Electronic – Combine two circuits

Here is a circuit using two 556 (dual 555) timers that I believe meets your needs. The 556 is available in a 14-pin DIP package.

The top half of the first 556 (IC1) acts as a Schmitt trigger to clean up any noise bounce from the pushbutton.

The bottom half of IC1 provides a 1/4 second output. It goes high when the pushbutton is pressed. It's output is fed to the solenoid K1. I have the solenoid connected to 5v, but you can connect the top end to 12v or whatever voltage is needed. When the output falls, it riggers the next timer.

The top half of the second 556 (IC2) provides a one-second delay between the release of the solenoid and the beginning of the pulsed output to the spark gap generator. The bottom half of IC2 provides a 1/4 second output. It goes high when the one second delay is over.

I didn't know what that interface to the spark gap generator looks like so I just show an output line ("To Spark Gap Gen"). You could add another NPN transistor interface if you like. The output to the spark gap generator is also set at 1/4 second.

As shown in the timing diagram, the pushbutton can be pressed either for less time than the 1/4 second delay for the solenoid, or greater; in the lattter case the 556 timer will not re-trigger.

The times are all easily adjustable by modifying either the resistor values or capacitor values or both. C1/R1 controls the duration of the output to the solenoid (currently 1/4 second); R2/C3 controls the delay between the ending of the pulse to the solenoid and the beginning of the pulse to the spark gap generator (currently 1 second); and R3/C4 controls the duration of the output to the spark gap generator (currently 1/4 second).

I used this calculator to figure out the values of the resistors and caps needed.

By making any of these fixed resistor trimpots, you can adjust the timing to whatever you need.

I suggest using tantalum caps instead of electrolytic since you can twice as good tolerance (5% vs. 10%).

You will also want to add 0.1 µF bypass caps between the Vcc (+5v) and the GND leads (14 and 7) of each IC.

But what happens if we do not add the resistor? ... The LED drops 2.5V, but the battery's potential is 9V. What about the "missing" 6.5V?

If you have a battery and LED with no current-limiting resistor, then you basically have the same scenario as when you just connect a wire across the two battery leads, but now with only 6.5 V instead of 9 V.

Basically, in this scenario you can no longer neglect the internal resistance of the battery or the forward resistance of the LED (and possibly the resistance of the wire) if you want to determine the actual current. A very large current is produced.

What happens to the battery, and what to the LED?

Like in the short-circuited battery case, the battery heats up. If the LED is only rated for 20 mA, and you put 100 mA or 1 A through it like you might in this situation, then very quickly the light-emitting diode becomes a smoke-emitting diode (at least momentarily) and then it usually becomes an open circuit.

What is the actual current in the circuit?

As I mentioned above, you need to consider the internal resistance of the battery and the equivalent resistance of the LED, and maybe the wire resistance. From these you can estimate the actual current. You probably can't calculate it exactly because the numbers depend on temperature, which will be rapidly changing as the circuit heats up and then cools down after the LED pops.