... I guess I should charge each one separately at 2.5V. A small "charger circuit" would be nice, that lets a led flash once 2.5V are reached, something like that. Will look into that.

In my opinion charging them in series isn't that bad. As a protection you could simply add a zener diode of 2,5V across each cap. Or 2,6V if you don't have a decent current limiting in series / use a charging circuit that may take them sliiiiightly over 5V.

Thank you so much for this video! So the author bought this super caps board which he thinks looks like this:

The moment one cap reaches 2.7V the voltage detector c73x will kick in and turns on the mosfet (it is connected to the mosfet's gate). The mosfet will than make sure the caps can discharge over two 10 ohm resistors in parallel (= 5 ohm). In the video the mosfet is labeled "l13" + a character I cant manage to read.

Do you think this is the right sketch? This would be very interesting for project. I would like to build this with a volt detector that can provide more Amps (~4 Amps at 5V).

The "C73X" was a bit confusing - turns out that is actually a Microchip SMD part, the TC53 spec'd for CMOS, 2.7V, no delay. Data sheet for the part is HERE (the last couple of pages show how to decode the "C73X" number on the top of the chip.)

The device actually is a Torex XC61C but it looks the same as the Microchip.

But did you also watch the second video? Although it can work with that device the Chinese made a huge design error. The Hysteresis is ON TOP of that 2,7V. So it will only start to discharge at 2,7X x 1,05 = 2,84V... They should have used the 2,5V (or even the 2,4V version) version which will start to discharge at 2,5V x 1,05 = 2,63V

But that device is a bit hard to get... What about a zener?

I found 2.4V zener diodes. 2.4 and 2.7 seems to be common. Is this a solution?
EDIT: I am confused. I thought about this and I think this doesn't make any sense, I did not come up with a good idea. I mean when you substitute the volt detector with a 2.4V zener diode there will always be a constant voltage through the zener diode. I think we need some kind of detector that will sense the volatage change when more than 2.4V is on the caps.
EDIT: I guess using this volt detector (2.5V) solves the problem.

The first resistor (at the caps) is necessary to load the caps with 4A at 2.5V (ohm's law: 2.5V = 4A * 0.625 ohm), the second one is for discharging. You think this will be ok?

Then you think about in wrong. Place the zener parallel to the cap but with the anode to the negative side of the cap. The zener will only START to conduct when the zener voltage is reached. Before that it's just high impedance and does nothing.

Mm, kind of. But place the zener only across the cap. And you can do with just one resistor for the whole bank.

And yes, the zener should be capable to handle the power. But if you connect them only across the caps, not the resistor, they will max dissipate I x 2,4V = P. So if you charge them with a constant current of 1A it becomes 1A x 2,4V = 2,4W.

But you are NOT charging them with constant current but with a resistor. You calculated a resistor of 5Ohm to limit the charge to 1A max (that’s only at the start when the full 5V is over the resistor). So, worst case we have one full cap (aka, zener wants to conduct) and one empty cap (0V over it). This leaves 5V - 2,4V = 2,6V for the resistor. 2,6V / 5Ohm = 520mA. So the zener will only get 520mA so it will only produce 2,4V x 520mA = 1,25W.

And that’s worst case! In real live you try to charge the caps the same so when one zener starts to conduct (cap is full) the other cap will be almost full. So let’s say the voltage on the other cap is 0,3V lower then the full cap. Then we have 2,4V + 2,1V = 4,5V over both caps. This leave 5V - 4,5V = 500mV for the resistor. 500mV / 5Ohm = 100mA will flow into the caps. The zener will have to burn 2,4V x 100mA = 240mW. So even a halve watt zener will do if you try to balance charge them and you use a resistor of 5Ohm to charge them