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Please help transistor amplifier

by michael1978
Tags: amplifier, transistor
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michael1978
#109
Nov28-12, 03:38 PM
P: 133
Quote Quote by Jony130 View Post
I don't think so.

I use LTspice, but circuit simulations are not good for beginners.
Circuit simulators are worthless, or worse, if you don't understand your circuit pretty well before you simulate it. Still, it won't hurt you to play with it, so long as you always keep in mind that garbage in = garbage out, or, in the case of simulators, garbage in=sophisticated garbage out.
but i dont understand, i have to do experiments, and i dont have all instruments, in electronics simulator you have all instruments, but me i try examples, but you know better i see you have experience.
thx for answer
michael1978
#110
Dec5-12, 03:26 PM
P: 133
Quote Quote by michael1978 View Post
thank you very much
joney is correct this amplifier, he give voltage gain of 44mv not 50mv
Attached Thumbnails
TRANSISTOR AMPLIFIER.jpg  
Jony130
#111
Dec5-12, 03:36 PM
P: 406
Quote Quote by michael1978 View Post
joney is correct this amplifier, he give voltage gain of 44mv not 50mv
I don't know. You need to tell me the bias point collector current and BJT model that you use.
michael1978
#112
Dec5-12, 04:01 PM
P: 133
Quote Quote by Jony130 View Post
I don't know. You need to tell me the bias point collector current and BJT model that you use.

--------
this your example i just simulate (do you see the foto the value) in quote 69

Vb = Ve + Vbe = Ic*(Re1+Re2)+ Vbe = 5mA* (180 + 14) + 0.65V = 1.62V

If we assume Hfe = 150 I (I use transistor 2N3903 I donít know other there are a lot of types)

Ib = Ic/hfe = 5mA/150 = 34μA

R1 = (Vcc - Vb)/( 11*Ib) = 22KΩ

R2 = Vb/(10*Ib) = 4.7KΩ
Re1 = Rc/Av - re = 1K/50 - 5.2Ω = 20 - 5.2Ω = 14Ω
Jony130
#113
Dec5-12, 04:20 PM
P: 406
But your circuit is different than mine. Also you can read from simulation DC collector current.
And your circuit has a voltage gain equal to:

Av = Vout/Vin = 44.33mV/2mV = 22.165[V/V]

My example look like this:



And has a voltage gain Av = 45.8V/V But we can easily change the voltage gain by changing the Re1 resistance.
Attached Thumbnails
a1.PNG  
michael1978
#114
Dec6-12, 06:24 AM
P: 133
Quote Quote by Jony130 View Post
But your circuit is different than mine. Also you can read from simulation DC collector current.
And your circuit has a voltage gain equal to:

Av = Vout/Vin = 44.33mV/2mV = 22.165[V/V]

My example look like this:



And has a voltage gain Av = 45.8V/V But we can easily change the voltage gain by changing the Re1 resistance.
---------
can you explain how you can change the voltage gain by changing Re1
------------------
Joney do you remember i ask you to make one amplifier with voltage gain of 50, only we Re1 Without Re2,

and you show me this example, look at quote 66, IF YOU HAVE TIME.
Re1 = Rc/Av - re = 1K/50 - 5.2Ω = 20 - 5.2Ω = 14Ω, and i think you did not show me complete example, becuse i am searchin here but i cant find,
do you remember now Re1 = Rc/Av - re = 1K/50 - 5.2Ω = 20 - 5.2Ω = 14Ω
for the rest value i think of maybe i misunderstand , are the same like this circuit the last one.... if not ? PLEASE CAN YOU MAKE ONE EXAMPLE ONLY WITH Re1 PLEASE THNX
yungman
#115
Dec6-12, 11:12 AM
P: 3,883
I did not read the whole thread. Remember when I left off, I showed you how to calculate the gain? It's the impedance seen at the collector divided be the impedance seen by the emitter? Adding C2 don't change this, the total emitter resistance is

r'e+ Re1+(Re2//Xc2) where [itex] X_{C2}=\frac 1 {j\omega C}[/itex]

You change the Re2, you change the impedance on the emitter side and change the gain.

You need to get the solution manual of Malvino and work through the problems one by one. I thought I left you in good hands already. There comes a point of time you just work on the problems one by one and struggle through it. You are spending too much time writing posts here instead of working through the problem in the book. These questions are in the book.
michael1978
#116
Dec6-12, 11:38 AM
P: 133
Quote Quote by yungman View Post
I did not read the whole thread. Remember when I left off, I showed you how to calculate the gain? It's the impedance seen at the collector divided be the impedance seen by the emitter? Adding C2 don't change this, the total emitter resistance is

r'e+ Re1+(Re2//Xc2) where [itex] X_{C2}=\frac 1 {j\omega C}[/itex]

You change the Re2, you change the impedance on the emitter side and change the gain.

You need to get the solution manual of Malvino and work through the problems one by one. I thought I left you in good hands already. There comes a point of time you just work on the problems one by one and struggle through it. You are spending too much time writing posts here instead of working through the problem in the book. These questions are in the book.
thnx for reply, i dont know where to get solution manual by malvino, til now i am to transistor amplifier, but i dont get nothing what joney explain me til now
yungman
#117
Dec6-12, 06:22 PM
P: 3,883
Quote Quote by michael1978 View Post
thnx for reply, i dont know where to get solution manual by malvino, til now i am to transistor amplifier, but i dont get nothing what joney explain me til now
Ha ha!!! I always gone on the internet and look for free download. You have to do some leg work. I yet to encounter a book that I had not manage to download the instructor or solution manual free yet. This is such a popular book. It is very important to have the solution manual to learn, they show you the steps to get the answer. Now put in your effort and try working out the answer before you peek into the solution manual!!!!

Make sure you get the correct edition, if you manage to download a version you don't have, go on Amazon and find a used text book of that version. They are very cheap used. I went on Amazon to look for one for you just now and can't find one cheap at the moment. In fact I just ordered a copy of Malvino a few minutes ago just to keep it in my library collection because it's that good. I only paid $US 8.00 including shipping!!! But you can go on Amazon later and see whether they have a copy cheap.
michael1978
#118
Dec7-12, 06:02 AM
P: 133
Quote Quote by yungman View Post
Ha ha!!! I always gone on the internet and look for free download. You have to do some leg work. I yet to encounter a book that I had not manage to download the instructor or solution manual free yet. This is such a popular book. It is very important to have the solution manual to learn, they show you the steps to get the answer. Now put in your effort and try working out the answer before you peek into the solution manual!!!!

Make sure you get the correct edition, if you manage to download a version you don't have, go on Amazon and find a used text book of that version. They are very cheap used. I went on Amazon to look for one for you just now and can't find one cheap at the moment. In fact I just ordered a copy of Malvino a few minutes ago just to keep it in my library collection because it's that good. I only paid $US 8.00 including shipping!!! But you can go on Amazon later and see whether they have a copy cheap.

i cant find, but is safe amazone site? can you order it with facture? of only with card
michael1978
#119
Dec7-12, 06:20 AM
P: 133
Quote Quote by michael1978 View Post
---------
can you explain how you can change the voltage gain by changing Re1
------------------
Joney do you remember i ask you to make one amplifier with voltage gain of 50, only we Re1 Without Re2,

and you show me this example, look at quote 66, IF YOU HAVE TIME.
Re1 = Rc/Av - re = 1K/50 - 5.2Ω = 20 - 5.2Ω = 14Ω, and i think you did not show me complete example, becuse i am searchin here but i cant find,
do you remember now Re1 = Rc/Av - re = 1K/50 - 5.2Ω = 20 - 5.2Ω = 14Ω
for the rest value i think of maybe i misunderstand , are the same like this circuit the last one.... if not ? PLEASE CAN YOU MAKE ONE EXAMPLE ONLY WITH Re1 PLEASE THNX
hi joney, can you answer me please, because i learn it in serie with Re2, and also in parallel, just with one resistor no i did not learn it, can you take a little time to show me this example, voltage gain 50 with one reistor,like other examples
yungman
#120
Dec7-12, 11:05 AM
P: 3,883
Quote Quote by michael1978 View Post
i cant find, but is safe amazone site? can you order it with facture? of only with card
I use Amazon ALL the time, The used books are so cheap. I buy so many text books, 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.
michael1978
#121
Dec7-12, 11:29 AM
P: 133
Quote Quote by yungman View Post
I use Amazon ALL the time, The used books are so cheap. I buy so many text books, 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 cant buy with invoice
yungman
#122
Dec7-12, 11:42 AM
P: 3,883
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 [itex] X_C=\frac 1 {j2\pi f C}[/itex].

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 [itex] Re2=\frac 1 {2\pi f C}[/itex], 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 [itex] \frac 1 { 2\pi fC}=Re2[/itex], 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 [itex] \frac 1 { 2\pi f C}=R_{e1}[/itex] is usually called [itex]f_c[/itex] where [itex] \frac 1 { 2\pi f_c C}=Re1[/itex] or [itex]f_c=\frac 1 {2 \pi R_{e1} C_2}[/itex]. 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
#123
Dec7-12, 11:45 AM
P: 3,883
Quote Quote by michael1978 View Post
WHY you cant 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
#124
Dec7-12, 12:33 PM
P: 133
Quote Quote by yungman View Post
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 [itex] X_C=\frac 1 {j2\pi f C}[/itex].

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 [itex] Re2=\frac 1 {2\pi f C}[/itex], 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 [itex] \frac 1 { 2\pi fC}=Re2[/itex], 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 [itex] \frac 1 { 2\pi f C}=R_{e2}[/itex] is usually called [itex]f_c[/itex] where [itex] \frac 1 { 2\pi f_c C}=Re2[/itex] or [itex]f_c=\frac 1 {2 \pi R_{e1} C_2}[/itex]. 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 wich show me joney,
michael1978
#125
Dec7-12, 12:41 PM
P: 133
Quote Quote by yungman View Post
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 wich card can accept amazone?
yungman
#126
Dec7-12, 03:19 PM
P: 3,883
Quote Quote by michael1978 View Post
thank for time, but i want just with r'e and re1 without re2 the voltage gain about 50mv, the procedure wich 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.


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