At first sight case A doesn't look like it's going to cause us trouble (but wait!). Back-of-envelope-calculation: the duty cycle is only 1 %, so the 25 mA will have to be compensated by a 250 µA, linear approximation. But the current isn't constant, and charging/discharging the capacitor over a resistor will go exponentially. You have only 1 V difference between the capacitor's 3 V and the LED's 2 V, and you don't want to drop the capacitor's too much before the 25 ms are over: not that fading will be noticeable as such, but the average brightness will be. So assuming a maximum allowed 300 mV drop in 25 ms will mean:

\$ 3 V \times e^{\dfrac{-25 ms}{R C}} = 2.7 V \$

then \$ R C\$ = 0.24 s.

For recharging we'll have to set an end voltage; if we would like to recharge to the full 3 V it would take an infinite time. So if we set our target at 99 % of 3 V we can write a similar equation:

\$ (3 V- 2.7 V) \times e^{\dfrac{-(2.5 s-25 ms)}{R C}} = 3 V \times 1 \% \$

then \$ R C \$ = 1.07 s.

Yes, that's different \$ R C\$ times because the \$ R\$ is different: for the discharge it's the LED's series resistor, for the recharging it's the resistor from the battery.

For the series resistor with the LED we can calculate

\$ R_1 = \dfrac{3 V - 2 V}{25 mA} = 40 \Omega\$

I'm cheating here: the 25 mA is the begin current, it will decrease during the 25 ms. (For the moment I'm going to ignore this, but the begin current should be higher if you want 25 mA average. I'll have to do some integration here.)

Since we know \$ R_1 C\$ and \$ R_1\$ we can find \$ C\$:

\$ C = \dfrac{0.24 s}{40 \Omega} = 6 000 \mu F \$

And now we can find the charging resistor too:

\$ R_2 = \dfrac{1.07 s}{6000 \mu F} = 180\Omega \$.

(more later. There's one snag: after the first 2.5 s cycle we have 2.97 V across the capacitor, where we started with a full 3 V. So next time we'll have 2.97 V \$\times\$ 99 % = 2.94 V, and so on. The capacitor seems to be drained slowly. Why doesn't this happen in reality? Because it's not only being discharged during the 25 ms on-time, but the battery keeps supplying current as well. I'll think it over and report back)

At first sight case A doesn't look like it's going to cause us trouble (but wait!). Back-of-envelope-calculation: the duty cycle is only 1 %, so the 25 mA will have to be compensated by a 250 µA charging current. That's for constant current, which varies the capacitor voltage linearly with time.

\$ C = \dfrac{t_1 \times I_1}{\Delta V} = \dfrac{25 ms \times 25 mA}{\Delta V} = \dfrac{625 \mu C}{\Delta V} \$

\$ C = \dfrac{t_2 \times I_2}{\Delta V} = \dfrac{(2.5 s - 25 ms) \times 253 \mu A}{\Delta V} = \dfrac{625 \mu C}{\Delta V} \$

So \$C\$ will be determined by the voltage drop you'll allow. If you would allow 200 mV drop, to 2.8 V, then you'd need a capacitor of 3100 µF.

But in most real-world applications current won't be constant, and charging/discharging the capacitor over a resistor will go exponentially. You have only 1 V difference between the capacitor's 3 V and the LED's 2 V, and you don't want to drop the capacitor's too much before the 25 ms are over; not that fading will be noticeable as such, but the average brightness will be. So assuming a maximum allowed 200 mV drop in 25 ms will mean:

\$ (3 V - 2 V) \times e^{\left(\dfrac{-25 ms}{R C}\right)} + 2 V = 2.8 V \$

then \$ R C\$ = 0.11 s.

For recharging we'll have to set an end voltage; if we would like to recharge to the full 3 V it would take an infinite time. So if we set our target at 99 % of 3 V we can write a similar equation:

\$ (3 V- 2.8 V) \times e^{\dfrac{-(2.5 s-25 ms)}{R C}} = 3 V \times 1 \% \$

then \$ R C \$ = 1.30 s.

Yes, that's different \$ R C\$ times because the \$ R\$ is different: for the discharge it's the LED's series resistor, for the recharging it's the resistor from the battery.

For the series resistor with the LED we can calculate

\$ R_1 = \dfrac{2.9 V - 2 V}{25 mA} = 36 \Omega\$

The 2.9 V is the average voltage during discharging, which allows us to calculate the average current. The begin current will be 27.5 mA, but that's not going to be a problem. I calculated the 2.9 V simply as the average between 3 V and 2.8 V, but that's quite OK, over this short time you can assume the discharge to be nearly linear. (I just did the calculation with the integral of the discharge curve, and that gives us 2.896 V average, which confirms that; the error is only 0.13 .)

Since we know \$ R_1 C\$ and \$ R_1\$ we can find \$ C\$:

\$ C = \dfrac{0.11 s}{36 \Omega} = 3100 \mu F \$

And now we can find the charging resistor too:

\$ R_2 = \dfrac{1.30 s}{3100 \mu F} = 420\Omega \$.

Note that the capacitance is the same as with our constant current charging and discharging. That's because the short discharge can be approximated well as linear, like we saw earlier, and also I rounded the values.

At first sight case A doesn't look like it's going to cause us trouble (but wait!). Back-of-envelope-calculation: the duty cycle is only 1 %, so the 25 mA will have to be compensated by a 250 µA, linear approximation. But the current isn't constant, and charging/discharging the capacitor over a resistor will go exponnentially. You have only 1 V difference between the capacitor's 3 V and the LED's 2 V, and you don't want to drop the capacitor's too much before the 25 ms are over: not that fading will be noticeable as such, but the average brightness will be. So assuming a maximum allowed 300 mV drop in 25 ms will mean:

\$ 3 V \times e^{\dfrac{-25 ms}{R C}} = 2.7 V \$

then \$ R C\$ = 0.24 s.

For recharging we'll have to set an end voltage; if we would like to recharge to the full 3 V it would take an infinite time. So if we set our target at 99 % of 3 V we can write a similar equation:

\$ 2.7 V \times e^{\dfrac{-(2.5 s-25 ms)}{R C}} = 3 V \times 1 \% \$

then \$ R C \$ = 1.07 s.

Yes, that's different \$ R C\$ times because the \$ R\$ is different: for the discharge it's the LED's series resistor, for the recharging it's the resistor from the battery.

For the series resistor with the LED we can calculate

\$ R = \dfrac{3 V - 2 V}{25 mA} = 40 \Omega\$

I'm cheating here: the 25 mA is the begin current, it will decrease during the 25 ms. (For the moment I'm going to ignore this, but the begin current should be higher if you want 25 mA average.)

Since we know \$ RC_1\$ and \$ R_1\$ we can find \$ C\$:

\$ C = \dfrac{0.24 s}{40 \Omega} = 6 000 \mu F \$

And now we can find the charging resistor too:

\$ R_2 = \dfrac{1.07 s}{6000 \mu F} = 180\Omega \$.

(more later. There's one snag: after the first 2.5 s cycle we have 2.97 V across the capacitor, where we started with a full 3 V. So next time we'll have 2.97 V \$\times\$ 99 % = 2.94 V, and so on. The capacitor seems to be drained slowly. Why doesn't this happen in reality? Because it's not only being discharged during the 25 ms on-time, but the battery keeps supplying current as well. I'll think it over and report back)

At first sight case A doesn't look like it's going to cause us trouble (but wait!). Back-of-envelope-calculation: the duty cycle is only 1 %, so the 25 mA will have to be compensated by a 250 µA, linear approximation. But the current isn't constant, and charging/discharging the capacitor over a resistor will go exponentially. You have only 1 V difference between the capacitor's 3 V and the LED's 2 V, and you don't want to drop the capacitor's too much before the 25 ms are over: not that fading will be noticeable as such, but the average brightness will be. So assuming a maximum allowed 300 mV drop in 25 ms will mean:

\$ 3 V \times e^{\dfrac{-25 ms}{R C}} = 2.7 V \$

then \$ R C\$ = 0.24 s.

For recharging we'll have to set an end voltage; if we would like to recharge to the full 3 V it would take an infinite time. So if we set our target at 99 % of 3 V we can write a similar equation:

\$ (3 V- 2.7 V) \times e^{\dfrac{-(2.5 s-25 ms)}{R C}} = 3 V \times 1 \% \$

then \$ R C \$ = 1.07 s.

Yes, that's different \$ R C\$ times because the \$ R\$ is different: for the discharge it's the LED's series resistor, for the recharging it's the resistor from the battery.

For the series resistor with the LED we can calculate

\$ R_1 = \dfrac{3 V - 2 V}{25 mA} = 40 \Omega\$

I'm cheating here: the 25 mA is the begin current, it will decrease during the 25 ms. (For the moment I'm going to ignore this, but the begin current should be higher if you want 25 mA average. I'll have to do some integration here.)

Since we know \$ R_1 C\$ and \$ R_1\$ we can find \$ C\$:

\$ C = \dfrac{0.24 s}{40 \Omega} = 6 000 \mu F \$

And now we can find the charging resistor too:

\$ R_2 = \dfrac{1.07 s}{6000 \mu F} = 180\Omega \$.

(more later. There's one snag: after the first 2.5 s cycle we have 2.97 V across the capacitor, where we started with a full 3 V. So next time we'll have 2.97 V \$\times\$ 99 % = 2.94 V, and so on. The capacitor seems to be drained slowly. Why doesn't this happen in reality? Because it's not only being discharged during the 25 ms on-time, but the battery keeps supplying current as well. I'll think it over and report back)