multistage amplifier

Thread Starter

amangupta1219

Joined Oct 18, 2012
19
for the circuit shown , if i want to calculate gain of the second stage then do i need to consider the output resistance of first stage as the source resistance for the second stage ?
first stage is common collector and second is common base
and voltage gain for common base is
Av = (gm/RS)(RcllRL)(re ll RE ll RS)
If yes then the output resistance of first stage (Common collector) will be
RO1 = re ll RE ll ro = .02513 KΩ
RS = Ro1
then gain equation will become AV2 = gm2(RC)(re2llRS) = 76.92
but in the solutions they have ignored the effect of RS and used AV2 = gm2(RC)
i.e they ignored the effect of stage 1 on stage 2 to find Vo2/Vo1 but have considered the effect of stage 2 on stage 1 to find Vo1/Vi
please clarify this
 

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Jony130

Joined Feb 17, 2009
5,487
For me the voltage gain is equal to:

Av1 = Vo1/Vi = (re2||RE)/( re1 + (re2||RE)) = (26Ω||1K) /(26Ω + 26Ω||1K ) = 0.493V/V

Common base stage voltage gain is

Av2 = Vo2/Vo1
gm* Rc Rc/re2 = 4KΩ/26Ω = 152.846V/V

And overall voltage gain is equal

Av = Av1 * Av2 = 75.3V/V

I ignore ro because Early effect voltage is equal to ∞.

Why you don't use a small signal analysis?



As you can see the circuit is quite simple
 

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Thread Starter

amangupta1219

Joined Oct 18, 2012
19
okay i am getting the correct answer from your method.
But if we use the standard formula of voltage gain for common base then do we consider the output resistance of 1st stage as the internal resistance of voltage source for the 2nd stage i.e common base ?
 

The Electrician

Joined Oct 9, 2007
2,971
okay i am getting the correct answer from your method.
But if we use the standard formula of voltage gain for common base then do we consider the output resistance of 1st stage as the internal resistance of voltage source for the 2nd stage i.e common base ?
In your first post you say:

"then gain equation will become AV2 = gm2(RC)(re2llRS) = 76.92
but in the solutions they have ignored the effect of RS and used AV2 = gm2(RC)"

How did you get from here "AV2 = gm2(RC)" to here "AV2 = gm2(RC)(re2llRS)"?

The second formula isn't dimensionally correct.

I would have thought you would want to let the output resistance of the first stage form a voltage divider with the input resistance of the second stage, which you haven't done correctly.
 

Thread Starter

amangupta1219

Joined Oct 18, 2012
19
In your first post you say:

"then gain equation will become AV2 = gm2(RC)(re2llRS) = 76.92
but in the solutions they have ignored the effect of RS and used AV2 = gm2(RC)"

How did you get from here "AV2 = gm2(RC)" to here "AV2 = gm2(RC)(re2llRS)"?

The second formula isn't dimensionally correct.

I would have thought you would want to let the output resistance of the first stage form a voltage divider with the input resistance of the second stage, which you haven't done correctly.
sorry
in the first post i also wrote the formula for voltage gain of common base as Av = (gm/RS)(RcllRL)(re ll RE ll RS)
and for my circuit the second stage is common base for which RE is
and no RL . So gain expression will become Av = (gm2/RS)(Rc)(re ll RS)
i forgot to divide by Rs. But still answer in the textbook for gain of second stage is 153.8 and the factor of 2 is coming because they ignored Rs.
This is a question from electronic circuits analysis & analysis by donald neamen
 

The Electrician

Joined Oct 9, 2007
2,971
The gain of the second stage without taking into account any loading effects due to the output resistance of the first stage would be Av = gm2*Rc

Now the voltage divider effect due to the output resistance of the first stage (Rs) and the input resistance of the second stage (re) would add a factor of re/(Rs+Re), or, as you have it, (re || Rs)/Rs. The value of this quantity is .5085, which is where the factor of 2 comes in.

Your confusion comes from the fact that you are accounting for the loading effects twice.

The input resistance of the second stage loads the first stage and reduces the first stage gain from .97466 (which is what it would be if the first stage wasn't loaded by the second stage), to .4936 with the loading of the second stage.

Now, if you include the effect of Rs in the gain of the second stage, as you have done, the loading of the second stage on the first stage is being included in the overall gain twice, and that's not what you want to do.

Furthermore, the problem asks you for Av2 = Vo2/Vo1, which is the ratio of the output voltage Vo2 to the voltage that actually appears at the node Vo1. The voltage at Vo1 is the voltage that comes after the voltage divider effect of Rs and re2, so you shouldn't include the loading effect in the gain of the second stage; it has already been included in the first stage gain.
 
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