Please help transistor amplifier

[michael1978]

can somebody teach mm
how to find desired voltage gain of ce a transistor amplifier
which is formula, i know the formula is rc:rl/re

but i want for example a voltage gain of 50
for example

input desired data:

desired voltage gain 2.0 which i want to know
f_min = 10 HZ
z_in = 50 K
Z_OUT = 2k
power suppply 12V


now how is the formula to find
r1 r2 i know to find
but rc and re i don't know to find, that is my problem

 

[yungman]

can somebody teach mm
how to find desired voltage gain of ce a transistor amplifier
which is formula, i know the formula is rc:rl/re

but i want for example a voltage gain of 50
for example

input desired data:

desired voltage gain 2.0 which i want to know
f_min = 10 HZ
z_in = 50 K
Z_OUT = 2k
power suppply 12V


now how is the formula to find
r1 r2 i know to find
but rc and re i don't know to find, that is my problem

You want output to be 2K, your rc has to be 2K. You want voltage gain of 2, so you want your re≈1K. Gain=rc/(re+r'e) where r'e= 1/gm≈Vt/Ic. Vt≈25mV at 25deg C. For Ie=1mA, r'e≈25Ω. So if you use 1mA bias current, re=1K is close enough.

If you choose a NPN with β>100, input impedance of the transistor is βXre≈100K, this is going to be a little tricky getting input impedance of 50K as the input impedance of the transistor is as low as 100K or higher if β is larger.


What is r1 and r2? is this a voltage divider to bias the base of the BJT? If so, you want Zin=50K, then you want r1//r2=100K so you get about 50K when parallel with the input impedance of the transistor. But this is not reliable.

You need to relax some requirement, if you can accept output impedance of say 20K, then re=10K and your input impedance of the transistor can by up to 1MΩ. Then you can ignore the input impedance of the transistor and make r1//r2= 50K.

For fmin, C=1/(2πRf) where R=50k input impedance, f=10Hz.

 

[michael1978]

thnx for reply very good
but can i ask you one question if you can explain in similary way
what for example, i want to desigin a amplifier with gain of 50, how can i design?
for example me i design a amplifier in the end i get other gain, and i want a gain of 50
can you teach me how to do it, to get desired voltage

 

[NascentOxygen]

thnx for reply very good
but can i ask you one question if you can explain in similary way
what for example, i want to desigin a amplifier with gain of 50, how can i design?
for example me i design a amplifier in the end i get other gain, and i want a gain of 50
can you teach me how to do it, to get desired voltage

For a high gain (e.g., 50) there are advantages in using two stages, each using one transistor. The first stage could have a voltage gain of x10, and the second a gain of x5.

But whatever voltage gain you need, for each stage you still use the equations that yungman provided, viz.,

Gain A_V=rc/(re+r'e) where r'e= 1/gm≈Vt/Ic. Vt≈25mV at 25deg C. For Ie=1mA, r'e≈25Ω.

 

[michael1978]

For a high gain (e.g., 50) there are advantages in using two stages, each using one transistor. The first stage could have a voltage gain of x10, and the second a gain of x5.

But whatever voltage gain you need, for each stage you still use the equations that yungman provided, viz.,

yes i know, can you show the first stage gain for example to be x10, which steps i have to take? WICH FORMULA? i know the gain is rcllrl/re , but i want for example x10 OF x5 ANY DESIRED VOLTAGE GAIN CAN YOU SHOW ME EXAMPLE PLEASE FOR DESIRED FIRST VOLTAGE GAIN

 

[michael1978]

You want output to be 2K, your rc has to be 2K. You want voltage gain of 2, so you want your re≈1K. Gain=rc/(re+r'e) where r'e= 1/gm≈Vt/Ic. Vt≈25mV at 25deg C. For Ie=1mA, r'e≈25Ω. So if you use 1mA bias current, re=1K is close enough.

If you choose a NPN with β>100, input impedance of the transistor is βXre≈100K, this is going to be a little tricky getting input impedance of 50K as the input impedance of the transistor is as low as 100K or higher if β is larger.


What is r1 and r2? is this a voltage divider to bias the base of the BJT? If so, you want Zin=50K, then you want r1//r2=100K so you get about 50K when parallel with the input impedance of the transistor. But this is not reliable.

You need to relax some requirement, if you can accept output impedance of say 20K, then re=10K and your input impedance of the transistor can by up to 1MΩ. Then you can ignore the input impedance of the transistor and make r1//r2= 50K.

For fmin, C=1/(2πRf) where R=50k input impedance, f=10Hz.

oh sorry i forget, if is rl load connected with rc, how you calculate can you teach like in first example please, show me example please thank you

 

[NascentOxygen]

michael1978, you might find this thread useful, including the URL I provide of a reference for common emitter amplifier calculations : https://www.physicsforums.com/showthread.php?t=644679

 

[yungman]

oh sorry i forget, if is rl load connected with rc, how you calculate can you teach like in first example please, show me example please thank you

If you connect rl to rc, then the gain will change like you said = (rl//rc)/re. The calculation I show was direct answer to your original request Zout=2K which is rc. Gain without load rl, equal 2.

For example, say if you keep Zout=2K, rc has to stay at 2K. Say if rl=2K, then rc//rl=1K. To get a gain of about 2, re=475Ω ( as for Ie=1mA, r'e=25Ω). But then, you have to worry about Zin as Zin=β(re+r'e)≈50K! That is not a reliable value.

 

[michael1978]

If you connect rl to rc, then the gain will change like you said = (rl//rc)/re. The calculation I show was direct answer to your original request Zout=2K which is rc. Gain without load rl, equal 2.

For example, say if you keep Zout=2K, rc has to stay at 2K. Say if rl=2K, then rc//rl=1K. To get a gain of about 2, re=475Ω ( as for Ie=1mA, r'e=25Ω). But then, you have to worry about Zin as Zin=β(re+r'e)≈50K! That is not a reliable value.

sorry and have you get re = 475OHM? LIKE THIS 1000/2? is 500OHM

 

[yungman]

sorry and have you get re = 475OHM? LIKE THIS 1000/2? is 500OHM

Remember gain=(rc//rl)/(re+r'e)

As I assume Ie=1mA, r'e=25Ω. rc//rl=1K. For gain of 2, re+r'e=500Ω so re=475Ω.

 

[michael1978]

Remember gain=(rc//rl)/(re+r'e)

As I assume Ie=1mA, r'e=25Ω. rc//rl=1K. For gain of 2, re+r'e=500Ω so re=475Ω.

ah so r'e=25Ω 1ma and re 475 so in total 500, you mean like this? but how you get r'e=25Ω?
what i have to do to get r'e=25Ω, like this 25m:ie 1ma = 25ohm

 

[yungman]

ah so r'e=25Ω 1ma and re 475 so in total 500, you mean like this? but how you get r'e=25Ω?
what i have to do to get r'e=25Ω, like this 25m:ie 1ma = 25ohm

r'e=1/gm=Vt/Ic. Just trust me with Vt=25mV at 25 deg C. That gets into semi conductor physics. I am not expert in it and the approximation I gave is quite good already. That's all you need to know unless you really want to dig into it.

Remember this formula r'e=25mV/Ic. If you use 2mA, r'e become 12.5Ω. If you use 0.5mA, r'e become 50Ω. You get the drift?

Don't try to get precision gain using transistor, it drift a lot with temperature and different device even of the same name.( even using 2 different 2N2222 give you slightly different gain). That's the reason, using approximation is very good already. You want precision, you need closed loop feedback like an op-amp.

 

[michael1978]

r'e=1/gm=Vt/Ic. Just trust me with Vt=25mV at 25 deg C. That gets into semi conductor physics. I am not expert in it and the approximation I gave is quite good already. That's all you need to know unless you really want to dig into it.

Remember this formula r'e=25mV/Ic. If you use 2mA, r'e become 12.5Ω. If you use 0.5mA, r'e become 50Ω. You get the drift?

Don't try to get precision gain using transistor, it drift a lot with temperature and different device even of the same name.( even using 2 different 2N2222 give you slightly different gain). That's the reason, using approximation is very good already. You want precision, you need closed loop feedback like an op-amp.

is that 25mv/ie of 25mv/ic, i learn till now r'e=25mV/Ie not ic

 

[yungman]

is that 25mv/ie of 25mv/ic, i learn till now r'e=25mV/Ie not ic

I believe it is supposed to be Ic, but then again, it is an approximation. Ie≈Ic in all respect if β>100. There are so many variables in these formulas you just use approx number. Here is the page explain in a lot more detail, actually they use Vt=26mV, it's close enough!

http://en.wikipedia.org/wiki/Bipolar_junction_transistor

Unless you really get into semi conductor physics, 25mV is good enough. I pretty much designing IC( actually all transistor circuits inside) using this approximation those days and it worked.

One thing I did not mention, you don't want to get gain of 50 out of one stage. Remember I show you how to calculate the Zin and Zout? You want Zout=2K, for gain of 50, the re+r'e=40Ω, that is low, then your Zin=βX(re+r'e)≈4000Ω. You cannot get Zin = 50K! Even if you can get the impedance you want, there are more limiting factor that you have not deal with, one namely Miller Effect that the circuit slow down as the gain goes up. These are a lot more important in real life than the Vt. If you want gain of 50, divide the gain into 2 separate stages of about 7 each.

 

[Jony130]

If you want to be utterly pedantic about r'e

\Large re =\frac{dVbe}{dIe} = \frac{Vt}{Ie}

And Vt we assume 25mV or 26mV
http://en.wikipedia.org/wiki/Boltzmann_constant#Role_in_semiconductor_physics:_the_thermal_voltage

And transconductance gm

\Large gm= \frac{Ic}{Vt}=\frac{Ic}{Ie}*\frac{Ie}{Vt}= \frac{\beta}{\beta+1}*\frac{1}{re}

 

[michael1978]

thanx for help, is enough for begin
do you know any good book about electronics transistor
because now i am reading electronics principles by malvino

 

[yungman]

thanx for help, is enough for begin
do you know any good book about electronics transistor
because now i am reading electronics principles by malvino

That's the best book...bar none! Everything I posted is in the Malvino! I gone very far as an engineer using nothing more than Malvino. Yes, I got into BJT IC design with just the knowledge of Malvino, it's that good! Of cause I learn a whole lot more since then, but it's good enough to get you started as an engineer.

You better read the part about Vt again, it's all there.

 

[michael1978]

That's the best book...bar none! Everything I posted is in the Malvino! I gone very far as an engineer using nothing more than Malvino. Yes, I got into BJT IC design with just the knowledge of Malvino, it's that good! Of cause I learn a whole lot more since then, but it's good enough to get you started as an engineer.

You better read the part about Vt again, it's all there.

so its good book just to read good and to understand thnx for advice

 

[michael1978]

That's the best book...bar none! Everything I posted is in the Malvino! I gone very far as an engineer using nothing more than Malvino. Yes, I got into BJT IC design with just the knowledge of Malvino, it's that good! Of cause I learn a whole lot more since then, but it's good enough to get you started as an engineer.

You better read the part about Vt again, it's all there.

can i ask you something which symbols is this || for exaple rc||rl
is not rc divide rl, i thought || is divide / but is not
can you tell how you calculate with this symbol ||
thnx

 

[Jony130]

If you have a resistors connect in parallel you use this two parallel lines "||" to express it
R1||R2 = (R1*R2)/(R1+R2)

 

[yungman]

I use // as parallel. And Jony130 is correct.

 

[michael1978]

I use // as parallel. And Jony130 is correct.

§thank you ver much

 

[yungman]

You're welcome, keep digging into Malvino, it's worth your time. I hold this book is such high regard I did try to order a newer edition last year, but I ordered the experimental manual by mistake. I lost my original book long long time ago...since the early 80s. I want to keep a copy in my library collection.

Another part Malvino is very good is the op-amp. I still use what I learn from the book. Of cause you need more on Bode Plot stability design, but this really give you a strong start. I designed many closed loop feedback system, I use the Bode Plot starting from what I learn from Malvino.

 

[michael1978]

You're welcome, keep digging into Malvino, it's worth your time.

thanx man

 

[michael1978]

may i ask you something
FIRST
is this formula OK
for example rc=(3.600X10000)/(3600+10000)=2650
desired voltage 0.5, 2650/0.5=5300 so re=5300
voltage gain is 2650/5300=0.5
is this correct formula? if is this is correct formula why my simulator is 43mV not 50mv
where is wrong my formula of my simulator? can you exaplian me please thanx

SECOND how if you connect one re in serie with r'e, how decrase voltage gain, can you explain in similary ways please

 

[Jony130]

Why you built a amplifier with gain smaller then one?

 

[michael1978]

i am trying to learn like in book, transistor ce class a, and i do experments with software, but like you see my result are not correct, i do somewhere mistake

 

[michael1978]

thanx man

jjjjj

 

[michael1978]

I use // as parallel. And Jony130 is correct.

may i ask you something
FIRST
is this formula OK
for example rc=(3.600X10000)/(3600+10000)=2650
desired voltage 0.5, 2650/0.5=5300 so re=5300
voltage gain is 2650/5300=0.5
is this correct formula? if is this is correct formula why my simulator is 43mV not 50mv
where is wrong my formula of my simulator? can you exaplian me please thanx

SECOND how if you connect one re in serie with r'e, how decrase voltage gain, can you explain in similary ways please

 

[Jony130]

Show as a diagram and all components values.
And tell as how long you have been learn electronics?

 

[michael1978]

Show as a diagram and all components values.
And tell as how long you have been learn electronics?

long enough,
rc=3.600
rl=10000
rc=(3.600X10000)/(3600+10000)=2650
re bypass 5300, only 1 resistor in parallel with capacitor
voltage divider
r1=10000
r2 = 2.200
power supply 10v
input voltage 2mv at frequnecy 20K

 

[Jony130]

If I understand you correctly your circuit look like this:

and Vcc = 10V
If so the collector current is equal to

Ic ≈ Ve/RE ≈ 1.1V/5.3KΩ ≈ 0.2mA and Vc = Vcc - Ic *Rc = 10V - 0.2mA * 3.6K = 10V - 0.72V = 9.28V .

And the voltage gain

Av = (Rc||RL)/re = 2.65KΩ/130Ω = 20[V/V]

And

Vout = 20* Vin = 20 * 2mV = 40mV

But your amplifier Q point (bias point) was not chosen properly.
See this post
https://www.physicsforums.com/showthread.php?p=4058469#post4058469

Where
re = 26mV/Ic
Ve = (Vcc* R2/(R1+R2)) - Vbe

And next time try use Engineering notation instead of 3.600

 

[michael1978]

If I understand you correctly your circuit look like this:

and Vcc = 10V
If so the collector current is equal to

Ic ≈ Ve/Re ≈ 1.1V/5.3KΩ ≈ 0.2mA and Vc = Vcc - Ic *Rc = 10V - 0.2mA * 3.6K = 10V - 0.72V = 9.28V .

And the voltage gain

Av = (Rc||RL)/re = 2.65KΩ/130Ω = 20[V/V]

And

Vout = 20* Vin = 20 * 2mV = 40mV

But your amplifier Q point (bias point) was not chosen properly.
See this post
https://www.physicsforums.com/showthread.php?p=4058469#post4058469

Where
re = 26mV/Ic
Ve = (Vcc* R2/(R1+R2)) - Vbe

And next time try use Engineering notation instead of 3.600

sorry i am a beginner
my circuits look like that,

but when you start to build an amplifier, which are the first steps to take
and how to get a desired voltage,
i so my amplifier is bad, but is was good the re was 1k but me i changet to experiment to get desired voltage

what do you think, do you have time to build one amplifier with desired gain, can you show step by step and to explain in similary way please

 

[michael1978]

If I understand you correctly your circuit look like this:

and Vcc = 10V
If so the collector current is equal to

Ic ≈ Ve/RE ≈ 1.1V/5.3KΩ ≈ 0.2mA and Vc = Vcc - Ic *Rc = 10V - 0.2mA * 3.6K = 10V - 0.72V = 9.28V .

And the voltage gain

Av = (Rc||RL)/re = 2.65KΩ/130Ω = 20[V/V]

And

Vout = 20* Vin = 20 * 2mV = 40mV

But your amplifier Q point (bias point) was not chosen properly.
See this post
https://www.physicsforums.com/showthread.php?p=4058469#post4058469

Where
re = 26mV/Ic
Ve = (Vcc* R2/(R1+R2)) - Vbe

And next time try use Engineering notation instead of 3.600

and how you get 130Ω? Av = (Rc||RL)/re = 2.65KΩ/130Ω = 20[V/V] PLEASE CAN YOU TELL ME YOUR CALCULATION

 

[michael1978]

i am not an enigner, i am total beginner, i try to leran self

 

[Jony130]

If you want to learn you need first understand how this amplifier work.
Also you need to understand how BJT work? And I don't ask about how BJT really work from physical point of view.

 

[michael1978]

If you want to learn you need first understand how this amplifier work.
Also you need to understand how BJT work? And I don't ask about how BJT really work from physical point of view.

i know a little bit, i know a biasing, i know to find a amplifier voltage, i speak about ce amplifier, but i don't know how to get desired voltage, that is my big problem, like you find 40mv

 

[Averagesupernova]

First things first. Can you tell us what the approximate DC voltages are at various points in the circuit and why? If you can then you are well on your way to understanding it. I realize Jony has already posted things of this nature but I feel you may not know how to arrive at this. You say you know biasing but I feel you are coming up short in that area. Do you know how to select the collector resistor to get the collector voltage where you want it based on a few other things in the circuit? And, what are those things and why they affect collector voltage? For that matter do you know where the collector voltage should be? Nothing wrong with not knowing this stuff. Not trying to make you out to look dumb or anyting. We're here to help. Where would you like to start specifically?

 

[michael1978]

First things first. Can you tell us what the approximate DC voltages are at various points in the circuit and why? If you can then you are well on your way to understanding it. I realize Jony has already posted things of this nature but I feel you may not know how to arrive at this. You say you know biasing but I feel you are coming up short in that area. Do you know how to select the collector resistor to get the collector voltage where you want it based on a few other things in the circuit? And, what are those things and why they affect collector voltage? For that matter do you know where the collector voltage should be? Nothing wrong with not knowing this stuff. Not trying to make you out to look dumb or anyting. We're here to help. Where would you like to start specifically?

look i am reading one book, and i learn about transistor, fundamental biasin, amplifier
but he show me how to find av, but not how to buid amplifier with desired voltage, like here they say formula is rc//rl/re, i know this formula from the book, but desired voltage he don't tell me, i can build one amplifier but not with desired vooltage, for example i want av about 50 and input 2mv = 100mv, i don't know how to do it, for that i ask help here, in similary way

 

[Averagesupernova]

You need to know the load impedance if before you go much farther, that is assuming you are driving a load. I will assume you don't know what the DC collector voltage should be. To start with we will set it at about half the supply voltage, so 5 volts. What do you know to do from here? I don't plan on designing the whole amp for you. Doesn't matter to me if it is not homework, I will treat it as so and make you learn it.

 

[NascentOxygen]

and how you get 130Ω?

That's the value of r_e, which is the slope (i.e., incremental or AC or small-signal resistance) of the B-E junction. The B-E junction has an average current (= bias current) of I_E, and for Si junctions the slope of the characteristic exponential graph V vs. I at any value of current is: slope = ΔV/ΔI = 0.026/I. So evaluate this fraction for the value of I_E bias to arrive at this value for r_e.

And tell as how long you have been learn electronics?

long enough,

Righto!

 

[michael1978]

You need to know the load impedance if before you go much farther, that is assuming you are driving a load. I will assume you don't know what the DC collector voltage should be. To start with we will set it at about half the supply voltage, so 5 volts. What do you know to do from here? I don't plan on designing the whole amp for you. Doesn't matter to me if it is not homework, I will treat it as so and make you learn it.

do you see
vcc 10V
collector 10
IC 10ma
Vce midpoint
Bdc 100

now what to do, how to find desired gain

 

[michael1978]

That's the value of r_e, which is the slope (i.e., incremental or AC or small-signal resistance) of the B-E junction. The B-E junction has an average current (= bias current) of I_E, and for Si junctions the slope of the characteristic exponential graph V vs. I at any value of current is: slope = ΔV/ΔI = 0.026/I. So evaluate this fraction for the value of I_E bias to arrive at this value for r_e.

Righto!

so you mean to decrase of increase re til i get the desired gain of something else? can you show me example, because i don't understand you so good my english is also not so good ;-)
do not exist some formula, how i have to do with calculator can you tell me please thnx

 

[Averagesupernova]

Thats not exactly what I had in mind but it is a start. I think if you actually knew how to get those resistor values and voltages you would not need to ask how to figure the gain. Did you plug some numbers into a simulator or something to get this? I mentioned to you that we would put the collector voltage at 5 volts. You have it at 6 volts which is close enough. AC voltage gain of this amplifier is 4 UNLOADED. How can you not know that? You have previously posted that you know the formula for gain which is (RL ||RC)/RE. Zout of this amplifier is 400 ohms.
-
Next. Zin. Originally you wanted a Zin of 50K. Do you see how that is next to impossible with transistor betas typically around 100 using components you have selected?

 

[michael1978]

Thats not exactly what I had in mind but it is a start. I think if you actually knew how to get those resistor values and voltages you would not need to ask how to figure the gain. Did you plug some numbers into a simulator or something to get this? I mentioned to you that we would put the collector voltage at 5 volts. You have it at 6 volts which is close enough. AC voltage gain of this amplifier is 4 UNLOADED. How can you not know that? You have previously posted that you know the formula for gain which is (RL ||RC)/RE. Zout of this amplifier is 400 ohms.
-
Next. Zin. Originally you wanted a Zin of 50K. Do you see how that is next to impossible with transistor betas typically around 100 using components you have selected?

i know the formula, but for example i want a gain of 5 and 2mv input is 100,how to say i want to be other gain, how to say i want self a gain to design, what i have to change, where i have to start, when somebody design an amplifier how they start from begin, and how they change a gain thnx

 

[michael1978]

You want output to be 2K, your rc has to be 2K. You want voltage gain of 2, so you want your re≈1K. Gain=rc/(re+r'e) where r'e= 1/gm≈Vt/Ic. Vt≈25mV at 25deg C. For Ie=1mA, r'e≈25Ω. So if you use 1mA bias current, re=1K is close enough.

If you choose a NPN with β>100, input impedance of the transistor is βXre≈100K, this is going to be a little tricky getting input impedance of 50K as the input impedance of the transistor is as low as 100K or higher if β is larger.


What is r1 and r2? is this a voltage divider to bias the base of the BJT? If so, you want Zin=50K, then you want r1//r2=100K so you get about 50K when parallel with the input impedance of the transistor. But this is not reliable.

You need to relax some requirement, if you can accept output impedance of say 20K, then re=10K and your input impedance of the transistor can by up to 1MΩ. Then you can ignore the input impedance of the transistor and make r1//r2= 50K.

For fmin, C=1/(2πRf) where R=50k input impedance, f=10Hz.

HI
can you explain me of design one amplifier with gain of 50, rc 1k, power supply 12 and the other select you please in similary way, like you explain in first example

 

[Averagesupernova]

Let me ask you this michael: How did you select the resistors in the schematic that you posted in post #42?

 

[Jony130]

When we start design any circuit we need to know circuit specification.
If you have a 2mV input voltage and 100mV at output you need a amplifier with gain
Av = 100mV/2mV = 50[V/V] and Zin = 50K and Rload >2K.
Is not so easy to meet all this requirements whit this simple amplifier.
So I change them to
Voltage Gain= 50
Load Resistance= 10k ohm
Vce= 5V

First we need select BJT I choose BC546C with typical hfe = 520 and Hfe_min = 420

I start selection from Rc resistor.

Rc < 0.1Rload = 1KΩ

Additional I assume Ve = 1V

So

Ic = (Vcc - Vce - Ve)/Rc = (10V - 5V - 1V)/1KΩ = 4mA

next

Re1 = Ve/Ic = 1V/4mA = 250 but I chose 220Ω

Vb = Ic*RE + Vbe = 4mA * 220Ω + 0.65V = 1.53V (voltage at base)

Ib = Ic/Hfe_min = 4mA/420 ≈ 10μA (base current)

R2 = Vb / ( 5 * Ib) = 30K

R1 = ( Vcc - Vb) / ( 6 * Ib) = 150KΩ

So know if we want voltage gain 50V/V

Av = 50 = (Rc|| RL) / ( re + (Re1||Re2) )

( re + (Re1||Re2) ) = (Rc|| RL) / 70 = 909Ω/50 = 18Ω

re = 26mV/Ic = 26mV/4mA = 6.5Ω

18Ω = (re + (Re1||Re2)) = ( 6.5Ω + (220||Re2) )

Re1||Re2 = 18Ω - re = 11.5Ω

Re2 = ( 220 * 11.5 ) / (220 - 12.5) = 12.1 = 12Ω.

And now we have a circuit that we can change gain quite easily.

From
Rc/Re1 = 1K/220 = 4.5[V/V] if we remove Re2 and C2

to

Rc/re = 1K/6.5 = 153[V/V] if we short Re1.And normally to meet all your requirements we need to use more practical amplifier circuit or op amp.

 

[michael1978]

when we start design any circuit we need to know circuit specification.
If you have a 2mv input voltage and 100mv at output you need a amplifier with gain
av = 100mv/2mv = 50[v/v] and zin = 50k and rload >2k.
Is not so easy to meet all this requirements whit this simple amplifier.
So i change them to
voltage gain= 50
load resistance= 10k ohm
vce= 5v

first we need select bjt i choose bc546c with typical hfe = 520 and hfe_min = 420

i start selection from rc resistor.

Rc < 0.1rload = 1kΩ

additional i assume ve = 1v

so

ic = (vcc - vce - ve)/rc = (10v - 5v - 1v)/1kΩ = 4ma

next

re1 = ve/ic = 1v/4ma = 250 but i chose 220Ω

vb = ic*re + vbe = 4ma * 220Ω + 0.65v = 1.53v (voltage at base)

ib = ic/hfe_min = 4ma/420 ≈ 10μa (base current)

r2 = vb / ( 5 * ib) = 30k

r1 = ( vcc - vb) / ( 6 * ib) = 150kΩ

so know if we want voltage gain 50v/v

av = 50 = (rc|| rl) / ( re + (re1||re2) )

( re + (re1||re2) ) = (rc|| rl) / 70 = 909Ω/50 = 18Ω

re = 26mv/ic = 26mv/4ma = 6.5Ω

18Ω = (re + (re1||re2)) = ( 6.5Ω + (220||re2) )

re1||re2 = 18Ω - re = 11.5Ω

re2 = ( 220 * 11.5 ) / (220 - 12.5) = 12.1 = 12Ω.

And now we have a circuit that we can change gain quite easily.

from
rc/re1 = 1k/220 = 4.5[v/v] if we remove re2 and c2

to

rc/re = 1k/6.5 = 153[v/v] if we short re1.


And normally to meet all your requirements we need to use more practical amplifier circuit or op amp.

so this is the steps if i need to build an amplifier thnx
gain rc//rl/re, but i see there in resistor in serie with capacitor
how come now gain RC//RL/RE2+RE1 am i correct, if not can you tell me
i don't get gain of 50

 

[michael1978]

Let me ask you this michael: How did you select the resistors in the schematic that you posted in post #42?

i think first i select voltage divider, and after rc and re