Please help transistor amplifier

[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

 

[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 voltage divider and after rc and re

 

[Averagesupernova]

i think first voltage divider and after rc and re

You went about it backwards. R1 and R2 are of course significant in order to get the proper base voltage but they are also important to determine input impedance. So their values need to be kept in mind for this. A low output impedance will reflect way back to a lower input impedance. The amplifier Jony posted has a Zin of about 1000 ohms. Can you tell why a single stage amplifier with the gain and Zout that you want cannot have a Zin of 50K?

 

[michael1978]

You went about it backwards. R1 and R2 are of course significant in order to get the proper base voltage but they are also important to determine input impedance. So their values need to be kept in mind for this. A low output impedance will reflect way back to a lower input impedance. The amplifier Jony posted has a Zin of about 1000 ohms. Can you tell why a single stage amplifier with the gain and Zout that you want cannot have a Zin of 50K?

ooo man i don't know how to deterimine zin and zout ;-) how do you select zin and zout, i am just a begginer, is zin R1 and R2 and zout rc and rl, if is yes, how you determine how you select calculate

 

[Averagesupernova]

Collector resistor is Zout.
Zin is R1||R2||(beta*Re). In the schematic that Jony posted Re becomes Re1||Re2.

 

[Jony130]

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

Simply voltage gain of a CE amplifier is always equal to

Av = ( resistance seen from collector toward the output) / (re + (resistance seen from emitter to gnd)

So we have

Av = (Rc||RL)/ ( re + Re1||Re2) = 909Ω/( 6.5Ω + 220Ω||12Ω ) ≈ 909/18.5Ω = 49.13[V/V]

Zin = R1||R2||( Hfe+1 *(re+Re1||Re2) ) = 25KΩ||( 421 * 18.5Ω) = 25K||7.8KΩ = 5.9KΩ
And
Zout ≈ Rc ≈ 1KΩ

 

[michael1978]

Simply voltage gain of a CE amplifier is always equal to

Av = ( resistance seen from collector toward the output) / (re + (resistance seen from emitter to gnd)

So we have

Av = (Rc||RL)/ ( re + Re1||Re2) = 909Ω/( 6.5Ω + 220Ω||12Ω ) ≈ 909/18.5Ω = 49.13[V/V]

Zin = R1||R2||( Hfe+1 *(re+Re1||Re2) ) = 25KΩ||( 421 * 18.5Ω) = 25K||7.8KΩ = 5.9KΩ
And
Zout ≈ Rc ≈ 1KΩ

thank you,
you show me a good example
but i have one more question, if somebody want to desigin a amplifier, they do it like your example, of another design, i mean for ce claas a amplifier
one more time thnx for good example

 

[michael1978]

Collector resistor is Zout.
Zin is R1||R2||(beta*Re). In the schematic that Jony posted Re becomes Re1||Re2.

yes he show me a good example, but is so much important zin and zout in amplifier,
did you show jony example, this steps i have to take all time if i design an amplifier
thnx for answer

 

[michael1978]

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

jony i start to look at you amplifier,
but i don't understand this ( re + (re1||re2) ) = (rc|| rl) / 70 = 909Ω/50 = 18Ω

where did you take 70? 909?50? can you explain me please...

 

[Jony130]

jony i start to look at you amplifier,
but i don't understand this ( re + (re1||re2) ) = (rc|| rl) / 70 = 909Ω/50 = 18Ω

where did you take 70? 909?50? can you explain me please...

There is a error instead-of 70 it should be 50. The voltage gain we want is equal to 50[V/V].
OE amplifier voltage gain is equal to
Av = Rc/re so to find desired re value I use this equation:

re = Rc/Av

And in my example the gain I want is equal to 50[V/V] and Rc = 1K; RL = 10K
So form this

Rc||RL = 909Ω

re = Rc||RL/Av = (1K||10K)/50 = 909Ω/50 = 18Ω

It is clear now ?

 

[michael1978]

There is a error instead-of 70 it should be 50. The voltage gain we want is equal to 50[V/V].
OE amplifier voltage gain is equal to
Av = Rc/re so to find desired re value I use this equation:

re = Rc/Av

And in my example the gain I want is equal to 50[V/V] and Rc = 1K; RL = 10K
So form this

Rc||RL = 909Ω

re = Rc||RL/Av = (1K||10K)/50 = 909Ω/50 = 18Ω

It is clear now ?

yes thank you