Please help transistor amplifier

[michael1978]

I use Amazon ALL the time, The used books are so cheap. I buy so many textbooks, but I never bought them new. When they said the condition is good, it is good. I have two VISA/Mastercard, I use one for everyday, the other JUST for online order like Amazon. So I can track transaction clearly every month. I never have problem.

WHY you can't buy with invoice

 

[yungman]

Regarding the circuit. This circuit is a voltage divider bias using R1 and R2 to set up the bias voltage of 10(4.7K/26.7K)=1.76V for Q1 . The emitter of Q1 is about 1.06V. Read this in Malvino. R5 together with R4 is to set up the DC of about 5mA current through Q1 and re' is about 25/5=5ohm. But without C2, gain of the stage is Rc/(Re1+Re2+re')=5. That's very low to be useful.

C2 is to provide a low impedance path to bypass Re2 at higher frequency. With the C2, the gain of the stage is Rc/(re'+Re1+(Re2//Xc)). The impedance of C2 is X_C=\frac 1 {j2\pi f C}.

But this is complicated for you. So you can use approximation.

1) At very low frequency, Xc is very high, so you can ignore it. So the gain is Rc/(re'+Re`1+Re2).

2) At frequency where Re2=\frac 1 {2\pi f C}, the total resistance of Re2// C2 decrease and the gain of the stage start to rise as show in the graph. It is not important to know the exact frequency as the final gain is usually the important one.

3) As frequency goes higher, Xc is getting lower and lower. Re1<<Re2, you can simplify by ignore Re2. At frequency where \frac 1 { 2\pi fC}=Re2, the effect of C2 start to level out. At even higher frequency, C2 can be approximated to be a short circuit( 0 ohm). So at much higher frequency, the gain of the stage is Rc/(re'+Re1) only, as C2 is a short circuit and Re2 is shorted out.

4) the transition frequency where \frac 1 { 2\pi f C}=R_{e1} is usually called f_c where \frac 1 { 2\pi f_c C}=Re1 or f_c=\frac 1 {2 \pi R_{e1} C_2}. You can see the point of fc where the graph of the gain start bending horizontal to level out.

Hope this help. This is an approximation. For more accurate calculation, you really have to use complex number, but believe me, it's good enough. I use this all these years for my own design at work.

 

[yungman]

WHY you can't buy with invoice

I never try that and sounds more complicate. You enter the credit card info the first time to set up the account, after that, you just login and it will do it for you. You never have to enter the credit card info again.

 

[michael1978]

Regarding the circuit. This circuit is a voltage divider bias using R1 and R2 to set up the bias voltage of 10(4.7K/26.7K)=1.76V for Q1 . The emitter of Q1 is about 1.06V. Read this in Malvino. R5 together with R4 is to set up the DC of about 5mA current through Q1 and re' is about 25/5=5ohm. But without C2, gain of the stage is Rc/(Re1+Re2+re')=5. That's very low to be useful.

C2 is to provide a low impedance path to bypass Re2 at higher frequency. With the C2, the gain of the stage is Rc/(re'+Re1+(Re2//Xc)). The impedance of C2 is X_C=\frac 1 {j2\pi f C}.

But this is complicated for you. So you can use approximation.

1) At very low frequency, Xc is very high, so you can ignore it. So the gain is Rc/(re'+Re`1+Re2).

2) At frequency where Re2=\frac 1 {2\pi f C}, the total resistance of Re2// C2 decrease and the gain of the stage start to rise as show in the graph. It is not important to know the exact frequency as the final gain is usually the important one.

3) As frequency goes higher, Xc is getting lower and lower. Re1<<Re2, you can simplify by ignore Re2. At frequency where \frac 1 { 2\pi fC}=Re2, the effect of C2 start to level out. At even higher frequency, C2 can be approximated to be a short circuit( 0 ohm). So at much higher frequency, the gain of the stage is Rc/(re'+Re1) only, as C2 is a short circuit and Re2 is shorted out.

4) the transition frequency where \frac 1 { 2\pi f C}=R_{e2} is usually called f_c where \frac 1 { 2\pi f_c C}=Re2 or f_c=\frac 1 {2 \pi R_{e1} C_2}. You can see the point of fc where the graph of the gain start bending horizontal to level out.

Hope this help. This is an approximation. For more accurate calculation, you really have to use complex number, but believe me, it's good enough. I use this all these years for my own design at work.

thank for time, but i want just with r'e and re1 without re2 the voltage gain about 50mv, the procedure which show me joney,

 

[michael1978]

I never try that and sounds more complicate. You enter the credit card info the first time to set up the account, after that, you just login and it will do it for you. You never have to enter the credit card info again.

yes but maybe in once some hacker can take your credit, but do you know which card can accept amazone?

 

[yungman]

thank for time, but i want just with r'e and re1 without re2 the voltage gain about 50mv, the procedure which show me joney,

As I explained, at frequency above fc, the reactance of C1 is very low and you can just calculate using r'e and Re1.

For example, if C1 is 100uF, at 1KHz, Xc=1/(2\pi f C)=1.59ohm. 1.59 ohm is so much lower than 180 of Re2. When parallel with Re2, Re2 disappeared. So you only have Re1 and r'e left.

 

[yungman]

yes but maybe in once some hacker can take your credit, but do you know which card can accept amazone?

Well, you always are taking a chance. But if you have a good password, that would protect you. Email to Amazon and ask them. Now a days, if you don't use online, you are missing the world.

 

[Jony130]

yungman you made a small mistake. I suspect a typo error.
If we assume that C1 and C3 >> C2 and Re1 << Re2
So now we can find corner frequency as:

\Large Fc\approx \frac{0.16}{(re+Re1)*C2}

 

[yungman]

Yes, in 4), it should by Re1. I change the 2 to 1 already.

 

[michael1978]

As I explained, at frequency above fc, the reactance of C1 is very low and you can just calculate using r'e and Re1.

For example, if C1 is 100uF, at 1KHz, Xc=1/(2\pi f C)=1.59ohm. 1.59 ohm is so much lower than 180 of Re2. When parallel with Re2, Re2 disappeared. So you only have Re1 and r'e left.

YOUNGMAN, i learn from joney step by step procedure, its is difficult to understand yours because joney he make in very similary way, biasing gain , step by step, do you see his examples, but he don't answer me anymore i don't know why
do you know to build one amplifier like joney step by step without re2

 

[michael1978]

yungman you made a small mistake. I suspect a typo error.
If we assume that C1 and C3 >> C2 and Re1 << Re2
So now we can find corner frequency as:

\Large Fc\approx \frac{0.16}{(re+Re1)*C2}

yes he make mistake you see he teach me bad hahaha i am just joking, joney why you don't answer me? WHATS WRONG?,

 

[Jony130]

joney why you don't answer me? WHATS WRONG?,

Because I and yungman gave you all information you need to know to design simple BJT amplifier with or without Re2 resistor.

 

[michael1978]

Because I and yungman gave you all information you need to know to design simple BJT amplifier with or without Re2 resistor.

no joney rely, i understand all, with re re1 re2 in serie, and also in parallel re1 and re2, but this only with re1 without Re2 , i rely don't understand, you show every example step by step, but with re1 i can't find it, i am searching in quote from the beginin but i can find, RELY otherwise i with not ask you , i ask you in qoute 5, to make one example but you did not complete of me i don't understand, rely i with not ask you anymore if i know it, because you show me complete procedure of amplifier with re1 re2 in parallel and in serie and i understand, only this one no, only with re1, for that i am asking so long and i am whaiting answer from you, like i see you make it very quickly, just one more without re2
AND IS DIFFICULT TO UNDERSTAND SOMEBODY ELSE WITH A LOT OF EQUATION, please can you make one more thnx for reply

 

[Jony130]

But why you don't want to try and use this information we have give you and start design amplifier without Re2 resistor. Once at least TRY to answer your own question and show SOME work, instead of expecting people here to be your no-effort-required-on-your-part answering service

 

[michael1978]

But why you don't want to try and use this information we have give you and start design amplifier without Re2 resistor. Once at least TRY to answer your own question and show SOME work, instead of expecting people here to be your no-effort-required-on-your-part answering service

that is right Joney but like i ask all time for example voltage gain of 50, is just to know steps how to make an amplifier and to do experiments in simulation, with other value, for example gain of 20 of 10, of other value of biasing etc, but i see the best was first to learn only with one resistor with re1, and after in serie and in parallel, for that i ask you, because this what you show me til now, i don't have in my book electronics principle 7 by malvino, and i read also some other book about microelectronics, there don't exist your example, i search it a lot in internet but pblackblack nothing i find, is possible for you just one more and after i with start experiments PLEASE

 

[michael1978]

I never try that and sounds more complicate. You enter the credit card info the first time to set up the account, after that, you just login and it will do it for you. You never have to enter the credit card info again.

but is that possible of not

 

[yungman]

no joney rely, i understand all, with re re1 re2 in serie, and also in parallel re1 and re2, but this only with re1 without Re2 , i rely don't understand, you show every example step by step, but with re1 i can't find it, i am searching in quote from the beginin but i can find, RELY otherwise i with not ask you , i ask you in qoute 5, to make one example but you did not complete of me i don't understand, rely i with not ask you anymore if i know it, because you show me complete procedure of amplifier with re1 re2 in parallel and in serie and i understand, only this one no, only with re1, for that i am asking so long and i am whaiting answer from you, like i see you make it very quickly, just one more without re2
AND IS DIFFICULT TO UNDERSTAND SOMEBODY ELSE WITH A LOT OF EQUATION, please can you make one more thnx for reply

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]

but is that possible of not

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]

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?

yes i try to download online, via google

 

[michael1978]

that is right Joney but like i ask all time for example voltage gain of 50, is just to know steps how to make an amplifier and to do experiments in simulation, with other value, for example gain of 20 of 10, of other value of biasing etc, but i see the best was first to learn only with one resistor with re1, and after in serie and in parallel, for that i ask you, because this what you show me til now, i don't have in my book electronics principle 7 by malvino, and i read also some other book about microelectronics, there don't exist your example, i search it a lot in internet but pblackblack nothing i find, is possible for you just one more and after i with start experiments PLEASE

ok JONEY SORRY, i read in post 77, i read and i read and you show me example good, but i don't understand why you add re2, and you don't 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]

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.

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 which you buy in amazone they are used books not new but not expensiv

 

[michael1978]

no i try, i can't 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 can't make an amplifier
but if you add for example to be Ie = 1mA, ic=5K*1mA = 5V after the gain going to change it can't be which i want

 

[yungman]

no i try, i can't 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 can't make an amplifier
but if you add for example to be Ie = 1mA, ic=5K*1mA = 5V after the gain going to change it can't be which i want

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]

 

[michael1978]

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]

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

 

[michael1978]

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.

did you see the post 146 of Joney, maybe i am mistake but malvino he don't 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]

did you see the post 146 of Joney, maybe i am mistake but he don't 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,

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]

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

yes but i say to joney i going to 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]

yes but i say to joney i going to 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

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.