# Determine gain for values of x

#### CNC682

Joined Jan 23, 2015
27
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

#### dannyf

Joined Sep 13, 2015
2,197
particularly for when x=0 gain tends to infinity which is what the equation proves, anyone back that up?
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.

#### CNC682

Joined Jan 23, 2015
27
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?

#### CNC682

Joined Jan 23, 2015
27
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.

#### Jony130

Joined Feb 17, 2009
5,457
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

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#### CNC682

Joined Jan 23, 2015
27
Er, any other idea? #### MrAl

Joined Jun 17, 2014
10,606
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.

#### dannyf

Joined Sep 13, 2015
2,197
V1 is easy to calculate, V1 = - 10 Vin.

You can go from there.

#### CNC682

Joined Jan 23, 2015
27
V1 is easy to calculate, V1 = - 10 Vin.

You can go from there.
How did you calculate that?

#### MrAl

Joined Jun 17, 2014
10,606

#### Jony130

Joined Feb 17, 2009
5,457
Er, any other idea? 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.

#### dannyf

Joined Sep 13, 2015
2,197
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.

#### dannyf

Joined Sep 13, 2015
2,197
For x=0.5, I have a feeling that Vout/Vin ~= -19x. You can check your answers against that.

#### dannyf

Joined Sep 13, 2015
2,197
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.

#### CNC682

Joined Jan 23, 2015
27
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

#### CNC682

Joined Jan 23, 2015
27
One way to avoid doing the full nodal analysis is to employ a Thevenin equivalent as suggested by the diagram below.
View attachment 100131
How would you represent the load resistor?

#### The Electrician

Joined Oct 9, 2007
2,936
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
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."