Exit Light Schematic Help

atm I'm trying to figure out how this exit light schematic works.

As far as the circuit goes I understand tha ballast part that drives the lamp but beyone that I'm a little unsure.

Here are my thoughts anyway.

I think that Q5 is used to turn (Q6-Q7) on and off. Basically when Q5 is on the system is either in a powered state, or the battery level is ok. When no power is applied the light gets powered from its battery. When the battery voltage drops too low, Q5 turns off isolating the battery from the ballast.

I'm confused how Q1 and Q2 operate and where most of the charging current for the 2.4V battery comes from.

Any tips or pointers on this schematic would be helpful.

oddbudman

q1 and q2 work as a current mirror with the diode

and the diode is a dual action
1} to make a bias to turn q1 and q2 on while the supply volts from the psu is .6 v higher than the cell
assume the cell is lead acid it will need a constant voltage and will source its own charge current unlike nicad that requires a constant current to charge it up

2} protects the main psu from frying backward conduction

q1 and q2 are just pass transistors arranged like a darlington to share the current
and drop another 1.2 volts
to create a mirror the charge operation is controlled by the oscillator psu
mirror to the collector of q1 q2

so as q1 q2 supply current they will need a feedback of voltage
the same level as the battery
to keep the emiter supplying current to the cell

so it is self regulating

:. the current comes only from q1 and q2
base / emiter voltage differential setup by the cells voltage level and the feedback from the oscillator psu of the voltage level of the battery to the collector via the system monitor leds etc

so.. the whole thing will fail to light when also the battery is dud ...!!!!

this is also quite a good design and i am sure it was approved



a good addition to any exit light is an oscillator that supplys a none regular pulsing tone so you can find the exit in smoke or dust etc...