Please help transistor amplifier

In summary: So, if you choose Ie=1mA, you only have 25Ω to play with, which means you want rl to be very large value, which means you want to use a Darlington transistor.In summary, if you want to design a CE amplifier with a gain of 50, you can use two stages, each with a gain of x10 and x5 respectively. Alternatively, you can use one stage with a gain of 50, but it may be more difficult to design and may not be as reliable. It is important to consider the input and output impedances of the circuit in order to achieve the desired voltage gain.
  • #36
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.
 
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  • #37
Jony130 said:
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
 
  • #38
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?
 
  • #39
Averagesupernova said:
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
 
  • #40
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.
 
  • #41
michael1978 said:
and how you get 130Ω?
That's the value of re, 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 IE, 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 IE bias to arrive at this value for re.
Jony130 said:
And tell as how long you have been learn electronics?
michael1978 said:
long enough,
Righto! :approve:
 
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  • #42
Averagesupernova said:
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.

attachment.php?attachmentid=53149&stc=1&d=1353416898.jpg


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

now what to do, how to find desired gain
 

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  • #43
NascentOxygen said:
That's the value of re, 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 IE, 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 IE bias to arrive at this value for re.

Righto! :approve:

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
 
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  • #44
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?
 
  • #45
Averagesupernova said:
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
 
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  • #46
yungman said:
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
 
  • #47
Let me ask you this michael: How did you select the resistors in the schematic that you posted in post #42?
 
  • #48
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.

attachment.php?attachmentid=53164&stc=1&d=1353439458.png


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.
 

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  • #49
jony130 said:
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.

attachment.php?attachmentid=53164&stc=1&d=1353439458.png


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
 
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  • #50
Averagesupernova said:
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
 
  • #51
Averagesupernova said:
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
 
  • #52
michael1978 said:
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?
 
  • #53
Averagesupernova said:
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
 
  • #54
Collector resistor is Zout.
Zin is R1||R2||(beta*Re). In the schematic that Jony posted Re becomes Re1||Re2.
 
  • #55
michael1978 said:
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
 
  • #56
Jony130 said:
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
 
  • #57
Averagesupernova said:
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
 
  • #58
michael1978 said:
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...
 
  • #59
michael1978 said:
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 ?
 
  • #60
Jony130 said:
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
 
  • #61
Averagesupernova said:
Collector resistor is Zout.
Zin is R1||R2||(beta*Re). In the schematic that Jony posted Re becomes Re1||Re2.
is so important to know resitance zin and zout, i am a begginer, how do you select zin zout is easy zout=RC but zin? you select, but zin how do you select
 
  • #62
michael1978 said:
yes thank you

oo joney sorry i have to ask you this
is this correct re2 = ( 220 * 11.5 ) / (220 - 12.5) = 12.1 = 12Ω.
of this one re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.1 = 12Ω.
 
  • #63
Michael, are you asking why Zin and Zout are important to know?
 
  • #64
michael1978 said:
of this one re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.1 = 12Ω.
This one is correct of course.
 
  • #65
Jony130 said:
This one is correct of course.

jony can i ask you something, is possible for you to make one another amplifier but without Re2, with gain av 50, i know you maket very good the one, i understand all, but how to make without re2, you can select self the value but desired voltage to be 50, can you do it the same as in first example , if you want thank you PLEASE
 
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  • #66
OK
Av = Rc/re = gm*Rc = Ic/26mV*Rc = (Ic*Rc)/26mV

50[V/V] = (Ic*Rc)/26mV

from this

(Ic*Rc) = 1.3

I choose Rc = 1K

So

Ic = 1.3/1K = 1.3mA

But as you can see this approach is not very good because give as low voltage swing.

We need to use Re1 or use a different type of a circuit. But lest as add Re1 resistor.
Now Av = Rc/(re+Re1)

And for Rc = 1K we can use this

Ic = 0.5Vcc/Rc = 5V/1K = 5mA

So re = 26mV/5mA = 5.2Ω

And

Re1 = Rc/Av - re = 1K/50 - 5.2Ω = 20 - 5.2Ω = 14Ω

This circuit is also not very good because low Re1 means that Hfe spread and temperature change will have significant impact on bias point stability. This is why I use Ve = 1V and add Re2 to reduce the voltage gain to 50[V/V]
 
  • #67
jony130 said:
ok
av = rc/re = gm*rc = ic/26mv*rc = (ic*rc)/26mv

50[v/v] = (ic*rc)/26mv

from this

(ic*rc) = 1.3

i choose rc = 1k

so

ic = 1.3/1k = 1.3ma

but as you can see this approach is not very good because give as low voltage swing.

We need to use re1 or use a different type of a circuit. But lest as add re1 resistor.
Now av = rc/(re+re1)

and for rc = 1k we can use this

ic = 0.5vcc/rc = 5v/1k = 5ma

so re = 26mv/5ma = 5.2Ω

and

re1 = rc/av - re = 1k/50 - 5.2Ω = 20 - 5.2Ω = 14Ω

this circuit is also not very good because low re1 means that hfe spread and temperature change will have significant impact on bias point stability. This is why i use ve = 1v and add re2 to reduce the voltage gain to 50[v/v]

thank you very much
 
  • #68
Also we can split Re resistor with a capacitor. And use Re1 = 14Ω and Re2 = Ve/Ic - re1 = 1V/5mA - 14Ω ≈ 180Ω

attachment.php?attachmentid=53268&stc=1&d=1353703873.png


And now we can select R1 and R2

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

If we assume Hfe = 150

Ib = Ic/hfe = 5mA/150 = 34μA

R1 = (Vcc - Vb)/( 11*Ib) = 22KΩ

R2 = Vb/(10*Ib) = 4.7KΩ
 

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  • #69
Jony130 said:
OK
Av = Rc/re = gm*Rc = Ic/26mV*Rc = (Ic*Rc)/26mV

50[V/V] = (Ic*Rc)/26mV

from this

(Ic*Rc) = 1.3

I choose Rc = 1K

So

Ic = 1.3/1K = 1.3mA

But as you can see this approach is not very good because give as low voltage swing.

We need to use Re1 or use a different type of a circuit. But lest as add Re1 resistor.
Now Av = Rc/(re+Re1)

And for Rc = 1K we can use this

Ic = 0.5Vcc/Rc = 5V/1K = 5mA

So re = 26mV/5mA = 5.2Ω

And

Re1 = Rc/Av - re = 1K/50 - 5.2Ω = 20 - 5.2Ω = 14Ω

This circuit is also not very good because low Re1 means that Hfe spread and temperature change will have significant impact on bias point stability. This is why I use Ve = 1V and add Re2 to reduce the voltage gain to 50[V/V]

AND value of r1 and r2 is r1=22k, and r2=4.7K and power supply 10V
 
  • #70
michael1978 said:
AND value of r1 and r2 is r1=22k, and r2=4.7K and power supply 10V
Yes for Vcc = 10V and hfe = 150
 
<h2>1. What is a transistor amplifier?</h2><p>A transistor amplifier is an electronic device that uses transistors to amplify an electrical signal. It is commonly used in electronic circuits to increase the strength of a weak signal.</p><h2>2. How does a transistor amplifier work?</h2><p>A transistor amplifier works by using a small current to control a larger current. This is achieved by applying a small input voltage to the base of the transistor, which then allows a larger current to flow from the collector to the emitter. This amplifies the signal.</p><h2>3. What are the benefits of using a transistor amplifier?</h2><p>Transistor amplifiers offer several benefits, including high gain, low distortion, and high input impedance. They are also small in size and require low power, making them ideal for use in portable electronic devices.</p><h2>4. What are the different types of transistor amplifiers?</h2><p>There are three main types of transistor amplifiers: common emitter, common base, and common collector. Each type has its own unique characteristics and is used for different applications.</p><h2>5. How do I choose the right transistor amplifier for my project?</h2><p>When choosing a transistor amplifier, it is important to consider factors such as the desired gain, frequency range, and input and output impedance. It is also important to select a transistor with a suitable power rating for your project.</p>

1. What is a transistor amplifier?

A transistor amplifier is an electronic device that uses transistors to amplify an electrical signal. It is commonly used in electronic circuits to increase the strength of a weak signal.

2. How does a transistor amplifier work?

A transistor amplifier works by using a small current to control a larger current. This is achieved by applying a small input voltage to the base of the transistor, which then allows a larger current to flow from the collector to the emitter. This amplifies the signal.

3. What are the benefits of using a transistor amplifier?

Transistor amplifiers offer several benefits, including high gain, low distortion, and high input impedance. They are also small in size and require low power, making them ideal for use in portable electronic devices.

4. What are the different types of transistor amplifiers?

There are three main types of transistor amplifiers: common emitter, common base, and common collector. Each type has its own unique characteristics and is used for different applications.

5. How do I choose the right transistor amplifier for my project?

When choosing a transistor amplifier, it is important to consider factors such as the desired gain, frequency range, and input and output impedance. It is also important to select a transistor with a suitable power rating for your project.

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