BJT - understanding its operation

V2 is the dc power supply.

I'm still learning about more general models, myself.

Well, they don't show one of them. But it's there. It's just usually not symmetrically built, so reversing the emitter and collector leads won't get you the same behavior. It's supposed to go in one way. But I understand that there are rare times when you might prefer to reverse it and use it the other way. Not that I've done so for any design purpose, though.

It's good to hear that the book is okay.

Yes.

It does. It affects the voltage seen at the collector when the BJT is "off" and there is effectively no Ib current. However, when the Ib current is active and substantially driving the BJT into saturation, the collector will be close to or less than a single diode drop above the emitter. This is because, in saturation, the base-collector diode is generally no longer reverse-biased and is starting to conduct in the forward direction. Whatever collector current is required to reach that point will flow, assuming the BJT's base drive is sufficient and maintained.

How that takes place is another story. You might take a look at John Popelish's recent contribution here to a thread titled, "transistors: so confusing!!" on his 'drunken bum on a crazy street' theory.

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Yes it is. It certainly affects the current flowing out the emitter.

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Like I said, it does.

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Look at it like this: V1 will cause 5mA to flow thru Rb, into the base and subsequently out the emitter to ground. Rc will allow up to 100mA to flow thru it and into the collector and out the emitter to ground (assuming no current is being drawn from Vo). As long as the DC beta (Hfe) is at least 20, a 5V pulse from V1 will cause the transistor to turn on and shunt all current available from V2 to ground. This has the measurable effect of lowering the voltage at Vo to something very close to ground (0.2V in your case). You should think of the BJT transistor in terms of current and not voltages as that's what it is all about. Transistors amplify the current applied to base by the beta of the transistor and will allow that amount of current to flow thru the collector. Something like a water valve allows you to apply a small amount of force to regulate a much larger force.

Sorry if it's not a very clear explanation. I'm sure others will do much better. :-)

Yes indeed, that's about all you need to know to figure out how this circuit works. Never mind about the internal pn stuff. V1 controls the transistor. So with V1 high, the transistor on, the transistor grounds out the node where you see Rc, the collector, and Vo all connected and thus you see essentially zero volts (actually 0.2) at Vo.

Now put V1 low. That turns the transistor off. Now with the transistor's collector/emitter terminals looking like an open circuit, the voltage at Vo equals V2 and Vi (why did they label it twice?), and you have your sample. This assumes that whatever you are using to detect the voltage at Vo has a high impedance, so it doesn't affect things by loading down the circuit.

You could build this circuit on a breadboard. Connect voltmeter leads to Vo and ground. This is your sample detector. Use a switch to turn the 5 volts on and off (the circuit drawing omitted that part). Use some low voltage at V2, like a twelve volt battery. The voltage on your meter's screen will either equal V2 or read very low (just a few tenths of a volt), depending on what's going on at the transistor's base.

V0 is what you want to sample.

Here's a link to John's drunken-bum description:

One thing I didn't realize, until many years had passed, is that the emitter has much higher doping than the other junctions, so nearly all the current in a transistor is from charges that originate in the emitter.

Mark