Hi, first fuel cell (any type) is never "discharging". When the fuel is fed in, the potential difference arises on the leads and when the circuit is closed, electric current result. Thus, as electric circuits theory says, the current flow will depend on the dV between SC and FC. You may as a first step to substitute FC with an equivalent battery and apply standard algorithms.

However, I suspect your SC would be never discharged in your case, as any discharge will lower its potential, so it will "absorb" more electric energy from the FC. It may only make sense when the FC is shut down, than SC supplies energy to the load. But this works with any battery/capacitor combination, like in a camera flash.

So what is your point?

Hi, thank you for answering. I understand how fuel cell works when it is the only source to the load.

I actually did the FC parallel with SC experiment these days. They both supplied to the load as long as the SC's voltage is higher than FC's voltage. I found the phenomenon is not as same as the electric circuits theory since my result is they supplied at different time, different rate(Please see my attachment). According the theory, the higher voltage source suppose to supply more current flow than others. However, my result of SC voltage is always higher than FC voltage. I found it strange and i'm wondering anything else cause the phenomenon?

The point is that I want to write a power consumption simulation when FC and SC parallel supplying. So i'm trying to find their relationship.

Hi Rei, now I see your point. There are several things to check.

First, there is a transient process. Is your load Ohmic and linear, as the resistance seem to change? If yes, how did you measure V and I? Including meters in the circuit sometimes make tricks as one cannot know what is inside them.

"the higher voltage source suppose to supply more current flow than others" - this is only valid if you have a proper circuit and equivalent resistances. Look at your currents: SC gives you 9 A at some point and FC 1 A, and later they interchange. They even will heat the leads differently (I^2*r*tau !). For a copper wire, passing 10A and 1A will definitely change its resistance (you may check it by sticking a thremocouple to insulation and record temperatures).

Second, when you diisconnect the load, as FC and SC still online (i.e. does FC "charge" SC)?

I recommend you to check transient formulation for power switches, there are milliseconds which this process is observed, but in your case you may have a large time constant.

di/dt of the super capacitor is  very high and tends to raise exponentially when you attempt to connect in parallel with any voltage source,  under no initial charge in the SC.This condition is undesirable. To control the charging current on the SC we use DC -DC converter to restrict the charging and discharging current .