Hi all,

    This challenge seems hard to solve.

    Well, search no longer ! The solution is here !

    This is the circuit :


    How does it work ?

    After powering on, C1 will be charged through R3 and the closed relay contact.

    Since S1 is open, T1 is blocked and the relay is in rest position.

    Now, when we push S1, C1 discharges through R1 and quickly turns on T1.

    The relay is energized and its contacts are reversed. We see also that T1 is kept conducted by the relay contact through resistors R2 & R4 in series.
    No matter how long we keep pressing S1, the situation does not change since the transistor is kept conducting by R2 & R4.

    Since the lamp is in series with the relay coil, it is lit.

    Remember that we talked about the relay in earlier posts ? It has a coil made for 6V. So we need to protect it against over voltage, thus we put the lamp in series.

    So, either we put the lamp in series or we can also put in in parallel with the relay coil. In this case we need a 120 Ohm resistor in series. To make assembly easier, I’ve chosen to put the lamp in series with the relay coil.

    This has also another advantage ! In rest position (lamp is not lit), C1 will be charged up to 9V. This is because the relay coil is not energized and thus not acting as a voltage divider.
    This higher voltage across C1 will make T1 conduct very quickly.

    Now, the top relay contact is also energized and we see that C1 is now discharged through R3 to Ground.
    Do you think this is needed ?
    You might think not, but if we would only briefly press S1, the capacitor might not be discharged completely and the 2nd action might not work !

    The 2nd action :
    When we press S1 again, the grounded potential of C1 opens T1 and the relay falls off.
    The situation is now again where we started.

    Here comes the magic !
    The circuit will only work when the proper values of the resistors are chosen.

    R3 (100K Ohm) is too large to make T1 conduct.
    R1 (47 Ohm) is very small so a briefly large current can flow to the base of T1. At it’s peak, it will be 176mA and decaying very fast. T1 cannot be harmed this way.
    R2 & R4 keep T1 activated, but are large enough to block T1 when it receives the grounded potential of C1.

    I hope you enjoyed this brief explanation of this neat and small circuit.
    By using only 1 transistor, we can build some great working circuits, less is more !

    In the setup photograph, you’ll notice that we can connect the meter in series with a 100K Ohm resistor at the connected cross wire on top to see the charging and discharging of C1 in action !