I understand 99.9%, except one thing. I don't understand how you got the resistor value, to get 0.7v for the base.
If you knew the voltage of the PSU, which is 10vdc, and knew that the base needed 0.7vdc, then it's 10 - 0.7 = 9.3vdc to drop acrost the resistor.
When finding the resistor value for LED's, you need the LED's forward current, and the amount of voltage to drop across the resistor to find the resistor value. Not knowing what the current is for the base of the NPN Transistor, it seems impossible. But cheating, and using what you've given me, 0.000093A, is pretty easy then.
Then it's 9.3vdc / 0.000093A = 100,000Ω
Where did the 93μA come from?
Other than that, I understand what your talking about.
View attachment 100303
First off, we are looking at biasing the transistor input, which means we want to get current to flow through the base emitter. That's all we want to do right now, and we dont care as much how much current it is, as long as it is not too high for now. We assume that we are using a small transistor, for now, like the 2N2222A for example.
Second, knowing transistors like this, we know the base emitter diode drops about 0.7 volts at 'reasonable' current values. Even if it does not drop exactly 0.7v at 93ua, we take it as truth that it does, at least for now, and that is where we start the analysis. It may in real life drop 0.6v to 0.8v, but we ASSUME, without proof, that it does drop exactly 0.7v for now. Later in the study, we may want to modify this assumption to better fit real life data, but for now we accept the 0.7v as that gives us a starting point. In the end we will find that it's not that far off anyway for what we intend to do and that is to design the simplest amplifier using one NPN transistor of a variety similar to the 2N2222A. Another example would be the 2N4400 (or 2N4401) NPN transistor.
We can later look at what happens if the base emitter voltage changes later, but we dont need to do that yet because it does not affect the very basic operation we need to talk about first too much.
Also, the 100k is a somewhat random selection, for now. We just need to get some current into the base so we have a starting point. Later we may decide to change that, but when we get to that point we will then know why we want to change it. Right now we set up a simple operating point, then see what happens later when we start to apply a signal.
Note we have not even gotten to the output circuit yet, which involves the collector. That will be next. Also, we will assume a gain of the transistor but we have to look at one more thing before we get to that.
You mention an LED and getting that to light up by knowing the forward voltage for the LED. That is similar to the base emitter diode. We dont really know the exact LED forward voltage, but we can use some nominal value like 2.2v or 3.5v to get started. We selected a resistor based on this forward voltage and the supply voltage, then go from there. That is similar to the base emitter diode where the forward voltage is assumed to be a nominal 0.7 volts. So if you understand the LED circuit then you understand the base emitter diode circuit. The reason for trying to get a particular current through the base emitter is similar to trying to get a certain current through the LED. In the LED we want to light it at a certain brightness, in the transistor we want a certain bias current which will set the DC operating point. The DC operating point gets the transistor circuit ready for amplification. It can not amplify very well until it is biased correctly.
For now however, we are using a somewhat random bias point: 93ua through the base emitter, but later we may change this once we see the overall operation and how this current affects the operation in a more grande way.
We can not jump in with every possible detail of the circuit all at once or it will become a little confusing.
If you have any more questions about the base emitter biasing feel free to ask. To save a little time i'll introduce the next two concepts but we dont have to move on until you are ready.
The next idea is the idea of a constant current source. I am not sure if you had encountered one yet that's the only reason i mention this first. If you have that's great, but if not we'll look at this next.
A constant current source has two leads like a battery and puts out a current that is ALWAYS exactly the same, regardless of what we connect to it. If we have a 1 amp constant current source (sometimes called a CC source) and we connect a 1 ohm resistor across it, then we have 1 volt across that resistor because of Ohm's Law:
That's simple enough right?
If we had a 2 ohm resistor across it, we would have:
One additional thing though, and that is if we had a 2 ohm resistor in SERIES with the current source, we would also have 2v across that resistor because we assume that the constant current always flows, and so the 2 ohm resistor gets 2 amps even though it is in series with the CC source. This seems a little strange, but here we assume that the circuit is complete, that somehow the current gets back to the source just like in real life if there was a complete circuit. That's because in our analysis we will almost always have a complete circuit.
Think about this and get back to me if you have any problems with this idea of a CC. It's important to understand these basics before we move on, but i'll mention one more thing in the interest of time. We dont have to move this fast though, we can always stop and come back to any of these ideas, and we probably will anyway, but if you have any questions about this dont hesitate to ask.
The next idea is about a controlled current source.
The controlled current source has four leads: two for output and two for input.
There are at least two types:
1. Voltage controlled current source.
2. Current controlled current source.
The input is either a voltage or another current, and the output is a current. The output current is related to the input by a formula which is just usually a gain 'A' such as 1, 2, 10, 100, etc. This means that if the input is 1 then the output is A*1. So the formula is:
If the input is a voltage, then we have a voltage controlled current source and the output is:
and if the input is a current then we have a current controlled current source and the output is:
Note in each case the input sets the output current through a gain factor of 'A'.
In the transistor we can use either type of source, but a common type to use is the current controlled current source which has output:
and here the factor "A" is often called the Beta.
The Beta is often listed on the data sheet, but we will start with a typical Beta and go from there. The Beta also varies quite a bit in real life, but we will assume, for now, a constant Beta. We can later look at what happens when the Beta varies, but we need not concern ourselves with that right now.
So the concepts we covered so far are:
1. Biasing the base emitter diode is like biasing an LED.
2. A constant current source puts out a constant current, has two leads like a battery.
3. A controlled current source puts out a current based on it's input, and for a current controlled type the input is another current which often comes from some other place elsewhere in the circuit. This is a four terminal device with two leads for input and two leads for output. The output for the current controlled current source is Iout=A*Iin. A typical value for A might be 100, so if we input 93ua we will get as output:
Iout=100*93e-6=9.3e-3 which is 9.3 milliamps (9.3ma).
Note we used just some simple algebra here to calculate the output current.
If any of these three is not clear now is the time to mention that or any questions you may still have. If you would like to make a diagram of the current source and the current controlled current source that would probably help, and i can update it in order to help explain certain things if needed. I mention this because you make some pretty good diagrams and that will help you internalize this too.
Believe it or not, we are already about 1/3 of the way there on the road to amplification
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