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I have seen a few examples of a low voltage warning LED using two NPN transistors, like this;

enter image description here

My question is, can this be simplified further using only one PNP transistor, if R2 were a value so that the PNP base current was low enough the transistor would be 'on' below a set voltage? Something like this;

enter image description here

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  • \$\begingroup\$ The original one is a self-purposing circuit - If you build it, you'll definitely need a low-battery warning because it draws up to 3.5 mA using the stated input range! \$\endgroup\$ – pipe Jul 27 '16 at 6:43
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No. For the PNP transistor to turn on, it's Base Voltage needs to be it's VB-E Forward Drop. For a typical Silicon Transistor, that's 0.6~0.7V. In your proposed circuit, the Base is held high through R2, and at no point will ever go lower than it. So it will not turn on.

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  • \$\begingroup\$ The advantage of the second circuit, however, is that is has much loew stand-by power consumption. \$\endgroup\$ – Curd Jul 27 '16 at 7:20
  • \$\begingroup\$ Ok I understand why this is a stupid question as it is... Can this be done using a PNP and resistors in any configuration? \$\endgroup\$ – JoshNZ Jul 27 '16 at 9:02
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Current needs to flow into the emitter and out of the base to make a PNP turn on. This can never happen in your proposed circuit. With no current flowing through R2, the base and emitter will always be at the same voltage and the transistor off.

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