The circuit is designed to connect to a 4 to 6 ohm load.
Hello,
If you want someone to know how your circuit works you at least have to tell them what kind load it is to drive. Again you throw a curve in declaring it to be a "4 to 6 ohm load". From the other posts i can see now it is supposed to be a solenoid. That's not a 4 to 6 ohm load, but has lots of inductance too.
To fully understand this however, it would be nice to know the typical operating current of the solenoid, the current to pull in and the current to hold, etc. Knowing the coil resistance is 4 to 6 ohms helps too.
Also in a cloud of obscurity is the chosen title for the circuit, which immediately makes the circuit sound like something totally different. Note:
1. It does not discharge a capacitor, it drives a solenoid, so "Capacitor Discharge Circuit" is not very appropriate for the name of the circuit.
2. Capacitor discharge circuits are used to discharge external capacitors, usually either high voltage or of large capacity or both.
Since it was made for and is used for driving solenoids, it should have been called "Solenoid Driver" at the least, or "Safe Solenoid Driver", or "Protected Solenoid Driver" or something like that. People would have a MUCH better idea what it was that way without going drudging through all the hedges first.
It does not matter what load the circuit is supplying, the operation of the circuit can be described when supplying a LOW, MEDIUM or HIGH impedance (or resistance) load and the features of the circuit can be explained.
Not really. A solenoid is not a resistor. It contains inductance which greatly impedes the flow of current during the first instants of time, and generates back emf that must be accounted for in the design.
Now you see why i asked what the load really is.