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I have been trying to understand how an NPN transistor works, specifically in the case of a photo-transistor.

As far as I can tell, it should be something like this:

NPN Transistor diagram

I don’t understand how current can flow between the Base and the Collector since the Collector is n type and is connected to positive terminal, which would surely make the BC reverse biased?

The base in the diagram would be connected to a photo-diode, allowing current to flow when light levels reached a certain intensity.

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    \$\begingroup\$ what did the site you grabbed this image from say and what part did you not understand? \$\endgroup\$
    – old_timer
    May 16 at 1:29
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    \$\begingroup\$ Is there a separate photo diode or is the transistor itself a photo transistor? You mention both in your question. If there is a separate photo diode, please include that in the diagram. \$\endgroup\$
    – AJN
    May 16 at 2:24
  • \$\begingroup\$ I have edited the question so it is more specific. Thanks. \$\endgroup\$ May 16 at 9:12
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Electrons (majority carriers) travel from the n-type emitter region to the p-type base region because of the forward biased BE junction. Once in p-type base region these electrons become minority carriers. Because the base is thin and lightly doped only a few of these electrons recombine to produce the base current, the majority are swept across the reverse biased CB junction attracted by the positive potential at the collector. It is because the electrons are minority carriers in the base region that they can travel across the reversed biased CB junction as it is quite possible for minority carriers to travel across a reversed biased PN junction.

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