Electronic – Behaviour of capacitor discharging through a transistor’s base

If capacitor is connected to a closed circuit it will either charge or discharge, depending on the voltage "forced" on it (the equilibrium or steady state voltage, that is reached when there is no current flow through the capacitor). If this voltage is higher than the capacitors \$Q/C\$, such as with case when it is connected to a power supply, it will charge to reach this voltage. If there is no voltage source, or it's voltage lower than \$Q/C\$, the capacitor will discharge to reach this voltage as well. In case of passive elements only this voltage would be \$0\$. So the capacitor will act as a power source, and the current will "go" into capacitor itself (in order to equalize the charges on the capacitor plates) through the passive components while the energy is dissipated as heat and/or light and/or other types of energy, depending on components.

You have not given C1 a means to discharge well below the point where it turns on Q1. Therefore, it will quickly reach that point when a little current is dumped onto it thru R2.

Even if C1 was kept at 0 V before SW1 is closed, you still don't have enough of a time constant, and the transition will be fairly abrupt. (10 kΩ)(100 µF) = 1 s. However, this will trip when the base voltage gets to about 700 mV out of 6 V, which is about .12 time constants or 120 ms.

Here is a crude circuit that has a better topology for what you are trying to do. It's not a great way to do this, but I tried to use about the same parts.

By using the transistor as a emitter follower, the current thru the LED will vary less abruptly. However, this still doesn't take the logarithmic response of the human visual system into account.

C1 and R1 are the timing components to turn the LED on, and C1 and R3 for off. This means you can adjust R1 and R3 separately for different on and off responses. R2 is sized assuming D1 is a typical green LED with about 2.1 V drop when on, and a current rating of 20 mA.