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My setup is simple: a NPN transistor is connected to a DC motor and a battery. The series goes like this: positive terminal - motor - collector; emitter - negative terminal. The transistor base is a loose, long wire.

I would assume that shorting the base and collector would turn the transistor into a diode, allowing current to pass. However, this does not happen. Instead, shorting the base and emitter makes the motor turn. Is it possible that this is not a NPN transistor and some other type instead? Or am I mistaken somewhere else?

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    \$\begingroup\$ What is the part number on your transistor? \$\endgroup\$
    – Dave Tweed
    Commented Apr 4, 2013 at 21:24
  • \$\begingroup\$ Easy to tell NPN from PNP (and positively identify the base) with a multimeter. \$\endgroup\$
    – user16324
    Commented Apr 4, 2013 at 22:19
  • \$\begingroup\$ The part number is КТ361Б. I salvaged it from a Soviet alarm clock. Google says that it is, in fact, a PNP. Silly me. \$\endgroup\$
    – Moshanator
    Commented Apr 5, 2013 at 7:59

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Maybe it's PNP and upside-down (or right way up): -

enter image description here

As soon as you conect the base, the current flows from emitter (or collector) through the base. In other words it's acting like a diode

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  • \$\begingroup\$ "like a diode" but not quite. If the collector was disconnected, then you have only the base -> emitter diode, through which current will flow, with a normal diode drop of say 0.7V. If you connect as in the diagram, collector current is beta times the base current. With beta of 100, that means only 1% of the current flows through the base, so the base voltage can be lower, perhaps 0.6V. If the base and collector were not shorted together Vce could go as low as 0.2V, "saturation", but here Vbe (which is tied to Vce) can't be that low as that would turn off the transistor. \$\endgroup\$
    – gwideman
    Commented Jul 31, 2020 at 21:59

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