Comparator
Vo = 1 if V+ > V -
Vo = 0 if V + < V-
For H = 5V,
V+ = 3.4V (due to voltage divider)
If current through R = 0.140 A, Vbe = 0.7V, and V+ = 0
thus the comparator output is driving the led driver transistor between 0.5 and 0.7 V.
If the the feedback transistor turns on completely, V+ will be pull below V- , shutting down
down the output of the comparator.
At least that's how I see it. If I am wrong, please correct me.