I'm not understanding how NPN transistors really work in this scenario. In this image, I have highlighted what I understand to be the collector and emmiter. My understanding is that power flows from the collector, TO the emitter right?

In this example, the base either turns the circuit on or off. I get that the circuit is open when the transistor isn't being fed positive power. But, why is the emitter running to ground? Shouldn't it be the other way, collecting and emitting to the LED?


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    \$\begingroup\$ I'm not really clear on what you're asking. What do you mean by "collecting and emitting to the LED"? \$\endgroup\$ – Stephen Collings Mar 21 '13 at 17:19

An analogy may help to visualize this:

Think of the transistor as a valve or faucet. The base is the knob, the water tends to flow from the positive side (storage tank) to the ground (drain), if you follow the normal "current flow" directions.

The LED is like a little transparent glass section in the pipe, with a small ball loosely held in that section.

When the faucet is opened, water will be allowed to flow, and the little ball will jump around due to the water's flow.

This will happen whether the LED is above or below the faucet section.

Now for the case of electron flow, as opposed to conventional current flow direction.

Consider the same pipe and faucet, but with the ground being a source for some gas, say natural gas at high pressure underground.

The Vcc is the open air, normal barometric pressure.

Again, as the faucet is opened up, the gas will flow up the pipe, the little ball will bobble around. Again, the glass pipe section (LED) could be before or after the faucet, it won't matter.

I hope this analogy helped.

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  • \$\begingroup\$ I really liked this analogy until I got to the transparent glass and ball. It didn't make any sense. But, your natural gas analogy made much more sense. So +1 from me. It's hard for amateurs like me to NOT think in conventional current flow. Especially when water analogies are used all the time. But the underground pressurized gas analogy was spot on. Thanks for that. \$\endgroup\$ – cbmeeks Jul 15 '15 at 13:07

But, why is the emitter running to ground? Shouldn't it be the other way, collecting and emitting to the LED?

It is, but the charge carriers are negative electrons, so they move (well, nudge each other along, really) in the opposite direction from what the idea of conventional current leads you to expect.

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  • \$\begingroup\$ So, does that mean it "pulls" to ground, thus completing the circuit? \$\endgroup\$ – JohnnyStarr Mar 21 '13 at 17:16
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    \$\begingroup\$ In the sense of conventional current, yes. But that doesn't help you with the names of the terminals which are with respect to the negative charge carriers in use. \$\endgroup\$ – Chris Stratton Mar 21 '13 at 17:20
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    \$\begingroup\$ @ChrisStratton and to add to the confusion, the arrows in the diagram are with respect to the conventional current direction \$\endgroup\$ – endolith Jun 24 '14 at 13:19
  • \$\begingroup\$ @endolith so, it sounds like it would be more beneficial (to amateurs like me) if the drawing would have the arrow pointing out of the COLLECTOR and not the EMITTER? Because that's the direction the current is really flowing right? \$\endgroup\$ – cbmeeks Jul 15 '15 at 12:59
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    \$\begingroup\$ @cbmeeks The conventional current is flowing into the collector and out of the emitter. The electron current is flowing into the emitter and out of the collector. I don't know why they're called "emitter" and "collector". It's best to just not think about electrons: physics.stackexchange.com/a/17131/176 \$\endgroup\$ – endolith Jul 15 '15 at 13:49

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