![]() We use that to do the remaining calculations: Now, the first transistor is only switched on when the battery is between 9V and 7.2V to keep the LED transistor from turning on. R(pull-up) = 100kOhm and R(base) = 33kOhm (makes 133kOhm together). Let's choose 133kOhm to stay well on the safe side: The current into the base of the LED's transistor then becomes 25uA at 7.2V and 20uA at 6V (divided by hfe, which is assumed 100). This again drives an assumption, as no specifications are made, I will assume the LED current is 2mA and that the red LED adheres to commonalities, needing about 1.9V at that current. a BC547C type may go up to 500 in the right circumstances.įirst off, you mention a low-power LED. Further, for ease of the example I am going to assume a hfe of the BC547 of 100. I am going to make the assumption you want it to work in the range of 6V to 7.2V, so we'll use those two as the extremes of our calculation. You want the LED to turn on at about 7.2V. ![]() I would propose a little math to find the best solution to your exact situation with lowest losses: ![]() This means some experimentation or graph-hunting may be required to find your value. Reverse biased zeners have their specified knee voltage at a specied current, your schematic allows for very small "leakage current", but this also influences the biasing of the zener. Your calculations based on the Vz of the zener are a little bit dependant on the zener current.
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