I have one Meanwell LED driver of 27V, 2300mA on hand and will be running 2 strings of 6 LEDs in parallel. Should I follow the same circuit with same component value or should I need to change the value of the resistors to my configuration `The addition of a small signal transistor as a current monitor protects the LEDs from being overd
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riven in the case of any LED failures. If LED1 LED5 fail open circuit, then the current in the second string falls to zero as before. However, if LED6 –LED10 fail, then the current increases in the first string until the voltage developed across the 1. 5 Ohm emitter resistor reaches around 0. 7V, thus turning on the BC337 transistor and pulling the base voltage of the power transistor to ground and limiting the current. With the component values given in the circuit, the measured current limit was 445mA with String 2 open circuit. " That circuit will NOT work at 1. whatever amps. Replace the 1. 5s with ~. 47s, the BD139`s with something bigger (anything with a DC current gain over like 80 at 1A and a maximum continuous current of 2A or more) and the 680s maybe drop to 470. Now that I think about it this circuit kinda sucks, if you get a transistor with a lower end gain its going to eat up a buttload of efficiency. That circuit will NOT work at 1. whatever amps. Replace the 1. 5s with ~. 47s, the BD139`s with something bigger (anything with a DC current gain over like 80 at 1A and a maximum continuous current of 2A or more) and the 680s maybe drop to 470. Now that I think about it this circuit kinda sucks, if you get a transistor with a lower end gain its going to eat up a buttload of efficiency. If built as shown, string 1 is going to be clamped at 460ma or so, read the blue paragraph as to exactly how. String 2 is going...
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