# Gain of an Amplifier using equivalent model for transistors

Discussion in 'Homework Help' started by AD633, Jul 19, 2013.

Jun 22, 2013
96
1
Hi,

I want to calculate the ampliefier circuit gain $Av = (Vo) / (Vi)$ using the equivalent model that i found for the circuit.

Therefore i need to know the expression for Vi and Vo
$

Vo=(hfeib2+ib2)R3

Vi=IRB+hie1*ib1

Vi=(V)/(R2)+hie*ib1

Av=vo/vi

AV=((hfeib2+ib2)R3))/((V)/(R2))+hie.ib1))

$

I need to eliminate the ib terms in the equation in order to be able to calculate the gain.How can i do that?Do i have to apply KCL in a specific node of the circuit?Something like Ib2=Ib1+I1.42kOhm
Thanks

(In AC_equi circuit E1 should be conected to the ground)

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• ###### AC_equi.PNG
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Last edited: Jul 19, 2013
2. ### LvW Active Member

Jun 13, 2013
674
100

at first, you have to determine the dc operating point.
This operating point determines the h parameters you are using.
More than that, you are not allowed to combine dc and ac values in one expression (as in the second equation).
Didn`t you realize that R1 is missing in your set of equations?

Jun 22, 2013
96
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The h parameters are given.h=200 for both transistors.

I did but i don't know how to introduce it in the equations.I could say that Ib2=Ib1+IR1 and then replace Ib2 ,then the equations would be
$

Vo=(hfeib2+(Ib1+IR1)R3

Vi=IRB+hie1*ib1

Vi=(V)/(R2)+hie*ib1

Av=vo/vi

AV=((hfeib2+Ib1+IR1)R3))/((V)/(R2))+hie.ib1))

$

I still don't know how to remove the voltage V,because the current that goes through R2 isn't equal to the base current of T1.

Thanks

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
A couple of observations ....

In your original schematic Q3 emitter is at 0V potential / "ground". Why have you not shown E1 in the equivalent circuit as also being grounded?

LvW already made the point that you can't combine DC and small signal parameters in the same equation - why are you persisting with this approach?

Start with working out the DC operating points and then do the small signal analysis.

5. ### LvW Active Member

Jun 13, 2013
674
100
AD633, if all the transistor parameters are given (h parameters) you have to care about small signal effects only. Thus, R2 plays no role at all.
As a most simple approach use the transconductance of each transistor because this allows a very simple calculation.
*Transconductance gm=hfe/hie.
*Common emitter stage: gain g1=-gm,1*Rc
with Rc=R1||r,in2 and r,in2=hie,2*(1+gm,2*R3)
*Comon collector stage: gain g2=gm,2*R3/(1+gm,2*R3)
This gain is often approximated to "1".

Jun 22, 2013
96
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If you notice i put a note on the first post:
I am persisting this aproach because i know that there is a current going through R2,which is different from ib1.Therefore altough i am doing a small singnal analysis, if i don't express that current in terms (V/R2) i don't it value..

Already done that before starting the small signal analysis.I got

Ic2=5mA and VCEQ2=12 V

Ic1=8mA VCEQ1=12,64 mA

Thanks

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
You might need to check the DC values. I get 5.2mA for Q3 collector current. Probably close enough to your 5mA. The rest I disagree with if HFE=200 and the values are as shown on the schematic.

Jun 22, 2013
96
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The value for HFE in DC is equal to 150.The hfe is only equal to 200 for small signal analysis.The values that i mentioned were for other values of the resistors,so you still may have find values wich are a litle different even if you use hfe=150.

Thanks

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Attached is my analysis for the "exact" voltage gain equation and its approximation for your amplifier.

Whilst one has to be careful, a usual "first pass" analysis might be to assume the second emitter follower stage has virtual unity gain and that it does not load the first common emitter stage to any significant degree.

The actual gain from the first stage would then be given to a reasonable approximation by ...

$\text{A_v=-\frac{R_{\small{C}}}{r_e}}$

with ...

$\text{r_e=\frac{26}{I_{\small{E}}} where I_{\small{E}} is the emitter current in mA}$

So if IE was 5mA and RC was 3kΩ, then the overall voltage gain would be approximately -577.

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10. ### LvW Active Member

Jun 13, 2013
674
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Why using this approximation for re when h parameters are known?
In this case: re=1/gm=hie/hfe (see my post #5).

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,497
1,267
From what I know almost no one use hybrid H parameters any more.

The Q3 transistor voltage gain is approximately equal to

Av1 ≈ (R1||Rin4)/re3

Where

Rin4 - Q4 stage input resistance = (hfe + 1)*(re4 + R3)

re4 ≈ 26mV/Ie4

re3 ≈ 26mV/Ie3

And voltage gain of a emitter follower

Av2 ≈ R3/(re4 + R3)

And overall vaoltage gain is equal to

Avs = Av1*Av2

As you can see all we need is hfe value. And then we can do small signal analysis by inspection

Jun 22, 2013
96
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What about the input and output impedances.

The input impedance can be determined by

$Rin=RB//hie1 =(3kOhm)//(1.2kOhm)=0.86kOhm$

For determing the output impedance i have to short the input source and add a virtual source in the output ritght?

$Rout=(hfeib2)/(ib2)=200 Ohm$ is this correct?Or the output impedance is simply equal to RE?

Thanks

13. ### LvW Active Member

Jun 13, 2013
674
100
Do you speak about r,in the fist stage?
In this case you have r,in=R2||hie,1

The output impedance is r,out2=R3||r,2

with r,2=1/gm,2 + r,out1/hfe,2
with
r,out1=R1 and
gm,2=1/re,2=hfe,2/hie,2.

Jun 22, 2013
96
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No i meant the input/output impedance of the entire amplifier circuit.

Thanks

15. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,610
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h parameters aren't the best parameters to use for analyzing a cascade connection of two transistors.

Converting the common emitter h parameters of the first transistor and the common base parameters of the second transistor to y parameters, Shekel's method can be used to obtain everything needed:

• ###### TwoTrans.png
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16. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The OP's problem apparently didn't specify values for hre and hoe. For a typical transistor, say a 2N3904, hoe will have almost no effect when Rc is 3k, but hre does have an effect. With an hre value of .00015, Av increases from -577 to about -628.

Most of these problems ignore hre because the various gain and impedance expressions become much more complicated if it's included; they become extremely complicated if hoe is included along with hre.

Last edited: Jul 21, 2013
17. ### LvW Active Member

Jun 13, 2013
674
100
Yes - that´s what I have given in post#13.

18. ### t_n_k AAC Fanatic!

Mar 6, 2009
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My approach on this particular amplifier would be to adopt the aforementioned rough approximation to estimate performance parameters such as gain and input impedance. Who in their right mind (albeit armed with all the h parameters) is going to use such a crude bias arrangement and naively expect to come up with a practical implementation that verifies the predicted values.

19. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,610
466
Because you don't have a full set of h parameters you don't need to short the input. The output impedance of a network only depends on what is connected to the input when there is reverse coupling. Since you have tacitly assumed that hre=0, the output impedance doesn't change whether you short the input or not.

For this network, if hre were .00015 rather than zero, the output impedance would change from 20.7525Ω to 21.944Ω with the input not shorted and then shorted.

20. ### LvW Active Member

Jun 13, 2013
674
100
Yes - I agree. It is an "awful" circuit, which will not be used in reality because there is no stabilization means in the first stage. However, I suppose it is something like an exercise only with the aim to become familiar with circuit calculations. You can find similar circuits (without current stabilization) in several basic textbooks.