the voltage gain of a common emitter circuit

Thread Starter

Ghina Bayyat

Joined Mar 11, 2018
139
(β+1) is a "transformation factor", any resistance at the base is seen (reflect it to the emitter) at the emitter as a β + 1 smaller resistance ( re = r_pi/(β + 1 ) ).
And any resistance at the emitter is seen at the base node as a (β + 1) large resistance r_pi = ( β + 1)*re
so it is a reflection like if we say
vre = re * ie is the same of vre = βre * ib and it depends just upon the node we're looking at the circuit from right ?
 

Jony130

Joined Feb 17, 2009
5,487
Av = β Rc / Rb and some times it is Rc / Re
How you define Rb and Re ?

The first equation is true only if you have a large Rb resistor in series with the base terminal (much larger than BJT's input resistance r_pi) or if you rename r_pi into Rb.
The second one is only true if you have a External Re resistor without the CE capacitor. And both of these formulas only give us approximate value (ballpark number).

vre = re * ie is the same of vre = βre * ib and it depends just upon the node we're looking at the circuit from right ?
vbe = re*ie = (β+1)re*ib
or
vbe = ib*r_pi = r_pi/(β +1) * ie
 

Thread Starter

Ghina Bayyat

Joined Mar 11, 2018
139
The first equation is true only if you have a large Rb resistor in series with the base terminal (much larger than BJT's input resistance r_pi) or if you rename r_pi into Rb.
The second one is only true if you have a External Re resistor without the CE capacitor. And both of these formulas only give us approximate value (ballpark number).
so if i have a ce capacitor i can't say that Av = Rc / Re the internal base emitter resistor and i have to use the equation Av = β Rc / Rb
vbe = re*ie = (β+1)re*ib
or
vbe = ib*r_pi = r_pi/(β +1) * ie
and so re = r-pi / (β+1)
thanks but here u are talking about Vbe and i meant Vre { i know i forgot to say that ie = (β+1) ib but it's not the problem } re and r-pi are not the issue here and i don't understand why u guys are talking about them
now we can say that
by looking from the emitter point of view Av = - rc / re if we have re external but if we don't we can't say that Av = rc / re internal
and if we don't we must look into the circuit from the base point of view and this will give us that Av = Vout / Vin and Vin will be vbe right ? that's why we say that
av = ic*rc / ib*rb ⇒( β*ib )* rc / ib *rb ⇒ Av = β Rc / Rb
but still my question : how could we have a different value of Av ?
just because re is exist or not so that the gain changes ??
 
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Thread Starter

Ghina Bayyat

Joined Mar 11, 2018
139
also all of this made me think of another question:
what about the current gain ?? does this mean that when we have a circuit with Re then we must say that Ai = iout / i in and i in is ie ?so the current gain will be equal to ALPHA ? { by the way since i learned about the ratio alpha and i don't know what do we actually use it for }
but if it's true then the existance of re will destroy the amplifiication
i know it affects the gain of the circuit but didn't know that it affects it that much
and from what i know the input resistor is the resistor connected in series with the base and the resistor that is connected in series is Re + RE
some times the base resistor isn't connected in series and if there is no RE then what do we do ?? there is no input resistance in this case if we can't consider Re internal as an input resistance so which equation do we have to use ?
 
Last edited:

Jony130

Joined Feb 17, 2009
5,487
so if i have a ce capacitor i can't say that Av = Rc / Re the internal base emitter resistor and i have to use the equation Av = β Rc / Rb
To stop further confusion let us exam this cases:

1 )

1.png

Av ≈ - Rc/re *β/(β+1) ≈ Rc/re ≈ Rc/(26mV/Ie) ≈ gm*Rc ≈ 40*Ic*Rc


2)

2.png

Av = - Rc/(re + RE) * β/(β+1) ≈ - Rc/(re + RE)


3)

3.png

Av = - R3/re * β/(β+1)≈ R3/(26mV/Ie) ≈ gm*R3 ≈ 40*Ic*R3
 

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Last edited:

Jony130

Joined Feb 17, 2009
5,487
Also, "little" re is sometimes called intrinsic emitter resistance. And to avoid confusion you should use capital letters for "real" resistors Rc, RE. And small letters for dynamic resistances r_pi, re.

As for Rin.

The BJT alone input resistance is d(Vbe)/d(Ib) = r_pi = gm/β = (β + 1)re

where

gm is a "gain" for any transconductance amplifier (voltage to current convertor)

And for BJT's the transistor Ic current is a controlled via input Vbe voltage.

And if you plot Ic = f(Vbe)



The gm is the slope of this curve

gm = d(Ic)/d(Vbe) = Ic/Vt = Ic/26mV
 

Thread Starter

Ghina Bayyat

Joined Mar 11, 2018
139
Also, "little" re is sometimes called intrinsic emitter resistance. And to avoid confusion you should use capital letters for "real" resistors Rc, RE. And small letters for dynamic resistances r_pi, re.

As for Rin.

The BJT alone input resistance is d(Vbe)/d(Ib) = r_pi = gm/β = (β + 1)re

where

gm is a "gain" for any transconductance amplifier (voltage to current convertor)

And for BJT's the transistor Ic current is a controlled via input Vbe voltage.

And if you plot Ic = f(Vbe)



The gm is the slope of this curve

gm = d(Ic)/d(Vbe) = Ic/Vt = Ic/26mV
yes i understand thank u
so there are different cases for the voltage gain depending on the resistors exist
and there is nothing wrong to say Av = rc / re internal if there is no base resistor connected in series with base nor RE resistors connected in series with the emitter
but can u please answer my question here
also all of this made me think of another question:
what about the current gain ?? does this mean that when we have a circuit with Re then we must say that Ai = iout / i in and i in is ie ?so the current gain will be equal to ALPHA ? { by the way since i learned about the ratio alpha and i don't know what do we actually use it for }
but if it's true then the existance of re will destroy the amplifiication
i know it affects the gain of the circuit but didn't know that it affects it that much
and from what i know the input resistor is the resistor connected in series with the base and the resistor that is connected in series is Re + RE
some times the base resistor isn't connected in series and if there is no RE then what do we do ?? there is no input resistance in this case if we can't consider Re internal as an input resistance so which equation do we have to use ?
 

Jony130

Joined Feb 17, 2009
5,487
hat about the current gain ?? does this mean that when we have a circuit with Re then we must say that Ai = iout / i in and i in is ie ?
For common-emitter amplifier the Ai = Iout/Iin And Iin is not equal to Ie.
Try solve for Ai yourself.

i learned about the ratio alpha and i don't know what do we actually use it for
alpha is simply Ic/Ie the current gain in common-base amplifer.

and from what i know the input resistor is the resistor connected in series with the base and the resistor that is connected in series is Re + RE
some times the base resistor isn't connected in series and if there is no RE then what do we do ?? there is no input resistance in this case if we can't consider Re internal as an input resistance so which equation do we have to use ?
Haven't I already talk about Rin ? See post #13 for example.
 

Thread Starter

Ghina Bayyat

Joined Mar 11, 2018
139
For common-emitter amplifier the Ai = Iout/Iin And Iin is not equal to Ie.
Try solve for Ai yourself.
i know that Iin =ib but i'm just asking IF i was looking at the circuit from the emitter point of view then would it mean that Iin = ie ? it can't be right i know but how can i look from the emitter point and say that ib is the input current ? this is my question maybe i couldn't explain well
alpha is simply Ic/Ie the current gain in common-base amplifer.
so it's got nothing to do with the common emitter { didn't have the common base lesson yet }
Haven't I already talk about Rin ? See post #13 for example.
yes yes i understand about that now i meant my question about the current gain
sorry
 

Jony130

Joined Feb 17, 2009
5,487
The current gain for common-emitter is equal to β = Ib/Ic for a ideal case.

For this amplifier

1.png
The current gain is less than the beta because some part of an input current flows into RB resistor. So the base current is not equal to Iin.

Ai = Io/Iin = RB/(Rb + r_pi)*β

Also adding the load resistance reduces the current gain

Ai = Io/Iin = RB/(Rb + r_pi)*β* Rc/(Rc+RL)
 

Thread Starter

Ghina Bayyat

Joined Mar 11, 2018
139
The current gain for common-emitter is equal to β = Ib/Ic for a ideal case.

For this amplifier

View attachment 151292
The current gain is less than the beta because some part of an input current flows into RB resistor. So the base current is not equal to Iin.

Ai = Io/Iin = RB/(Rb + r_pi)*β

Also adding the load resistance reduces the current gain

Ai = Io/Iin = RB/(Rb + r_pi)*β* Rc/(Rc+RL)
yes i get it
every thing is clear now thanks to u all
 
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