Exit sign circuit question

I recently acquired a working exit sign with backup light. After retracing the schematic (see attached image), I've hit a bit of a hard spot. I'm pretty sure Q1 is used as a cutoff to protect the NiCad battery pack, but I can't figure out what C1 and D6 are for. I'm also not sure about my diagnosis on Q1.

Quick summary of the circuit:

D1-D4 are individual diodes wired as a bridge rectifier.
P1 was from transformer, haven't checked voltage yet but I'm guessing around 9V.
P2 was the N.C. test switch to check the battery.
P3 is the "Sign running on AC power"/"OK" light on the exterior.
P4 is the battery hookup.
P5 is where the LED strips hooked in -- LED groups wired in parallel, each group with a resistor and 2 LEDs. I think that's why there is the separate rectifier; BR1 is a tiny can that I think would probably burn out.
P6 is an unsoldered terminal on the board (just pads/silkscreen) -- wired up the same way as the other LED strip, maybe for a different model.

So basically my question is: "What role does Q1 play and how do C1 and D6 influence that?"

Note: The attachment is a 352K PNG file that I drew at 3AM; it's now 4AM here but I wanted to hear from all of the great UK people...

Thanks for the input!

(Hmm...can't seem to attach the image. Posted on my ISP's space: **broken link removed** ...after attachments are working (will try again after I sleep) I'll re-post it here as an attachment. Enjoy!)

Q1 looks like it helps prevent the battery from deep discharging.

C1 and D6 don't make sense. Check them again.

Missed a trace from the + side of C1 to the + supply.

I think everything else is correct; the only reason why I missed it was that there were multiple pads really close; thought it was fiberglass and not copper.


Still any ideas?

It looks to me like BR-1 charges the battery when AC power is present, and BR-2 powers the leds when AC is present. C1 is just the filter cap for BR-1.
When AC power is on BR-1 charges the battery through D5 and R1 and also powers the led at P3. Diode D6 applies a voltage to Q1 this holds the transistor in cutoff so it doesn't conduct. When AC power is lost you will not get this voltage through D6 and R3 biases Q1 into saturation which then puts power to the leds.

OK I got it all figured out.

Thanks for your help everyone.