Oh, I can calculate the values very rough, not precise.
Say you have a voltage drop of 4 volts over both led. We must asume that it stay 4 volts, but in reality the voltage will drop when cap is discharging - so that make calculation less presice.
Say that diodes needs at least 15 mA to light. Lets also say that voltage over cap must bee at least 6V then. To get maximum resistance value:
Ur = 6-4 = 2V.
R = U/I = 2V/0,015A = 133 ohms.
Since it's hasrd to get a 133 ohm resistor I choose the closest avaiable E12 value: 150 ohm.
Then again, we must recalculate the lowest voltage over cap neccesary for leds to light:
U_cap_minimum = R_e12*I_led + U_leds = 150*0,015 + 4 = 6,25V
To make discharge time calculation easier, lets add 4 volts to all voltages. So new values become:
'Vcc= 9V
'V_low = 2.25V
Now, all we have to do is to calculate. The discharge formula says:
T = -R*C*ln(1 - Vcc/V_low)
So calculating C:
C = -T / (R*ln(1 - V_low/Vcc))
Inserting values:
C = -5*60 / (150*ln(1 - 2,25/9)) = 6,95 F
Asuming my calculations isn't extremly wrong, you'll ending up with a relly huge cap. You're really sure you're better off with a cap instead of a little extra battery?
[Edit]
I think a
jfet connected as a current source would be the best alternative. Still, 5 minutes is a long time for any cap.