Hi, the array contains 6 LEDs and supply is 5 volts.
by the way, if the condition for bottom LED failure is taken into account, and a divider is designed, then the voltages can be sampled and compared. Can i implement it by this way? I mean this will be the extreme condition right?
Dude, I am actually working on a street lamp which contains 2 channles of 6 LEDs in series. This damn thing powers up and delivers a optimal light at 5 volts itself. If you dont mind, here is the link to the datasheed of the LED
**broken link removed**
I am sorry if I caused any inconvenience.
By the way, which software you used for that simulation thing?
Here is the schematic I thought of:
View attachment 82504
Now my question is this: Is the above configuration reliable/feasible?
And my idea is to give the base each transistor to a MCU pin. And the program is to consecutively writing "1" on each pin in cyclic order thus turning on the tansistor which provides a short. Now, the short which again lights up the circuit can be taken as the place of defective LED. Can this be implemented?
Regards,
Anish
Yeah I can understand that dude. But for now, I am doing for the open circuit case alone.There is one major flaw in your reasoning though, you assume the faulty LED has gone open circuit, it is possible it has short circuit or even in rare cases, become a resistor and not emitting any light.
Yeah this approach also suggested by some friend I think. What do you mean by a "opto-coupler with current amplifier" ? Can you elaborate the circuit?One option would be to use an opto-coupler with current amplifier instead of the single transistor. The LED side of the opto would need the same voltage to turn it on regardless of where in the chain it sits.
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