Help with NPN Transistor Diagram?

"I don't get why there is a strait connection between the switch and the 0v (with the other 1K resistor between them). "

That's the voltage that energizes the base of the transistor creating the base current that flows through the emitter. This increases the conductance between the collector base junction and it all flows out the emitter to ground. The base current is the control current. A small current from the base to the emitter produces a much larger current from the collector to the emitter, and a small change in current from the base to the emitter produces a large change in current from the collector to the emitter. In this example they are using the transistor as a switch, like a relay. Think of the base current as the current that energizes the coil, and the collector to emitter current as the current that flow through the switch.

I don't like their circuit. It is contrived. They could accomplish the same thing with just the switch, the resistor, and the LED. It would illustrate the point they were trying to make better if the switch was in the base circuit rather than energizing both the base and the collector.

I like the diagram on this page better.

http://electronicsclub.info/transistorcircuits.htm

"Also, I am having a hard time understanding how many ohms my resistors should be."

That depends on the properties of the transistor and the LED.

Agreed, it's a somewhat odd diagram. The switch is in the supply, and turns *everything* on and off.

I'd have wired the switch shut, and inserted a switch in the wire between the supply and the junction of the two 1k resistors. That would just turn on the base current of the transistor (and hence the larger collector current, lighting the LED.) *All the following assumes such a circuit*.

We want to turn on the transistor hard enough so that the collector current is limited only by the components in the collector circuit- *not* by gain times base current. What is that collector current? The voltage is 9 (battery) minus LED voltage (about 2) minus the transistor collector/emitter voltage- negligible, if the transistor is turned on hard enough. Divide that voltage by 680 ohms; you get about 10mA.

The resistor going to the base must allow enough base current (9v - 0.7v Vbe)/Rbase so that the gain of the transistor times the base current is greater than the collector current. Even if the gain of the transistor is as low as 20, we need only 0.5mA in the base, and the base resistor can be 18kohms or so.

The remaining resistor just stops the base floating around when the switch is open. A bit arbitrary but 18kohms would be OK.

Does that help?

The schematic is stupid. You are right to wonder what the 1kΩ resistor is doing, the one that goes from the switch to ground (0V.) If it had been placed from the base of the NPN to ground, there might be a vague justification -- though probably not for 1kΩ -- perhaps as low as 10kΩ could be argued, though. One could also argue where it is located, I suppose (it's a resistive path from base to ground in case charge builds up somewhere.) But not 1kΩ. That 1kΩ is just a current path to waste power. If I were worried about giving a resistive path to ground from the base, I'd put it between the base and ground and use a much higher value. For the purposes of understanding the circuit, though, this 1kΩ resistor is in a separate circuit and has NOTHING to do with the NPN switch. So just delete it, mentally. It's not there. Ignoring it will change NOTHING about your understanding of the circuit.

The circuit is stupid in another way. The side of the 680Ω resistor that goes to the switch should probably be directly connected to the +9V battery connection if you are trying to focus on understanding the BJT as a switch. That way, the switch itself doesn't alter that part of the circuit and therefore add confusion. You are trying to understand the NPN switch as it may be found in practical circuits and in most cases where you are choosing to use an NPN as a switch you don't have the convenience of a mechanical switch that also gates power to the collector circuit. So you really should imagine that this side of the 680Ω resistor bypasses the switch and connects directly to the +9V side of the battery, instead.

Start with the NPN transistor -- it's the central feature of this circuit. (Well, so is the battery... but that one is easy to understand.)

In order to operate the NPN BJT as a switch, you want its collector to drop low -- very close to its emitter. You don't see it in the diagram, but the collector (back in the day) once included a diode symbol on it, as well. When the collector is very close to the emitter voltage, the base-collector diode is forward biased very much like the base-emitter. And it get it there requires a lot of base current compared to the collector current -- perhaps a base current that is ¹⁄₁₀th or ¹⁄₂₀th of the collector current. (For "normal" operation, the base current may be more like ¹⁄₁₅₀th or ¹⁄₂₀₀th or less, even. So this is a big difference.

So, we don't exactly know the voltage required by the LED, but if it is a red one than the voltage is probably no more than 2V. Assuming that the circuit is designed right, the collector will be very close to the emitter or about 0V. So the 680Ω resistor gets about (9V-2V) or 7V across it. This works out to about 10mA, if you compute 7V ⁄ (680Ω). The base current therefore needs to be 0.5mA to 1mA, or more. More is okay. But at 1mA you are already doing far more than enough in most cases.

Assuming the BJT is operating, the base-emitter junction voltage for silicon NPN BJTs is about 0.75V here. So the remaining voltage left over the "other" 1kΩ resistor (the one hooked to the NPN base) is (9V-0.75V) or 8.25V. This means more than 8mA. So... well, yeah. The NPN collector will certainly be very close to the emitter voltage with this much base current. But it's actually overkill. You could get by, probably, with a 10kΩ resistor there (825μA of base current.)

That circuit is a little whacky.

One is as initially current to base(shift current), to the 0 is for current without working of transistor hence as output of former chain.In emitter chain resistor to take U from it. It is complicated to count all this by myself. There can be filters and C for working with needed frequencies.