please help transistor amplifier

yungman

Read my post #122, I explained step by step why at high frequency, Re2 disappeared. The explanation is adding on top of what Jony130's diagram and all. You really need to stop and work it out step by step, put in some numbers and work it out.

BTW I still miss one on 3). Re2 should be Re1.

You need to show some effort, use the number Jony130 gave in the diagram and show us your calculation. Whether it is correct or not, show you try to solve the problem. I help you at the beginning, together with Malvino, you should be able to get started. But after 9 pages of question and answer, you are still pretty much on the same circuit with the addition of C2 and Re2!!!!

Work out some numbers and people are more than happy to help.

yungman

I don't know what else to say, you need resources, but you don't want to take any chances. I don't think you can just walk into a book store and find what you want and pay cash!!! At some point, you are going to have to trust something.

If you are worry, do what I am doing, get a credit card with low limits, use that only for online order. So if you get scam, you know right away and it will be minimal loss.

Did you try looking online for free download?

michael1978

yes i try to download online, via google

michael1978

ok JONEY SORRY, i read in post 77, i read and i read and you show me example good, but i dont understand why you add re2, and you dont have to calculate in gain? i will try to make one example

yungman

Read the bias scheme in Malvino. Re2 is to stabilizing the collector current. This is called voltage divider bias. R1 and R2 form a voltage divider to give about 1.7V at the base. The emitter is about 0.7V below so the emitter is at about 1V. You have Re1+Re2 about 200 ohm ( don't be picky), so you are setting up the emitter current ( collector current) of about 5 mA. This 5 mA will give r'e ≈ 5 ohm. The point of using voltage divider bias is to control the current through the BJT by the voltage drop across the Re2. This make it a lot more predictable. Read Malvino, it's all there.

The C2 is to bypass Re2 at frequency you want to amplify. At high frequency, Re2 is being shorted out by C2, so the resistance on the emitter side is r'e+Re1=19 ohm or 20 ohm.

No matter how the circuit looks like, in common emitter BJT circuit, the gain is the total impedance at the collector divided by the total impedance at the emitter. AND it is inverted. It is just that simple.

I just went back and looked at page 5, 6 7 of this thread, Jony130 really....REALLY spent the time explained to you on this. You really need to spend less time asking question here and more time reading the book over and over, work out the problems. I don't have the book, I think it has answers even if you can't find the solution manual. I learn all these from Malvino in 1979 without the solution manual. You are spending almost a month on this and I am sure it's cover in only about 4 to 5 pages in Malvino. I am sure anything I said at this point, HAD been totally covered by Jony130. Read the book over and over and over until you understand. Or read this thread over and over. Write out everything Jony130 wrote step by step to verify that you are following, don't just read, write it out.

michael1978

thanks for time, in malvino i have version 7, the book tell only how to desigin an amplifier, and joney he make quicky the amplifier, yes he spent a lot of time for me, the book wich you buy in amazone they are used books not new but not expensiv

michael1978

no i try, i cant make amplifier without adding Re1, because &
10vcc
R2=170Ω=1.7V
R1=830Ω=8.3V
Ve=1v but i need one resistor for Ic
Ic≈Ie
how can i select current to Ie i need to add resistor, and after to select Vce and Vc

is that , that we need to add resistor to Ve to select Ie, otherwise you cant make an amplifier
but if you add for example to be Ie = 1mA, ic=5K*1mA = 5V after the gain gonna change it cant be wich i want

yungman

So you know it is 1.7V at the base and 1V at the emitter.

YES, you set Ie with the emitter resistor ( Re1+Re2). This is the KEY of DC biasing. You first set up the operating current Ie of the transistor, this is the first and foremost thing. Collector don't set the current of the BJT, collector current is only follow the emitter current. AGAIN, this is in Malvino.

1) In the diagram, Re1+Re2 is about 200 ohm. You put 1V at emitter, you put 1V across the Re1 and Re2. SO you force 5mA through the resistors......Which, means you set the Ie to 5mA. This is how you set up the DC bias current.

2) After setting the Ie, then you start looking at the collector. You know β is high, so you know the Ic≈5mA. If you have Vcc=10V, and you have 1K resistor at the collector, then you know 5mA through 1K is 5V. So the collector is at +5V.

3) AGAIN, gain is determined by the impedance at the collector divided by the impedance at the emitter. We gone over this over and over and over already. Read the old posts AGAIN.

These are all in Malvino. Please read that before you ask any more question. If you don't get it, read it, write out the numbers, work through the problems before you post.

Jony130

We need Re2 resistor simply because it is almost impassible to build amplifier with desired low voltage gain and Ve > Vbe. Which is needed for good bias point stability.
But if you determined to use only Re1 resistor. We need to change DC bias method.
So we are not going to use voltage divider, instead of voltage divider we will use a collector-feedback bias circuit.
So for our example for Vcc = 10V and Rc = 1K;RL = 10K; Ic = 5mA and Av = 50.

re = 26mV/Ic = 5.2Ω

Re1 = (Rc||RL)/Av - re = 909Ω/50 - 5.2 = 12Ω

Vc = Vcc - Ic*Rc = 5V

Vb = Ic*Re1 + Vbe = 0.71V

And

Ib = Ic/hfe = 5mA/150 = 33.4μA

Rb1+Rb2 = (Vc - Vb)/Ib = 4.29V/33.4μA = 128KΩ

Rb1 = Rb2 = 128K/2 = 68KΩ



And simulation show that voltage gain is equal to AV = 51.8092[V/V]

Attached Thumbnails

   

michael1978

JONEY thank you for reply i am learning a lot from you, you are the beste;-) COMPLIMENT

michael1978

did you see the post 146 of Joney, maybe i am mistake but malvino he dont show you how to get desired voltage, i wil like to ask you why Jony, he put so big resistance of voltage divider i say for r1 22K and for R2 4.7k how he calculate,? me i put it R2=170, and R1=830, look how big difference, he show one example but me i put in other way,

yungman

We were talking about the diagram of post 114 all along until now.

The newer diagram is not as desirable. I believe it called self bias or something. It depend a lot on the beta of the transistor. Voltage divider bias is a better way to go.

michael1978

yes but i say to joney i gonna make one amplifier, and he help answer me thanks from him, ok why Joney select so big resistance of R1 22k and R2 4.7k
and he get the sam base almost 1.7V, how i have to do it big resistance like joney

yungman

You talking about post 114? The resistance is 22K//4.7K ( if I ignore the input resistance of the transistor). In fact the input resistance is quite low to me.

Usually you want the input resistance to be a little on the high side so the stage driving this input don't have to drive a low impedance. Think if you have a stage like in post 114, if the collector is driving the following stage with low impedance like 1K. The collector resistance is 1K//1K=500 ohm. You lower the gain of the driving stage.

In both diagrams, the input resistance of the transistor is β(r'e + Re1).=β(20Ω)≈2K assuming β=100. This is quite low. Another way to look at it is the transistor in both case require 5mA/β= 50uA. The impedance of the input biasing network has to be low enough so it is not loaded down by the input of the transistor. Jony got into this a little in post 79 about the base current.

michael1978

CAN I ASK YOU SOMETHING about impedance, my english is not so good, how you select input impedance, not to say in the end the input impedance was 50k of 10k etc, you select in the begin of how

yungman

For normal frequency ( not RF), you usually want it to be as high as practical. It is limited by the input requirement of the amplifier.........Like in your case, the transistor Ib of 50uA and transistor input impedance. Jony in post 79 explained this and I explained about the input impedance in the post 150.

In your case, with drawing in both post 114 and 146, the emitter resistor Re1 is eithe 14 or 12Ω, that is very low. That is really the gating factor of the circuit. Also you are trying to get gain of 50 out of one transistor, this force you to have such a low Re1 in order to be able to use a collector resistor of 1K.

In real life design, if I want to have a gain of 50, I would do either one below:

1) Using Re1=150, Rc=10K, Ic=0.5mA, then use an emitter follower transistor to buffer the output.
2) Use two stage of this and divide the gain between the two stages. This is the prefer way.
3) Use JFET instead of BJT.

michael1978

thank you for answer, and R1, R2? but first i am bussy with transistor, is difficult jfet to learn, and for the book, you can buy it chip used boooks in amazone, because i want to open one visa of mastercard