2

schematic

simulate this circuit – Schematic created using CircuitLab

I am using an NPN transistor (2N4401).

In the above schematic, charge is only coming in through the base, and not the collector.

My LED is dim, but still lit. Is this the expected behavior of a transistor, or am I doing something wrong?

My expectation was that there should be no charge flowing through the emitter since the collector is not connected. Obviously, when the collector has charge applied, but the base does not, then the LED does not light up. This is expected, but it's also happening the other way around.

Josh Beam
  • 325
  • 1
  • 2
  • 11
  • 1
    Your schematic says the transistor is a 2N3904 but your text says its a 2N4401. It doesn't matter in this case, but this sort of confusion and sloppiness can get your question downvoted and/or closed under different circumstances. – Olin Lathrop Jul 16 '14 at 12:32
  • 3
    If current is going in the base, where is it coming out? – Phil Frost Jul 16 '14 at 12:40
  • Also probably insightful: [Why would one drive LEDs with a common emitter?](http://electronics.stackexchange.com/q/57845/17608) – Phil Frost Jul 16 '14 at 12:41

2 Answers2

13

In such a circuit the emitter current equals the collector current PLUS the base current:

Ie = Ic + Ib

In normal situations the collector current is much larger than the base current, hence by good approximation

Ie = Ic

But in your case the collector current is zero, so the first equation degrades to

Ie = Ib

Which is exactly what you see: a very small current through the LED.

Another way to look at your circuit is to replace the b-e of the transistor with a diode. This gives the same results.

Wouter van Ooijen
  • 48,407
  • 1
  • 63
  • 136
9

The base-emitter junction of a transistor is just a diode. When the collector is disconnected, current can flow through R1, the base-emitter junction, D1 and R2, with a current of roughly 6.5V / 25270Ω = 260µA, which is enough to see.

When the collector is connected, R1 no longer limits the current to the LED, so you'll get more like 6.5V / 270Ω = 24mA through the LED.

Dave Tweed
  • 168,369
  • 17
  • 228
  • 393