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In the circuit below (schematic diagram), the command from a microcontroller allow to send the 5V to the output through two transistors command Q1 and Q2.

schematic

simulate this circuit – Schematic created using CircuitLab

My question is: what happen if the transistor Q1 is short circuit, the output can be energized even if the command from the microcontroller is off ?

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closed as unclear what you're asking by Olin Lathrop, Daniel Grillo, Ricardo, nidhin, Scott Seidman May 15 '15 at 23:43

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ Your circuit will never work - for many reasons. The primary one being that Q1 can't make Q2 turn on in that configuration. Then there's the lack of resistors, etc, etc, .... \$\endgroup\$ – brhans May 12 '15 at 14:04
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    \$\begingroup\$ Q2 needs to be a PNP transistor if you want it to turn on when pulled to ground. Also it is bad form to leave transistor gate floating like that (Q2). Agree with brhans on the resistors. Even theoretically this wouldn't work as Q2 would always be off. If you were to fix the circuit so Q1 being on turns Q2 on, then yes, obviously Q1 being shorted would turn Q2 on. \$\endgroup\$ – I. Wolfe May 12 '15 at 14:07
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    \$\begingroup\$ To turn Q2 on you must drive some current into its base, and this is not going to happen (not even in principle) with that circuit. If Q1 is ON, Q2 base is grounded, so Q2 is off. But when Q1 is OFF, Q2 base is floating and there is not path for any current to flow into it, so (as other have already said) it will remain OFF. \$\endgroup\$ – Lorenzo Donati May 12 '15 at 14:14
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    \$\begingroup\$ The emitter of Q2 must connect to the positive power, not the negative as you show. \$\endgroup\$ – Wouter van Ooijen May 12 '15 at 15:09
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    \$\begingroup\$ As for what seems to be your question: When Q1 conducts, whether because it is commanded so by the uC or because it has failed short, Q2 will be on and power will be provided to the output. \$\endgroup\$ – Wouter van Ooijen May 12 '15 at 15:11
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So lets assume you have a proper working circuit like below, where putting positive voltage at the signal input will turn Q1 on, which will pull Q2 to ground, turning Q2 on (means output is on).

schematic

simulate this circuit – Schematic created using CircuitLab

In this case, it does not matter if the signal is positive, or Q1 is shorted. If Q1 is shorted, Q2 is still pulled to ground, and the output will still be on.

So to answer your question, if Q1 is shorted, the output will always be on, regardless of the signal input. If Q1 is working properly, the output will be on/off as the signal is on/off.

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