Electronic – Capacitors charging in a DC circuit


I've built this circuit to charge 3 large, .47F capacitors:


The capacitors are really big, and they charge up slowly. The addition of R2 and D is for D to light up in increasing power as the capacitors charge, and when they are fully charged, D should light up in full power.

However, I didn't manage to get it working correctly. D lights up in seamingly full power when the capacitors are not even half full.

How can I create a V(t), I(t) and finally a R(t) functions for the capacitors? (It isn't an ohmic resistance but you get my point). Using these I'll be able to find proper values for R2 and D, and anyway I'm just intrested in how the voltage is built up.

Best Answer

Time constant \$= \tau = RC\$.

\$ R = 3 \cdot 470 mF\$ (based on one of two values you give) so:

$$\tau = 10 \cdot 0.47 F \cdot 3 \approx 15 \text{ seconds}$$

That is one exponential time constant or about 65% of full. Each successive time constant fills it about another 65% of the remaining voltage.

However, the LED (assume \$V_{operate} \approx 2V\$) has a current drain of:

$$I = \frac{V}{R} = \frac{V_{cap}-V_{led}}{R2} \approx 2 mA$$

about \$2mA\$ per volt on caps above 2 volts. So, about \$2 mA\$ at \$3V\$, \$6 mA\$ at \$5V\$, or \$10 mA\$ at \$7V\$.

Light level between \$3V\$ and \$7V\$ is about 10 mA : 2 mA = 5:1 but your eye sees it as a much lower range than that. You need a means of delinearising LED response or, easier, using some other indicator. A bar graph voltmeter or a small analog meter would do a better job for you.

LM3914 based bar graph voltmeter
And similar