The maximum voltage on the capacitor is determined by the comparator. I don't know where you got 1.45 from, but there are 3 resistors on the comparator that determine when it goes high and turns on the MOSFET, and that determines the maximum capacitor voltage. 1.45 isn't it, but it should be easy to figure out.

Beta is a parameter only BJTs have, not MOSFETs. Beta is current gain, and you can't have current gain when there is no input current.

I'll make a simulation and let you know my conclusions. I really don't care if I don't currently have the knowledge to understand. I already invested too much effort. I'm going to understand it even if it kills me.

 

The maximum voltage on the capacitor is determined by the comparator. I don't know where you got 1.45 from, but there are 3 resistors on the comparator that determine when it goes high and turns on the MOSFET, and that determines the maximum capacitor voltage. 1.45 isn't it, but it should be easy to figure out.

Beta is a parameter only BJTs have, not MOSFETs. Beta is current gain, and you can't have current gain when there is no input current.

Watch the attitude by the way. I tried the simulation and received 2V an the non-inverting node of the op-amp while the inverting node starts at 9V. How is that even possible?
I thought the inverting and non-inverting node should match voltages. I figured out the current is draining from the NMOS to charge the capacitor and being discharged quickly when the NMOS is turned on. Its not really off until right after discharge. There's something your not telling me. Why in the world are you being so evasive. I get the concept, just give me a source or somewhere I can go for help. I've been searching and found nothing. Stop beating around the bush and help understand this thing comprehensively. I'm putting in the effort, but your so stubborn because you don't want to give away the "answer" when all I'm asking for is guidance. What's your problem?

 

Here is a simpler version of a Current Mirror Linear Ramp that you might be able to understand. Q3 acts as a voltage sense, its triggering threshold is defined via D1, D2 and R3; Q4 helps Q3 discharges the capacitor as fast as possible. When the voltage at the capacitor rises beyond Q3's threshold, it starts turning on, sourcing current to Q4, turning it on; Q4 pulls Q3's base lower, thus turning Q3 on harder, Q3 sourcing more and more current to Q4, looping until Q3's emitor is no longer higher then its base, thus, turning it of.

Note: I cant upload the .asc file, for some reason, it does not appear in the 'Upload a file' browser?

 

Here is a simpler version of a Current Mirror Linear Ramp that you might be able to understand. Q3 acts as a voltage sense, its triggering threshold is defined via D1, D2 and R3; Q4 helps Q3 discharges the capacitor as fast as possible. When the voltage at the capacitor rises beyond Q3's threshold, it starts turning on, sourcing current to Q4, turning it on; Q4 pulls Q3's base lower, thus turning Q3 on harder, Q3 sourcing more and more current to Q4, looping until Q3's emitor is no longer higher then its base, thus, turning it of.

Note: I cant upload the .asc file, for some reason, it does not appear in the 'Upload a file' browser?

That's so clever! Your using the diode and a very large resistor to set the voltage at Vb3. Your using the diodes so the current can't go back up to the current mirror. By setting Vb, all you have to do is charge the capacitor until the emitter voltage is higher than the base voltage at Q3. That's a pnp transistor so I think the threshold is about 0.7V above Vb. Then the current drains and since the relationship between the base current and emitter current is ib*(B+1), the current drains even faster creating a short discharge time and a longer charge time. Did I get it right?

 

Your using the diodes so the current can't go back up to the current mirror

Nope! The diodes are there to make sure that Q3's base is 0.7V lower then its emitter/ setting the triggering threshold. You can use voltage divider resistors such as 3k/10k and it would still works. The main reason i'm using these diodes is to eliminate wasteful current running through the voltage divider resistors.

 

Nope! The diodes are there to make sure that Q3's base is 0.7V lower then its emitter/ setting the triggering threshold. You can use voltage divider resistors such as 3k/10k and it would still works. The main reason i'm using these diodes is to eliminate wasteful current running through the voltage divider resistors.

Still cool. Thanks for the correction.