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I do not think the connection to BD139 is correct; it is unlikely to work.
You should modularize the circuit; even if it means more components. You will see that modifications will be easier and maintenance negligible.
1. Lamp driver should have some current control (brightness) and switched on /off with a transistor that works like a switch.
2. Battery charger with over and under voltage cut-off. It should be permanently connected with trickle charge capability. The battery capacity, the load power consumption and the charger current should be properly evaluated.
3. Small LED indicators that show (i) battery charging (ii) replace battery (battery voltage lower than the under voltage cut-off value) (iii) mains power on (iv) battery down and mains power off (how the led will run then)
4. I suggest LEDs connected in series and the batteries too be connected in series.
I'm probably stating the obvious but looks like the idea is to have the pnp act as a nc switch. As long as mains power is on there is power to the base and it would act as open switch. Maybe I'm missing something, but I would not think this would work as intended.
I do not think the connection to BD139 is correct; it is unlikely to work.
I do not understand what you mean?
Do you mean the connection method for BD139 is wrong? or do you mean i have to replace that transistor with another one?
You should modularize the circuit; even if it means more components. You will see that modifications will be easier and maintenance negligible.
1. Lamp driver should have some current control (brightness) and switched on /off with a transistor that works like a switch. Do you mean change the 4,7 Ohms with a variable resistor?
2. Battery charger with over and under voltage cut-off. It should be permanently connected with trickle charge capability. The battery capacity, the load power consumption and the charger current should be properly evaluated. True
But i don't know how can i make that?
3. Small LED indicators that show (i) battery charging (ii) replace battery (battery voltage lower than the under voltage cut-off value) (iii) mains power on (iv) battery down and mains power off (how the led will run then) True
But i don't know how can i make that?
4. I suggest LEDs connected in series and the batteries too be connected in series When LEDs connected in series the voltage on LEDs will rise and when the batteries connected in series the li-ion battery charger can't be used in this condition and i have to use BMS board?!
Let me assume that the power transistor pin labels are accurate. Try to see how the LED can get current at all; It will be easier to understand, for you and also for all of us, if you make a schematic diagram.
Commonly used, base acts as the current pin that controls the flow of current through the collector and emitter. In your circuit, the load (LEDs) is connected via the base pin which is wrong because the current at the base in input and not the other way.
If both transistors are conducting at the same time, the battery will be shorted.
Let me assume that the power transistor pin labels are accurate. Try to see how the LED can get current at all; It will be easier to understand, for you and also for all of us, if you make a schematic diagram.
Commonly used, base acts as the current pin that controls the flow of current through the collector and emitter. In your circuit, the load (LEDs) is connected via the base pin which is wrong because the current at the base in input and not the other way.
If both transistors are conducting at the same time, the battery will be shorted.
Medhat972, you can see why conventional schematics are drawn based on components between nodes instead of by physical layout.
Look at the route the current has to take between the battery and the LEDs and you will see what is wrong. We can see what you are trying to do but if we told you exactly what was wrong you would learn nothing.
hint: follow the path from the battery, through the transistor, through the 4.7 Ohm resistor, through the LEDs and back to the battery.
The base of both transistors acts as the current key that controls the flow of current through the collector and emitter.
NPN (BD139) depend on PNP (BC557) ===> the first one will not woke without approve from the second one.
When the main power (220V) is on, the base of (BC557) prevent current to flow between the emitter and collector ===> The NPN (BD139) is off ===> the LEDs is Off.
When the main power (220V) is off, the base of (BC557) will allow the current to flow between the emitter and collector ===> The NPN (BD139) is on and the current will flow between the collector and the emitter ===> the LEDs is on
Going back to an earlier point, it was suggested you join the batteries in series and the LEDs in series. To connect the batteries in series you need a 7.4V charger so I wouldn't advise that but the LEDs should ideally be wired in parallel with a resistor in series with each one instead of a single resistor (4.7 Ohms) for all of them. The reason is that not all LEDs have exactly the same forward voltage so if you join them directly in parallel, there is as good chance that they will not share the current equally and one or more may be damaged. If you wire a resistor of about 22 Ohms in series with each LED it will balance the current while still giving the same brightness.
I have a similar device at my home where the electricity often fails but it only turns on if the electricity fails AND it is dark so the battery isn't used up if it is still daytime. It also recharges the battery from daylight as well as AC so it works even if the electricity has failed for long periods.
What purpose does the transistor serve? It looks like the battery would keep the leds on all the time the way I'm seeing it. I'm puzzled!!!
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I was thinking the emitter pin in his second schematic had to go to positive of battery to give it half a chance of working as intended. If this is not correct would you mind explaining why?
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Emitter of 139 transistor
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Maybe not realized but there are now three schematics and all connected differently.
If you are putting the LEDs in parallel, it is best to add a small resistor in series with each of them.
What it means that you connect a small resistor to the black leg of each LED before you connect them together and connect to the positive end of the battery.
I am surprised you did not ask WHY?
Even if you did not ask, I am going to tell you. You should use the same idea whenever you use a diode. LEDs are basically diodes and once they start conducting, they have a low impedance.
It simply means that the current increases by a large value for a small increase in the voltage (resistance is very small). The point (in voltage; the x-axis) where this happens is the knee voltage, that is the time the diode starts conducting appreciable current.
If we change the voltage by a small amount the current increases by a large amount the diode will get spoiled. Only way to limit the current is to have a series resistance.
If you connect two diodes in parallel, there is no guarantee that their knee voltages will be exactly same and both will share the load current equally. The diode having the slightly higher knee voltage will share less current.
If we put some series resistors with the diodes, they will be forced to share the load current equally (approximately equal) because the resistor will drop some voltage proportional to the current.
You can use the same idea for zener diodes too; they cannot be connected in parallel because it is very difficult to guarantee the exact same knee voltage.
You use the exact opposite idea for batteries; if you connect them in parallel, it is not guaranteed that they will have exactly same voltage and the cell with the higher voltage will force current into the other cell. That will somehow force both the cells to have the same voltage always but if one goes bad, the other one goes too!!
Now you connect the transistor (collector) to the other end of the LED array (together). For NPN transistors, the collector should be at a higher voltage than the emitter. So the emitter should be connected to the ground. The base is the handle which will be used to turn on or off the set of five LEDs. A small positive current into the base will turn on the transistor and the LEDs will light up. If you connect the base to the ground, the LEDs will turn off.
I believe the lights will still burn all the time according to your last schematic. If you notice the ground and positive from battery are directly connected to your led string. A suggestion would be to take the left side key/switch out and place it on ground on the opposite end of your led string. This could be used as a manual test switch for your leds. I'm also a newbie and I have a lot to learn yet and this thread probably helped me as much as it's helped you.
I believe the lights will still burn all the time according to your last schematic. If you notice the ground and positive from battery are directly connected to your led string. A suggestion would be to take the left side key/switch out and place it on ground on the opposite end of your led string. This could be used as a manual test switch for your leds. I'm also a newbie and I have a lot to learn yet and this thread probably helped me as much as it's helped you.
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