Active bleeder circuit design clarifications

Question

I was learning about bleeder circuits from this source.My question is about item number 5[An Alternative].

The circuit diagram is given below.

In the description you can see that "The major change is from a BJT to a MOSFET, and this allows C1 to be much smaller, which means you can use a film cap."

May I know how this is possible ?.

I have one more question about the voltage across C1. Please see my simulation result below.

Before the discharge starts you can see that one end of the capacitor is (Vout,Blue line) 44.4V and the other end (Vx,Red Line) is at 44.2V.

In this case Can I say the capacitor is charged to 200mV.(I mean for a capacitor if I connect one end of the capacitor to let it be 10V and other end to 5V,can I say the capacitor is charged to 5V) ?.

Also you can see that "Using two diodes as shown reduces the ripple voltage across C1 to about 620mV peak (vs. 1.3V with one diode), which is a better option."

How D1 and D2 reduces the ripple ?.

Answer 1

The ripple is reduced because you use two diodes D5,D6 (full-wave) instead of one (half wave). It can be said the two diodes reduce the cap size needed.

The 200mV is just diode drop difference between D1,D3 and D5,D6 because of different current level. The more loaded diodes D5,D6 have higher drop than light loaded D1,D3.

Btw, the L1&L2 inductances are too low for 50Hz.