batteriescapacitorledresistorstransistors
Now we have two circuits one with single npn lets take it as first circuit and another circuit with one npn and one pnp lets take it as second circuit.
when i tried out both the circuits. the first circuit has no output in the sense the capacitor does not discharges.
coming to second circuit the output is the led blink so it means the capacitor discharges
can anyone explain how the capacitor discharges in second circuit and why it does not discharges in the first circuit?
Initially, the voltage on the + plate of the capacitor is 0V.
Pressing S1 connects it to a 9V source; current flows, charging the capacitor.
On releasing S1, the voltage on the + plate of the capacitor remains at 9V.
Pressing S1 again connects a 9V source to a plate charged to a potential of 9V, so no current flows. The LED does not light again.
If you attach a highish value resistor (10k-100k) between the + plate and ground, it will discharge the capacitor between presses of S1, so it will light again for each press of S1.
What you've seen with two LEDs it is called an astable multivibrator.
It has two states that change whenever one of the capacitors is discharged.
Assume Q1 is ON. This means the positive side of C1 gets connected to ground. This leads to reverse charging of it (discharging), thus modifying the voltage at the negative side - it becomes positive and biases Q2 which turns ON. Q2 starts discharging C2 and it turns ON Q1 again. And this cycle repeats forever.
While either of the capacitors is charged, the base voltage of the transistor is negative, thus the transistor remains OFF. The frequency of this is related to the time needed for the capacitor to discharge (when it is connected in reverse polarity to ground via a resistor - e.g. C1-R2).
Source: Wikipedia - Astable Multivibrator
LED Flasher circuits usually work on the same principle.
Source: next.gr
When they have one LED, the other is usually replaced by a resistor.
Best Answer
I'll get you started. With any circuit analysis it helps to start from a known state, so let's assume that you've just turned the circuit on and C1 is fully discharged so we can consider it to be a short circuit. Let's also assume that the battery is giving out 6 volts.
Then R1 and R2 form a potential divider, and the voltage across R1 is
6*R1/(R1+R2) = 132uV
So with a Vbe as low as that, Q1 is turned off, no current flows through R3 and Q2 is also turned off; the LED is not lit.
The current flowing through R1 and R2 now starts charging C1. After a certain length of time, the voltage across C1 and R1 becomes high enough to turn on Q1 and current begins to flow through R3.
Let's see if you can take it from there.