I have to apologize to everyone who is following this. I made a algebraical error in my first attempt. Here are the calculations again.

The error I made was in estimating I_{o} = (V_{i} - V_{o}) / R_{t}, when in fact, I_{o} = (V_{i} - V_{o}) / (R_{t} - R_{c}) = (V_{i} - V_{o}) / (R_{i} + R_{s})

Given this, R_{s} = 50 is not longer valid as V_{o} will exceed 5.5V slightly, but for a lengthy period. So, doing the calculations again, this time correctly the differences are not huge, but....

For the following calculations, we define:

V_{c} = Final voltage across capacitor (to be calculated).

V_{o} = Voltage across capacitor at cutoff

V_{i} = Open circuit voltage of solar cells

I_{o} = Current at cutoff

R_{c} = internal resistance of capacitor = 30 ohms, from the capacitor datasheet (should be measured)

R_{s} = Added series resistance

R_{i} = Internal resistance of solar cells = 3 ohms (estimated)

R_{t} = R_{c} + R_{i} + R_{s}

V_{i} = 6.7V, measured

V_{o} = 5V, initial arbitrary value

I_{o} = (V_{i} - V_{o}) / (R_{i} + R_{s})

And finally V_{c} = V_{o} - (R_{c} * I_{o})

Case 1: High charge rate -> R_{s} = 75 ohms.

Then:

R_{i} + R_{s} = 78 ohms

I_{o} = (6.7 - 5) / 78 = 22ma

V_{c} = 5 - (30 * 0.02) = 4.35V

~~4.4V~~ 4.35V is not enough for the final capacitor voltage. Using a cutoff voltage of 5.4 will yield:

I_{o} = (6.7 - 5.4) / 78 = 16.7ma

V_{c} = 5.4 - (30 * 0.0154) = 4.9V

Now, for a quick charge rate use a 75 ohm series resistor, stop charging at 5.4V and use a 5.6V protection zener

Case 2: Medium charge rate -> R_{s} = 180 ohms.

Then:

R_{i} + R_{s} = 183

I_{o} = (6.7 - 5) / 183 = 9.3ma

V_{c} = 5 - (30 * 0.0093) = 4.7V

4.7V is not enough for the final capacitor voltage. Using a cutoff voltage of 5.2 will yield:

I_{o} = (6.7 - 5.2) / 183 = 8.2ma

V_{c} = 5.2 - (30 * 0.0082) = 4.95V

So, for a medium charge rate use a 180 ohm series resistor, stop charging at 5.2V and use a 5.6V protection zener

Case 3: Low charge rate -> R_{s} = 330 ohms.

Then:

R_{i} + R_{s}= 333

I_{o} = (6.7 - 5) / 333 = 5.1ma

V_{c} = 5 - (30 * 0.0051) = 4.85V

if 4.85V is not enough for the final capacitor voltage. Using a cutoff voltage of 5.1 will yield:

I_{o} = (6.7 - 5.1) / 333 = 4.8ma

V_{c} = 5.1 - (30 * 0.0048) = 4.95V

So, for a low charge rate use a 330 ohm series resistor, stop charging at 5.1V and use a 5.3V protection zener

Again, these are based on estimates for the internal resistance of the solar cells and the capacitor. Both should be properly determined before actual values can be calculated.