However, can there not also be small current that has passed through the green light which enters the base of emitter 2 (as the base voltage is 0.7V and there still exists a voltage differential across the 47k resistor) which will also allow current to flow from collector to emitter of transistor 2 and cause the red LED to glow. Attach the 9V battery to the battery clip.
The IC contains adjustable reference and accurate 10-steps divider. Lm3914 drives led’s, LCDs and vacuum fluorescents. The operating voltage of this IC is 3v to 25v DC. The circuit will work even if the battery voltage is 3V. I also understand that when the battery voltage is less than 6.9V the first transistor will not allow current to flow through to its emitter, and due to the 47k resistor there will be only a very small current through the green LED into the base of emitter 2 - not enough for it to visibly be turned on - which in turn allows current to flow through from collector to emitter of transistor 2 and causes red LED to glow. The circuit is designed to monitor 10V to 15V DC. This current must first pass through the green light, which turns it on. The LoBat low battery indicator circuit is designed for use in electric guitars and basses with active pickups and electronics that use 9 volt batteries. Can you please explain it to me? What I understand so far is that when the battery voltage exceeds the zener voltage + base emitter voltage of the BJT (6.2 + 0.7 = 6.9V) the zener will reverse breakdown and there will exist small current that is able to enter the base of the first transsitor, which allows current to flow from the collector to the emitter. Hi I haven't built the circuit but viewing the comments below and analysing the circuit myself by pen and paper I don't see how the red light will ever be switched off.
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