Preliminary linear regulator for power supply

Img D:
circuit.

This Vreg provides short-circuit protection, no overload protection.
The base current for T1 is supplied by T2.
The base current for T2 is provided by the output voltage of T1, which is initially 0V.
So T2 and T1 never open.
Therefore bootstrap current to T1 is provided by R3.
T3 steels base current for T2 if the output voltage rises too high, thus stabilizing the output voltage at it's base's threshold.
Base current to T2 is set to ~1mA at preferred output voltage. (1.5V)
T1 shall provide ≥2A with hFE ~50, so 40mA base current needed.
Uin shall be from 5V up, so a safety resistor R1 of at most 100Ω can be used: R = U/I = (5-0.9)/0.04.

T2 does not open before Uout reaches 0.55V, so a bootstrap current must be supplied to provide enough current for up to ~ 0.7V.
As the bootstrap current is also the short-circuit current, it should be as low as reasonable.
All current in the CPU finally runs through a transistor, so the CPU does not consume current (except leakage) before 0.55V. Estimation in diagram Img C, line Icpu.
The max. current supplied by the Vreg is similar, due to similar use of resistors and transistors. To pass over the critical point at 0.55 - 0.7V (dotted black circle) a high enough bootstrap current is needed.
Bootstrap current is set to ≥50mA @ 5V Uin with R3 ~ 4.7kΩ. With T1 hFE ≥ 50 this yields ≥ 50mA.
At Uin = 12V the bootstrap current is more than doubled, with T1 hFE ~ 100 it is again doubled: ≥ 200mA.
So bootstrap and short-circuit current are in range 50mA .. 200mA.

note: for driving an resistive load as in Img 9 & A, a high bootstrap current is required.

measured voltages:
	T3 base: 0.612V
	T2 base: 0.589V
	T1 base: 0.659V

Img C:
U/I diagram.
Icpu:
	expected current cunsumption for Uin = 0 .. 2V
	since every current runs through a resistor and a transistor, no current flows up to ~0.55V, except leaking. above 0.7V expected "approximately linear" increase.
@1kΩ:
	based on 3 values from former measurements. caveat: these are calculated based on assumptions!
I_PSUmax:
	the reulator circuit has similar characteristics as in the CPU design:
	up to 0.55V T2 cannot provide any current to the base of T1, so the regulator does not start up.
	therefore a bootstrap resistor supplies initial current.
	this also is the short-circuit current! so this current shall be low.

-----------------------

yellow = T3 collector
blue   = 1-6: T2 collector
         7-B: output

note:
the 2k2Ω resistor is used for bootstraping and n.c. in operation

Img 1:
output: 100nF, 110Ω - pulsing

Img 2:
input:  100nF
output: 100nF, 110Ω - pulsing

Img 3:
input:  100nF + 100µF
output: 100nF, 110Ω - pulsing

Img 4:
input:  100nF + 100µF
output: 100nF + 47µF, 110Ω - regulating

Img 5:
input: 2V
output: 1496mV

Img 6:
input: 12V
output: 1496mV --> 0.0% change

Img 7:
blue: output voltage ripple: AC

Img 8:
output 110Ω load: 47kΩ bootstrap resistor
10kΩ is too low to regulate a load of 110Ω at Uout=1.5V
47kΩ is too low to regulate a load of 220Ω at Uout=1.5V

Img 9:
output: 18Ω, 1485mV --> 11mV dropped

Img A:
output: 9Ω, 1472mV --> 24mV dropped

Img B:
using 5V USB power supply
less ripple in general, but regular small bursts