marchache said:
Excellent, thanks for the explanation.
Last question (I think). How does C1 discharge if there is no load placed on it by the LED ? Johnson777717 said something about discharging through the base emitter pair of Q1 but since Q1 is on doesn't the current flow through the collector emitter ?
Thanks
Marc
I had not studied the circuit in any detail before my last post.
When the power is first turned on, C1 is uncharged and so it starts to charge via the 100 Ohm, LED, 1k, pot and 100k.
When the voltage between the base and emitter of Q1 rises to around 0.5 ~ 06. Volt, Q1 collector current starts to flow and so Q2 receives base current. The collector voltage of Q2 thus starts to fall, this increases the Q1 base current (because of the charge on C1 increase the base/emitter voltage of Q1) hence its collector current rises further. Thus Q1 and Q2 are rapidly turned on by the regenerative action (positive feedback). Q2 saturates and the LED glows.
When Q2 saturates, its collector voltage will be about 0.1V. So the right hand plate of C1 is now at this voltage.
Since the capacitor was charged to about 0.6 V when the switching occurred, the left hand plate of C1 will be at about -0.5V, ie. 0.1 - 0.6 = -0.5V (the capacitor's charge cannot change instantaneously).
The base of Q1 will be at about 4.5 - 0.7 = 3.8V. Thus C1 starts charging (through the emitter/base of Q1) towards this potential. As the voltage across C1 rises, the current through it decreases and consequently, the base current deceases and the collector current of Q1 also decreases in proportion. When the Q1 collector current decreases to a small value, Q2 starts to come out of saturation and so the potential at the right hand plate of C1 starts to rise and, of course, the potential of the left hand plate also rises at the same rate. This starts a rapid regenerative action since it increases the base emitter voltage of Q1 thus driving it into cutoff. Hence the collector falls to zero and so Q2 also turns off.
Since the left hand plate of C1 has a positive charge of about 3V (with respect to its left hand plate), the base/emitter junction of Q1 is reverse biased thus holding it in cutoff.
The circuit is now back to its initial state with Q1, Q2 and the LED off. So C1 again starts to charge via the 100 Ohm, LED, 1k, pot and 100k.
You will notice that I have not mentioned the discharge of C1. This because charge and discharge are essentially the same. Current going into or out of a capacitor changes its state of charge and hence the voltage across it changes in proportion.
Capacitors charge towards the applied voltage (ie. the voltage you would measure if the capacitor was not there). In the description above, the capacitor changes from a positive charge to a negative one and then back to positive on each cycle. Thus at certain instants, its voltage (and charge) is zero. At these instants, the capacitor is (technically) discharged, but only at those instants.