Please help me understand how the emergency lighting circuit work (China)

[samy555]

This is the electronic circuit,,,,
When the main 220Vac is exist the transistor off

When the main 220Vac is OFF the transistor ON

I can not understand how the circuit works, what is the mystery in using D1 & D2?
It's amazing that the battery can not make the transistor On!
The other thing is that the main power alone control the state of the transistor and in a reverse manner ???
thank you

 

[Dan Mills]

With mains off, the base is biased hard on via R3 and all the diodes are effectively at either 0 or reverse bias and thus effectively out of circuit.

With the mains on, the current can flow in the bridge rectifier establishing ~4.7V or so across C2 which charges the battery via D2 (Putting Q2 emitter ~0.7V above the lower end of C2. D1 then serves to clamp the base of Q2 also at 0.7V above the bottom of C2, Q2 Vbe is thus ~0V and the led is off.

It is quite elegant, for all that the use of the ground symbol on the schematic is IMHO dangerous (ALL of the circuit is live when power is applied), and a modest series resistor in series with C1 would help reliability in the presence of transients.

HTH.

Regards, Dan.

 

[samy555]

With mains off, the base is biased hard on via R3 and all the diodes are effectively at either 0 or reverse bias and thus effectively out of circuit.

With the mains on, the current can flow in the bridge rectifier establishing ~4.7V or so across C2 which charges the battery via D2 (Putting Q2 emitter ~0.7V above the lower end of C2. D1 then serves to clamp the base of Q2 also at 0.7V above the bottom of C2, Q2 Vbe is thus ~0V and the led is off.

It is quite elegant, for all that the use of the ground symbol on the schematic is IMHO dangerous (ALL of the circuit is live when power is applied), and a modest series resistor in series with C1 would help reliability in the presence of transients.

HTH.

Regards, Dan.

Thank you very much Dan
It was an excellent answer and I understood well
Some simple questions remained:
(1) Why you said "hard" in "the base is biased hard on via R3 " ,, I think you mean something?
(2)

With the mains on, the current can flow in the bridge rectifier establishing ~4.7V or so across C2

I did my calculations and get 2.178V
I know I'm wrong!!!!

Thank you very much

 

[Dan Mills]

The usual rule of thumb for a bipolar transistor is that to get it switched hard into saturation requires that Ib be greater then about 1/10th Ic, which is clearly the case here as R3 < (10 * R4) even ignoring the LED voltage drop.

|Xc1| = ~ 3K ohms, and output voltage is clearly limited by the battery charging voltage (Which is small), so current in C1 is ~ 230V/3K = ~100mA which seems reasonable for charging the battery, the voltage is set by the fact that the battery is essentially a voltage source and will maintain a substantially constant voltage of around 4-5V or so over a wide range of charging currents, D2 is in series with the battery so you need to add 0.7V there. Hence 4.7V or so.

Your capacitive divider would be correct if the bridge rectifier was not present, but its presence means that C2 is a smoothing cap that cannot discharge back into C1 so it does not play a part in a voltage divider, instead it serves as a place to dump the small currents flowing in the diodes when the mains voltage dips to zero every 10ms. Once the mains is removed, it discharges via R3,D1 until the base voltage of Q2 rises sufficiently to turn on the transistor and light the led.

It is really quite elegant in a sort of maximally reduced BOM cost way.

 

[samy555]

Thank you Dan, its now OK