# Determine gain for values of x

Discussion in 'Homework Help' started by CNC682, Feb 3, 2016.

1. ### CNC682 Thread Starter New Member

Jan 23, 2015
25
1
Q. Determine the gain, Vo/Vin, of the amp for when :
x=0.3
x=0.4
x=0

My solution:
R1=10k ohms R2=100k ohms R3= 1k ohms Rp=10k ohms potentiometer

Vin=(-R2/R1)*V0

Vo=(R3+(xRp*R2/xRp+R2))/(xRp*R2/xRp+R2)

Vo/Vin= (-R2/R1)*(R3+(xRp*R2/xRp+R2))/(xRp*R2/xRp+R2)

When x= 0.3, Vo/Vin=-11.333

When x= 0.4, Vo/Vin=-11.25

When x= 0, Vo/Vin= -∞ (negative infinity)

I'm pretty happy with that, particularly for when x=0 gain tends to infinity which is what the equation proves, anyone back that up?

ham3388 likes this.
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
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No, there is something wrong. For example x = 0.4 the Rp form a voltage divider.

Rpa = 0.6*10kΩ = 6kΩ and Rpb = 0.4*10kΩ = 4kΩ

3. ### dannyf Well-Known Member

Sep 13, 2015
2,196
417
Probably not right,

You should calculate the voltage on the wiper, Vp = f(x, Vo).

Vo should be such that, through Vp, the voltage on the inverting pin is the same as the voltage on the non-inverting pin = 0v.

You solve for the gain from that equation.

4. ### CNC682 Thread Starter New Member

Jan 23, 2015
25
1
Ok if we split Rp into xRp and (1-x)Rp and use KCL on a node (We'll call it V1) between them and the R2 resistor and then solve for Vo

So I have (V1-V0)/xRp +(V1-0)/((1-x)Rp+R3) +(0-V1)/R2.

Do I now solve for Vo although there is the presence of V1?

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
I think that your nodal equation is wrong. Because your circuit looks like this

6. ### CNC682 Thread Starter New Member

Jan 23, 2015
25
1
Ok so instead we have (V1-V0)/(1-x)Rp +(V1-0)/(xRp+R3) +(0-V1)/R2=0.

So how do I eliminate V1 in the equation? Unless I replace it with -I*R2/(R2+R1) which gives me "I" to deal with.

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Well, you have two unknowns and only one equation. So you need a second equation.
Or you can assume some Vout value, and next solve for Vin and gain = Vout/Vin.
For example for x = 0.5, Rpa = Rpb = 5kΩ and Vout = 10V so we have
V1 = Vout * R2||(Rpb+R3)/(Rpa +R2||(Rpb+R3) ) ≈ 10V*5.66/(5+5.66) ≈ 5.3V
IR2 = IR1 = 5.3V/100kΩ =0.053mA = 53μA and Vin = -IR1*R1 = -0.53V and the gain 10/-0.53 ≈ - 18.8V/V

Last edited: Feb 4, 2016
8. ### CNC682 Thread Starter New Member

Jan 23, 2015
25
1
Er, any other idea?

9. ### MrAl Distinguished Member

Jun 17, 2014
2,540
511
Hi,

Did you try setting the voltage at the inverting terminal equal to zero? That means the current through R2 must equal the current through R1, and that means the output voltage Vo must be a certain value for a certain value of Vin.

10. ### dannyf Well-Known Member

Sep 13, 2015
2,196
417
V1 is easy to calculate, V1 = - 10 Vin.

You can go from there.

11. ### CNC682 Thread Starter New Member

Jan 23, 2015
25
1
How did you calculate that?

12. ### MrAl Distinguished Member

Jun 17, 2014
2,540
511
Current same in R1 and R2!

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
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As I said early you have two unknowns and only one equation, so you need to write one more nodal equation for Inverting node (Vn) and solve it for V1.
And then you will know why V1 = - 10 Vin.

14. ### dannyf Well-Known Member

Sep 13, 2015
2,196
417
Once V1 is known, you can do an analysis on the current flowing in / out of that point.

Flowing in:
from Vo: (Vout - V1) / [(1-x)Rp];
from gnd: (0 - V1) / (R3 + xRp).

Flowing out:
from V1 to virtual ground: (V1 - 0) / R2.

The sum of those current should be zero (with signs), or balance out (without signs).

That equation can then be expressed in Vout/Vin.

15. ### dannyf Well-Known Member

Sep 13, 2015
2,196
417
For x=0.5, I have a feeling that Vout/Vin ~= -19x. You can check your answers against that.

16. ### dannyf Well-Known Member

Sep 13, 2015
2,196
417
If you go through the thread, I have given you two ways to calculate the gain:

1. the correct way is calculate current through the node V1.
2. the incorrectly correct way is laid out in post #3. and I will expand it more here.

The circuit can be viewed as a feedback to a divider hang on the output. The divider is formed by (1-x)Rp as the upper arm, and xRp + R3 on the lower arm.

since R2 >> (1-x)Rp // (xRp + R3), R3's loading on the divider can be ignored (side note: you will find that the correct way to think of the lower arm is R2 // (xRp + R3)).

So V1 = (xRp + R3) / (Rp + R3) * Vout.

From the input side, we know that V1 = -R2 / R1 * Vin.

So

-R2/R1*Vin = (xRp + R3) / (Rp + R3) * Vout.

or Vout / Vin = - (R2/R1) * (Rp + R3) / (xRp + R3).

The correct value will be fairly close to that.

One easy way to check is to set x=0, making this a typical inverting amp, and its gain = -R2/R1 -> consistent with that approach above.

It also points out why R3 is needed.

17. ### RBR1317 Active Member

Nov 13, 2010
264
54
One way to avoid doing the full nodal analysis is to employ a Thevenin equivalent as suggested by the diagram below.

18. ### CNC682 Thread Starter New Member

Jan 23, 2015
25
1
I can see that with R3, x=0 is not equal to -R2/R1. Adding a resistor in series with a pot prevents gain tending to A.

If, Vout / Vin = - (R2/R1) * (Rp + R3) / (xRp + R3)
when x=0 Vo/Vin=-110

For the other two values of x
when x=0.3 Vo/Vin=-27.5
when x =0.4 Vo/Vin=-22

19. ### CNC682 Thread Starter New Member

Jan 23, 2015
25
1
How would you represent the load resistor?

20. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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You do realize that these Vo/Vin values are only approximately correct, don't you? As dannyf said "The correct value will be fairly close to that."