# Single NPN Transistor Amplifier Help

#### MrAl

Joined Jun 17, 2014
7,891
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?

View attachment 100303

Hi,

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:
V=I*R
V=1*1
V=1 volt.

That's simple enough right?

If we had a 2 ohm resistor across it, we would have:
V=I*R
V=1*2
V=2 volts.

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:
Iout=A*input

If the input is a voltage, then we have a voltage controlled current source and the output is:
Iout=A*Vin

and if the input is a current then we have a current controlled current source and the output is:
Iout=A*Iin

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:
Iout=A*Iin

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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#### hp1729

Joined Nov 23, 2015
2,304
You're approach in post #16 is called dangle-biasing, and it is very unstable. The base-emitter junction is a signal diode; you do not have to limit the current through it to develop 0.7 V across it. When driving LEDs you want the transistor to act as a saturated switch. This means intentionally driving the base with more current than necessary for the circuit's collector current. For example, if the LEDs in the collector are limited to 20 mA by their resistor, then a common rule of thumb says that the base current should be 1/10th of that for "hard" saturation. I've always thought this is a bit excessive for a small signal transistor, so I would go with 1/20th to 1/40th. Another approach is to look up the transistor's gain at the application's collector current, and drive the base with two to five times the necessary current.

ak
Of course you are most correct. You explain it to him in a way he will understand.

#### Guest3123

Joined Oct 28, 2014
404
If we take the current of the LED on the output, 20 mA, we know the collector current.
One characteristic of the transistor is the "beta", the current gain of the transistor. What the collector current will be with a given base current. Look at a data sheet for any NPN transistor. If the transistor has a gain of 200 at a collector current of 20 mA we divide collector current (20 mA) by the gain (200) and get a base current of about 0.1 mA.

Looking at page 2 of the attached data sheet we see the gain of the 2N3904 under different collector currents. At 10 mA we have a gain of somewhere between 100 and 300. We estimate that at 20 mA we might have a gain of 200. Precise calculations are impossible as you can see.

LED...?

Ok, so with the data sheet, it says the Collector Current — Continuous (Ic) is 200 mAdc.
That's 200 mA Direct Current right? That's what mAdc means..? I wasn't even taught about collector current yet, I'm still being taught about Emitter and base.. so..

I'm really sorry.. I'm afraid I'm just starting to learn about NPN Transistors.. I don't know what your talking about. I know very little about transistors, and I'm sorry, but I just don't understand what your talking about. I understand very little what your saying, but I honestly don't understand.

#### Guest3123

Joined Oct 28, 2014
404
I just seen the new post he posted.. after I posted my last comment.. I'm going to read it now.

#### MrAl

Joined Jun 17, 2014
7,891
LED...?

Ok, so with the data sheet, it says the Collector Current — Continuous (Ic) is 200 mAdc.
That's 200 mA Direct Current right? That's what mAdc means..? I wasn't even taught about collector current yet, I'm still being taught about Emitter and base.. so..

View attachment 100315

I'm really sorry.. I'm afraid I'm just starting to learn about NPN Transistors.. I don't know what your talking about. I know very little about transistors, and I'm sorry, but I just don't understand what your talking about. I understand very little what your saying, but I honestly don't understand.

Hi,

That's a good idea. Once you learn about the two sources you will be ready to learn about the basic role the collector plays, and then things will start to fall in place.

Also, i added a diagram to the previous post to show in more detail the connections we are talking about. Mainly, the emitter is grounded (at 0v) and the collector is open.

#### hp1729

Joined Nov 23, 2015
2,304
LED...?

Ok, so with the data sheet, it says the Collector Current — Continuous (Ic) is 200 mAdc.
That's 200 mA Direct Current right? That's what mAdc means..? I wasn't even taught about collector current yet, I'm still being taught about Emitter and base.. so..

View attachment 100315

I'm really sorry.. I'm afraid I'm just starting to learn about NPN Transistors.. I don't know what your talking about. I know very little about transistors, and I'm sorry, but I just don't understand what your talking about. I understand very little what your saying, but I honestly don't understand.

Good, yes. That is mA of DC, 200 mA is the suggested maximum to not exceed.

Maybe a hands-on exercise is needed. Do you have a btreadboard setup you can build a circuit on?

#### Guest3123

Joined Oct 28, 2014
404
Good, yes. That is mA of DC, 200 mA is the suggested maximum to not exceed.

Maybe a hands-on exercise is needed. Do you have a btreadboard setup you can build a circuit on?
As a matter of a fact, I do own a solderless breadboard.

#### hp1729

Joined Nov 23, 2015
2,304
As a matter of a fact, I do own a solderless breadboard.
Transistor exercise 1 and 2 for an NPN transistor.
Required stuff:
Power (anything under 15Volts or so)
A data sheet that tells you which pin is which.
Set of resistors of all values or an adjustable voltage source.
Volt meter
Calculator
graph paper or Microsoft Excel or any spreadsheet is handy

If you don't have these things I probably have a similar exercise already done for you to look at. The exercise can take hours, but it is an enlightening experience to do.

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#### MrAl

Joined Jun 17, 2014
7,891
As a matter of a fact, I do own a solderless breadboard.

Hello again,

I hope we didnt move too fast on this.

Yes, a breadboard might help, but just be careful that the connections are good. Solderless breadboard differ in quality and sometimes the connections add resistance to some of the connection points.

You could actually start by biasing the input and measuring the base emitter voltage, etc.
In the end you will have an actual transistor amplfiier

#### Guest3123

Joined Oct 28, 2014
404
Hello again,

I hope we didnt move too fast on this.

Yes, a breadboard might help, but just be careful that the connections are good. Solderless breadboard differ in quality and sometimes the connections add resistance to some of the connection points.

You could actually start by biasing the input and measuring the base emitter voltage, etc.
In the end you will have an actual transistor amplfiier
No.. Didn't move too fast.. I feel I'm just not ready for this yet.. I'd love to learn about the NPN Transistor, as it's been on my learning to do list for quite some time. There's nothing wrong with how your teaching, it's just that I really don't want to bother with this. Plus your sick and I just feel bad having you do this. If your up to it, maybe I'll learn about this during the summer, or when it gets warmer out. This doesn't have to be done right away, it can wait.

I'll probably buy a few transistors from mouser, if you'd give me a part number. Then at least we can go along with a transistor that you've picked out.

I have a decent DMM too, it's a UNI-T 6000 counts dmm, so it's decent.

#### MrAl

Joined Jun 17, 2014
7,891
Hi,

2N4401 NPN and 2N4403 PNP would get you started. Cheap too.