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While working on a design last night, I had a question pop up.

Is there any risk in bypassing a transistor, with a jumper/switch? Or shorting the Collector and Emitter pins with a wire? Is there a difference if the transistor is on or off?

For example, the transistor will carry a 166 mA load, driven by a 8 mA base current (through Appropriate Base Resistor, omitted in the schematic) to put it into saturation. If I short the collector and emitter pins so the load bypasses the transistor (which means there is no load for the transistor), but still drive the base with that 8mA, Will it overheat or die, or will it simply pass the base current without issue?

schematic

simulate this circuit – Schematic created using CircuitLab

The switch might be momentary, or it could be left on for hours, if that makes a difference.

Slightly related, would a Mosfet bypassed the same way also have any issues?

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    \$\begingroup\$ Note that almost exactly the same thing happens if you use another transistor for the switch, on the left, whereby you basically have an RTL gate \$\endgroup\$
    – Kaz
    Commented Nov 21, 2013 at 0:35
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    \$\begingroup\$ en.wikipedia.org/wiki/File:RTL_3-Input_NOR_Gate.svg \$\endgroup\$
    – Kaz
    Commented Nov 21, 2013 at 0:35

3 Answers 3

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Shorting collector to emitter causes no problem in either circuit. The current into the base is likely to be partly shared by the collector when shorted but this should not cause an issue.

Ditto source and drain on a mosfet.

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If you add base resistors to both transistors, there is no risk in bypassing the transistors in your diagram. You basically turn the transistor in a diode and you have to limit the base current / GPIO-pin current from your microcontroller as the base-emitter voltage will be about 0.7V.

You must use base resistors anyway, even without the bypassing circuit.

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  • \$\begingroup\$ The base resistors were omitted just to unclutter the diagram, as the real question is simply about the bypassing. Typically ~1kΩ. So would leaving the NPN base high with the 8mA sized resistor, will cause no issues when the C-E is bypassed? \$\endgroup\$
    – Passerby
    Commented Nov 20, 2013 at 21:24
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Base current is irrelevant to the collector current (the reverse is false though)

Anything entering base will have to exit somewhere, and that's the emitter even when no current is passing through collector (shorted in this case)

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