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#11
11-16-2011, 12:22 AM
 jegues Senior Member Join Date: Sep 2010 Posts: 734

I've started working on part c),

$R_{i} = R_{s} + r_{e1} + \frac{R_{11}}{\beta + 1} \quad ; \text{where} \quad R_{11} = R_{1}//R_{2}$

$R_{o} = R_{22}//R_{L} \quad ; \text{where} \quad R_{22} = R_{1}+R_{2}$

Are these expressions correct?

How does one go about solving for A?

Thanks again!
#12
11-16-2011, 03:10 PM
 Jony130 Senior Member Join Date: Feb 2009 Location: Poland/Wroclaw Posts: 2,467

Quote:
 Originally Posted by jegues $R_{i} = R_{s} + r_{e1} + \frac{R_{11}}{\beta + 1} \quad ; \text{where} \quad R_{11} = R_{1}//R_{2}$ $R_{o} = R_{22}//R_{L} \quad ; \text{where} \quad R_{22} = R_{1}+R_{2}$ Are these expressions correct?
It looks ok to me.

Quote:
 Originally Posted by jegues How does one go about solving for A?
Simply use classic method.
Replace BJT with his Hybrid-pi model and find the open loop gain.

And we can use "brutal force" nodal analysis to find the voltage gain.
I can write equation for Va node.

$\frac{Vout - Va}{r_{\pi 2}} = \frac{ Va - Vin}{ro1} + gm1*(r_{\pi 1} *\frac{0-Vin}{R1||R2+r_{\pi 2}})$

And for Vout node

$\frac{Vout}{R1+R2} +\frac{ Vout}{ro2} +\frac{Vout-Va}{r_{\pi 2}} = gm2*(r_{\pi 2} * - \frac{Vout - Va}{r_{\pi 2}})$

And when I solve this for Vout/Vin this is what I get
Additional I assume that ro1 = ro2 = ∞

$\frac{Vout}{Vin}=\frac{gm1*(R1+R2)*r_{\pi 1}*(1 + gm2*r_{\pi 2})}{R1||R2 + r_{\pi 1}}$

I also use another method. And I get similar solution.
The method I use is described here

From all this "methods" my "simplified method" (show in post 9 in thread) is the fastest one.
Attached Images
 23.PNG (13.1 KB, 25 views)

Last edited by Jony130; 11-16-2011 at 05:11 PM.
#13
11-16-2011, 06:32 PM
 The Electrician Senior Member Join Date: Oct 2007 Posts: 1,722

Quote:
 Originally Posted by jegues I've started working on part c), $R_{i} = R_{s} + r_{e1} + \frac{R_{11}}{\beta + 1} \quad ; \text{where} \quad R_{11} = R_{1}//R_{2}$ $R_{o} = R_{22}//R_{L} \quad ; \text{where} \quad R_{22} = R_{1}+R_{2}$ Are these expressions correct? How does one go about solving for A? Thanks again!
Be aware that the output impedance you have given is for the case where the source Vs is not connected to Rs; that is, with the left end of Rs not connected to anything. Assuming Vs is an ideal voltage source with zero output impedance, if Vs is connected to Rs, then Ro will be different. Which do you suppose the problem is asking for?

Also, if RL is connected, then the expression for Ri will include RL.
#14
11-17-2011, 12:54 AM
 jegues Senior Member Join Date: Sep 2010 Posts: 734

Quote:
 Originally Posted by The Electrician Be aware that the output impedance you have given is for the case where the source Vs is not connected to Rs; that is, with the left end of Rs not connected to anything. Assuming Vs is an ideal voltage source with zero output impedance, if Vs is connected to Rs, then Ro will be different. Which do you suppose the problem is asking for? Also, if RL is connected, then the expression for Ri will include RL.
How am I supposed to decipher which the problem is asking for?

Based on the examples in the text, Vs nor Rs are considered when calculating $R_{o}$.

Last edited by jegues; 11-17-2011 at 01:01 AM.
#15
11-17-2011, 01:56 AM
 The Electrician Senior Member Join Date: Oct 2007 Posts: 1,722

Quote:
 Originally Posted by jegues How am I supposed to decipher which the problem is asking for?
Since you are the student, ultimately it will be up to you to determine. You could ask your instructor. I was asking if you had any way to know.

Quote:
 Originally Posted by jegues Based on the examples in the text, Vs nor Rs are considered when calculating $R_{o}$.
That may be sufficient. However, since Vs would be driving the circuit when the load is connected, the output impedance with a source connected would seem to be relevant.

Did the other examples in the text you're referring to lack feedback? Perhaps they had output impedances that didn't change when the source was connected.
#16
11-17-2011, 05:05 AM
 Jony130 Senior Member Join Date: Feb 2009 Location: Poland/Wroclaw Posts: 2,467

We here in Polad use this definition for Rout
Rout = Vout_open / Iout_short when Vs and Rs is replaced by a short circuit.

 Tags amplifier, feedback, seriesshunt

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