# Flashing LED circuits

How does the current flow through this circuit and thus how do the components work together to cause the flashing LED?

I know the basics in that the current first charges the capacitor which consequently turns on the transistors, however I don't understand the role of the resistors or how the circuit works as a whole. Also, why does the current first reach the 2.2uF capacitor first?

Thank you

• 1st realize that 9V is a poor choice to flash a 2V or 3V LED then choose a better design using logic from 3.3 to 5V range Oct 26, 2017 at 8:13
• Operation is more complex than it appears. There should be a resistor in series with the capacitor to stop "brutal things" happening, evem while working. The lh transistor base gets driven below ground the BC557 turns off. Transistor bets (cirrent gain) and resistor values affect operation. Sep 30, 2022 at 9:47

It's not a particularly good circuit. The base current to the BC547 needs to be less than actually needed, but not by much, and setting this up is finicky. Too easy for it to simply go into a quiescent state that doesn't oscillate. I guess that's why the pot is added.

A better circuit that's similar but more manageable is:

simulate this circuit – Schematic created using CircuitLab

No pot required. But you can modify $R_6$ to change the timing between pulses. It's also fairly easy to understand. When first powered up, $C_1$ has no voltage across it so the emitter of $Q_1$ is at ground. The divider pair of $R_1$ and $R_2$ are sufficient to put $Q_1$ active and with its collector pulling towards ground. As it pulls a little, current in $R_4$ activates $Q_2$, which pulls up on $R_5$ adding to the base current in $Q_1$ and quite soon pulls $Q_1$ fully on and therefore also $Q_2$. This leaves the LED ON, of course. Now, as $Q_1$ continues to pull collector current, as well as the addition of current through $R_6$, this adds to the voltage on $C_1$ and the emitter of $Q_1$ rises upwards, eventually squeezing $Q_1$ off. When that happens, $R_3$ helps ensure that $Q_2$ is off and suddenly there's no current in $R_7$ or the LED and the collector of $Q_2$ goes towards ground. This causes $R_5$ to pull down still more on the base of $Q_1$ ensuring that it is off. Now $C_1$ leaks current via $R_6$ towards ground through $R_7$ and the LED (weak current, too low to see) and $C_1$ discharges. As it does, $Q_1$'s emitter voltage declines and evenually $Q_1$ becomes active again, pulling back up on the collector of $Q_2$ and repeating the cycle.

The resistors are there to slightly turn on the transistors so they will amplify. The gain of the circuit is well over unity and the capacitor provides a positive feedback path, so the circuit oscillates. Specifically, when the battery is first connected the resistors begin to pull the base voltage of the first transistor upward, bringing it closer to the "on" state. As it enter the "on" state, it pulls the base voltage of the second transistor downward, which causes it to start conducting (emitter to collector), as well. This conduction pulls the collector voltage upward, and that change in voltage is conducted (by the capacitor) to the base of the first transistor, which further turns it on. Once the transistors are saturated, their gain drops below unity and they begin to turn off: As the second collector voltage now begins to fall, that change is conducted (by the capacitor) to the first transistor's base, which further turns it off. The cycle repeats.

Also, why does the current first reach the 2.2uF capacitor first?

Assume the initial condition is that the capacitor is uncharged, i.e., has zero volts across and both transistors are off.

In that case, the only path for current is through the resistors, capacitor, and LED. This is a high resistance path and so it will take some time for a voltage to develop across the capacitor (the base of Q1 will become more positive as the capacitor charges).

Eventually, the voltage across the capacitor will be large enough to forward bias the base-emitter junction of Q1 which will then 'turn on' and provide forward bias to Q2 which will then provide a low resistance path for current to the LED from the collector of Q2.

The voltage at the anode of the LED rises toward its nominal forward bias level which actually turns Q1 on 'harder' (positive feedback)

However, the capacitor must eventually discharge through the base of Q1 and then charge to the point that the voltage at the base of Q1 falls enough to 'turn off' Q1 and thus Q2 and the LED.

Now, the only path for current is through the resistors, capacitor and LED as before.