The circuit does have a flaw; the transistor base bias assumes that the voltage across C1 goes 'low' when the power fails. But the only load on it is the no-load current drain of the LM317. I would add a 560 ohm 1/2-watt resistor across C1, and would change R15 to 270 as well.
And of course I would use the LED resistors as shown in the original circuit. Why leave out resistors? Do you prefer building things that don't work?
I would think T2 should be performing better than your figures indicate, since it is merely a switch in the circuit. Circuit description indicates it is capable of l.5 A with heat sink.
Of course, your "clocking" it up to 6 times the designed number of LEDs might have a bearing on the situation.
T2 needs a reasonable base bias to perform as a switch. With 1k in series with the no-load bias of the LM317, it's getting at most 2mA. At the worst case saturated gain of 10, this would supply 20mA to the load. Obviously the transistor being used is doing better than this.
This is why I suggested increasing the base current to 6mA. It still might not be enough, but I wanted to keep the resistors 1/2 watt for the convenience of the OP.
I planned to put just one 5.6ohm 5 watt resistor with 60 LEDs to step down from 6v to 3.2v to delive 1amp but the voltage automatty reduce without using any resistance and current drawn is only 450mA.
60 LEDs required 1amp as the description said BD140 could support upto 1.5amp so i thought increasing number of LEDs should not be a problem.
@Nigel: I didn't put resistor coz when LEDs are operating i don't know for what reason the voltage automatically drop form 6V to ~3V on the driver transistor.
@mneary: I've already put 1/2 watt resistors. How to increase the base current?
T2 needs a reasonable base bias to perform as a switch. With 1k in series with the no-load bias of the LM317, it's getting at most 2mA. At the worst case saturated gain of 10, this would supply 20mA to the load. Obviously the transistor being used is doing better than this.
This is why I suggested increasing the base current to 6mA. It still might not be enough, but I wanted to keep the resistors 1/2 watt for the convenience of the OP.
By changing the position of variable resistor the brightness of LEDs increase but on the other hand the charging voltage decrease to just 2 volts and vice versa.
What can to done so the charging voltage does not decrease and base of BD140 gets sufficient current?
You set the variable resistor to the correct charging voltage. If you have a 560 ohm resistor in parallel with C1, the variable resistor should not change the brightness.
I don't think it has been mentioned directly yet, but LEDs like to be driven from a Current Source, just like flourescent tubes, and the LED brightness is proportional to the current through it and not the voltage across it.
All those resistors simulate current sources.
As I understand it, the voltage across the LED for a specified brightness is probably not controlled by the manufacturer and he may not care what it happens to be.
By changing the position of variable resistor the brightness of LEDs increase but on the other hand the charging voltage decrease to just 2 volts and vice versa.
What can to done so the charging voltage does not decrease and base of BD140 gets sufficient current?
Setting the voltage via the variable resistor should have no (zero) effect on the LEDs, since that part of the circuit is inoperative when power to the mains is cut off. When power is off, T2, a PNP transistor, switches on and allows a completed circuit from battery through the LEDs and resistors, to ground. The rest of the circuit is dead, as per design.
When power is returned, then the LM317 circuitry returns to life, and replenishes the battery. (The PNP transistor is off, thus the LED circuit is broken).
You set the variable resistor to the correct charging voltage. If you have a 560 ohm resistor in parallel with C1, the variable resistor should not change the brightness.
I have designed a PCB including the changes as suggested: 560Ω/0.5W resistor across C1 and base resistor of T2 reduced from 1KΩ to 270Ω. Additionally I put four 47nF capacitors across each connection of the bridge rectifier. (makes the LM317 happy) The transformer used is a Gerth print transformer type 421.1-9 with max 556mA.
The entire design is purely single sided.
The design can be used as one single board or might be split into two as outlined in the dimension drawing. Connections between boards are marked A-A1 and B-B1.
Everybody interested in the layout (and schematic) is welcome to obtain it free via email. It is made using Eagle v. 4.16.
My knowledge in electronics is limited. I built this circuit.
I had a problem. The variable 2k2 has one point on negative and other point and middle point common with adj LM317 and collector of BC548. When i do this bridge, capacitor and LM gets really hot.
Then in changed and made on point of variable on negative other point on adj of LM317 and middle point on collector of BC548 and everything worked perfectly OK.
Am i reading the schematic wrong or what?
Secondly we are having 6-8 hrs loadshedding (power cut) per day in our country and my 6v 4.5AH battery gets fully charged in 20 hours, can i make some changes to it so that it can charge in just couple of hours?