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I am just now learning about how transistors work, so please let me know if I have a misunderstanding.

I ran a simulation of the circuit drawn below, and the results were that current flowed only in the bottom loop (with the resistor and switch). When the switch is open, the current flows to the base, thus turning the transistor on (and allowing current to flow through the upper loop with the transistor).

However, my question is: when the switch is closed, why doesn't current flow to the base of the resistor? Can't it pass through the base and through the right side emitter (in NPN transistor), thus creating a sort of a "parallel" circuit? Again, please forgive me if I have a misconception, and share your knowledge! :)

enter image description here

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In order for current to flow in the base of the transistor, there must be about 0.7 V (maybe 0.5 to 0.6 V for a small current) between the base and emitter.

But you have a short circuit between the base and emitter of the transistor.

If you had 0.7 V (or even 0.1 V) between the base and emitter, then in theory an infinite current would flow through that short circuit. (Practically it would be 10's of A until the wire melted) But your supply can't provide infinite current (or even 10's of A) because of the 1 kohm resistor.

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  • \$\begingroup\$ Thanks for the reply. So, if we cross the 0.7V, then the transistor turns on, right? But won't that allow infinite current to pass through the upper loop through the transistor (because there is no resistance in that path)? And wouldn't this mean that current would stop flowing through the resistor (since the other path has 0 resistance), turning off the transistor (and this repeats)? \$\endgroup\$ – F16Falcon Oct 11 '18 at 2:58
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    \$\begingroup\$ No, the transistor has a finite "current gain", so the collector current is limited to a finite multiple of the base current. \$\endgroup\$ – Peter Green Oct 11 '18 at 3:20

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