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Deriving gain from non-ideal Op-Amp

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magician13134

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Hi everyone, I've got a homework problem that's turning out to be quite difficult. We've been given an inverting opamp circuit and told to use the op-amp equivalent circuit to derive an expression for the gain, V_out/V_S.
**broken link removed**
(In this image, taken from Wikipedia, they use V_S instead of V_CC, we use V_S as the voltage connected to the negative terminal)

This is the circuit we're given
**broken link removed**
Again, V_in = V_S and R_in = R_S, but they can be called whatever you want.

Then we're asked to obtain an expression for the closed loop gain, G = V_out/V_s ONLY in terms of R_s, R_in (internal), R_out (internal), R_Load (not shown in the circuit, it simply goes from V_Out to ground), R_F and A (the amplification of this specific op-amp). I tried to use KCL at node n (negative op-amp terminal) and at V_Out, which is what he showed us in class, and I got two equations but I can't seem to get V_Out/V_S in only those terms, I have other terms like V_n (voltage at the negative terminal), V_p (voltage at the positive terminal, but that equals 0, right?).

Here are the equations I got:
[LATEX](V_n-V_s)/R_s + (V_n-V_p)/R_i+(V_n-V_o_u_t)/R_f=0[/LATEX]
[LATEX](V_o_u_t-A(V_p-V_n))/R_o_u_t + (V_o_u_t-V_n)/R_F+V_o_u_t/R_L=0[/LATEX]

Assuming V_p = 0:
[LATEX](V_n-V_s)/R_s + V_n/R_i+(V_n-V_o_u_t)/R_f=0[/LATEX]
[LATEX](V_o_u_t-A*V_n)/R_o_u_t + (V_o_u_t-V_n)/R_F+V_o_u_t/R_L=0[/LATEX]
But that still leaves me with a V_n that I can't seem to get rid of and I can't for the life of me do the algebra to solve for V_out/V_s. Remember, this op-amp is NOT ideal, so you cannot use infinity or 0 for any of the values.

Can anyone please help me?
Thank you!
 
Hello there,


I am posting a new diagram so we can all be sure of the circuit you are talking about.
Note that in the "Equivalent Circuit" shown in the diagram, the source "-A*vn" is possible because of the internal source A*vi and because vp=0 and vi=-vn.

You should be able to solve for both vn and Vout and then elminate vn, but in the mean time since vn is linearly related to Vin you should be able to force vn to equal 1v and then analyze Vout and Vin and then divide to find the gain G=Vout/Vin. Try that and see what happens if you like or else we can do the full analysis.
 

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Use Vo = GVin - Rout*Iout => Vin = (Vo + Rout*Iout)/G to eliminate Vin. Now guess what Vin is?
 
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Hi there Brownout,

Perhaps you can clarify a little. What is Iout?
Also, in the equivalent circuit the internal gain is A while the circuit gain is G. I guess you meant the internal gain by 'G'?
I drew up the equivalent circuit to make it easier to see a solution.

Isnt it simple enough to force vn to equal 1 volt and go from there? Setting vn=1 eliminates vn right away.
 
If we're not considering Rin, or Ri in your simplified drawing,then Iout is = (Vout - Vin)/Rf + Vout/Rload. If we consider Rin, then it's a little more complicated, but still managable.

We may also calculate Vn from Vout and the resistor network, considering Rin. I got Vn = (Vo/RF + Vin/RS)/(RF//RS//Rin).

Then plug the above result into Iout=Vo/RL + (Vo-Vn)/RF.

There are other ways to solve, I'm sure.
 
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magician13134,

Hi everyone, I've got a homework problem that's turning out to be quite difficult.

Yes, it is difficult. Not the equation setup, but all the tedious algebra necessary to solve those equations. Mr Al was kind enough to show the schematic. It seems to work well to use node analysis. As you can see in the attachment, when the opamp gain is very high, the closed loop gain equation simplifies to -Rf/Rs . The load Rl is not a factor provided the opamp can supply the power without dropping the high gain. See the attachment for the solution. You will notice that is is prudent to use a equation solver instead of doing the algebra manually, and increase your chances of making a mistake. Ratch
 

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Hello again,

I figured that setting vn to some constant would make the analysis easier because we would eliminate that one variable already, but i have to admit it still isnt super simple although it does mean not having to find a solution to a set of simultaneous equations. Here's the solution:

(for a simpler matrix solution see far below)

Assume vn=1 volt constant.

Make RoRL a resistor as Ro and RL in parallel.
Make RfRL a resistor as Rf and RL in parallel.

Now Vout is the sum of the contributions from vo and vn which is:
VoutA=RoRL/(Rf+RoRL)=(Ro*RL)/(Ro*RL+Rf*RL+Rf*Ro)
VoutB=-A*RfRL/(Ro+RfRL)=-(Rf*A*RL)/(Ro*RL+Rf*RL+Rf*Ro)

Note that the denominators are the same, so we get:
Vout=VoutA+VoutB=((Ro*RL)-(Rf*A*RL))/(Ro*RL+Rf*RL+Rf*Ro)

and this can be represented as N/D:
Vout=N/D

where N is the numerator of Vout and D is the denominator of Vout.

Now Iin is the current that flows through Ri and Rf, and with vn=1 we get:
Iin=1/Ri+(1-Vout)/Rf

and this expanded a little we get:
Iin=1/Ri+1/Rf-Vout/Rf


Now since Rs has this same current Iin the voltage across it is:
vRs=Rs*Iin

and so the input voltage Vin must be:
Vin=Rs*Iin+1

and now substitute for Iin with the equation above for Iin:
Vin=Rs*Iin+1=Rs/Ri+Rs/Rf-Rs*Vout/Rf+1

Now forming the gain Vout/Vin we get:
Vout/Vin=Vout/(Rs/Ri+Rs/Rf-Rs*Vout/Rf+1)

Now finding a common denominator for the terms in the denominator we find Ri*Rf, and applying that we get:
Vout/Vin=Vout/(Rs*Rf/(Ri*Rf)+Rs*Ri/(Ri*Rf)-Rs*Ri*Vout/(Ri*Rf)+Ri*Rf/(Ri*Rf))

and now multiplying top and bottom by Ri*Rf we get:
Vout/Vin=Vout*(Ri*Rf)/(Rs*Rf+Rs*Ri-Rs*Ri*Vout+Ri*Rf)

and now making Vout=N/D on the right side we get:
(Vout/Vin)=(N/D)*(Ri*Rf)/(Rs*Rf+Rs*Ri-Rs*Ri*N/D+Ri*Rf)

and now multiplying top and bottom by D we get:
(Vout/Vin)=N*(Ri*Rf)/(D*Rs*Rf+D*Rs*Ri-Rs*Ri*N+D*Ri*Rf)

and now substitute N and D with their equations we get:
Vout/Vin=(Rf*Ri*(Ro-Rf*A)*RL)/(Ri*Rs*(Ro*RL+Rf*RL+Rf*Ro)+Rf*Rs*(Ro*RL+Rf*RL+Rf*Ro)+Rf*Ri*(Ro*RL+Rf*RL+Rf*Ro)-Ri*Rs*(Ro-Rf*A)*RL)

and after expanding all terms and dividing top and bottom by Rf we get:
CircuitGain=Vout/Vin=
(Ri*RL*(Ro-Rf*A))/(Ri*Rs*A*RL+Ro*Rs*RL+Ri*Rs*RL+Rf*Rs*RL+Ri*Ro*RL+Rf*Ri*RL+Ri*Ro*Rs+Rf*Ro*Rs+Rf*Ri*Ro)

This turns out a little easier because we can do this all on paper without having to deal with exceptionally long equations and we dont have to find a solution to a multivariate simultaneous equation.

Of course using a matrix method leads to an even simpler set of three equations although we need a matrix solver then:
-E1/Rs+v2*(1/Rs+1/Ri+1/Rf)-v3/Rf=0
-v2/Rf+v3*(1/Rf+1/RL+1/Ro)-v4/Ro=0
v2*A+0+v4=0
where v3 is the output voltage Vout and then when solved the circuit gain is v3/E1.
I guess we could simplify right off by solving the last equation for v2 and that leaves two equations in two unknowns which could also be solved by hand.
 
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it's a very simple question.
as u already mention in ur diagram,Vp=0
then Vout = A(0-Vn);
Vout = -AVn
Vn = -Vout/A .....................eq(1)
now another equation we have
(Vs-Vn)/Rs = (Vn-Vout)/Rf
now put eq(1) in eq(2) and after rearrange we will get
Vout/Vs = -A/[1+(A+1)/K] where k=Rf/Rs
 
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it's a very simple question.
as u already mention in ur diagram,Vp=0
then Vout = A(0-Vn);
Vout = -AVn
Vn = -Vout/A .....................eq(1)
now another equation we have
(Vs-Vn)/Rs = (Vn-Vout)/Rf
now put eq(1) in eq(2) and after rearrange we will get
Vout/Vs = -A/[1+(A+1)/K] where k=Rf/Rs
Where did you account for Ri, Ro, and RL?
 
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