I'm trying to understand how a LDO topology works. It's in NXP IC's datasheet:
It has the control circuitry inside the IC with external transistor.

Parameters:
VDD output = 1.8V
Collector supply = 2.367V
NPN: Nexperia PBSS4620PA
L1: ferrite bead 26 Ohm @ 100MHz
C1 = 470nF
Rp = 1 Ohm
Rs = 0.15 Ohm
R1 = 49.9 Ohm
R2 = 18 Ohm
Voltage reference is supposedly a bandgap reference at 1.2V.
This topology isn't like the typical LDO, where the error amplifier feedback is taken from a voltage divider at Vout. In this circuit R1 and R2 don't have a ground path, but connect back to BJT base with AC coupling. How is this supposed to work?
I ran an LTSpice simulation to find the DC load curve. The result doesn't even make sense.
LTspice schematic:
(U1 is UniversalOpAmp generic model without input and output limits)

Vout vs RL waveform:

This doesn't make sense, because it's an LDO, Vout is expected to be always 1.8V with varying RL.
How should this work?
It has the control circuitry inside the IC with external transistor.

Parameters:
VDD output = 1.8V
Collector supply = 2.367V
NPN: Nexperia PBSS4620PA
L1: ferrite bead 26 Ohm @ 100MHz
C1 = 470nF
Rp = 1 Ohm
Rs = 0.15 Ohm
R1 = 49.9 Ohm
R2 = 18 Ohm
Voltage reference is supposedly a bandgap reference at 1.2V.
This topology isn't like the typical LDO, where the error amplifier feedback is taken from a voltage divider at Vout. In this circuit R1 and R2 don't have a ground path, but connect back to BJT base with AC coupling. How is this supposed to work?
I ran an LTSpice simulation to find the DC load curve. The result doesn't even make sense.
LTspice schematic:
(U1 is UniversalOpAmp generic model without input and output limits)

Vout vs RL waveform:

This doesn't make sense, because it's an LDO, Vout is expected to be always 1.8V with varying RL.
How should this work?