Hi David, here is the promised write-up.
If you look at a datasheet for the 555, it shows the IC is made up from dozens of transistors. Although it will detail the various parameters for each pin, it doesn’t help you to understand how it works. I think this block diagram I found on the ‘net explains it better.
![555 Basics.gif](./download/file.php?id=1805&sid=5fe92a8bf4276bd26fdcc8a41e2384af)
- 555 Basics.gif (6.85 KiB) Viewed 75866 times
You can see that the heart of the 555 is a flip-flop (F-F). It can be set by a negative pulse from COM1 output and will remain set until it is reset. That is done with a negative pulse from COM2 output, or by taking the reset pin (4) low.
When set, the output pin (3) goes high.
When reset, the output pin (3) goes low, and the discharge transistor is turned on, so that the discharge pin (7) is shorted to the negative supply pin (1).
Note there are three equal value resistors across the supply. With a 12V supply, this provides a 4V tap that is fed to COM1. So whenever the trigger pin (2) goes below 4V the F-F will be set.
Likewise, an 8V tap is fed to COM2 so that whenever the threshold pin (6) goes above 8V the F-F will reset.
My circuit uses the F-F in monostable mode to give a single pulse at the output when pin2 is triggered. Referring to my previous circuit, this is how it works. Note that I now realise that D2 and C2 are not needed and R5 can be wired directly to pin3.
At switch-on the F-F starts up in a random state (ie. either set or reset). If set, C1 will charge up through R4 until the voltage on pin6 gets to 8V. Then the 555 resets and pin7 discharges C1. The circuit now waits for a trigger pulse. Then the F-F will set, the output pin (3) goes high and C1 starts to charge via R4. When there is 8V across C1 the F-F will reset, the output will go low, and pin7 will discharge C1 again ready for the next cycle.
The length of each single pulse is determined by R4 and C1, and the time is equal to 1.1 x C (in microfarads) x R (in megohms) so in this case = 1.1 x .01 x .047 = 517µs The control pin (5) is only normally used to give fine frequency control when the 555 is wired up in astable mode to make an oscillator.
Now for the trigger from the ignition lead. The pickup of a few turns round the HT lead will form a capacitor of only about 10pf to 20pf, so the LH side of R1 will pulse high for a very short time. But the base of the NPN will have inherent capacitance hence the small value for R1 so that the base can rise quickly to conduct before the pulse ends. There will be a negative pulse when the voltage on the HT lead reduces to zero, and D1 protects the NPN from that negative base voltage. Note that in the circuitry from the pickup I am treading on virgin ground because I have no access to an ignition coil etc. on which to carry out tests, so the circuit may need some modification.