Solution to Op-Amp Circuit with Adjustable Resistor RΔ

[Sunny16]

Homework Statement

The op-amp in the circuit shown is ideal (please see attachment). The adjustable resistor RΔ has a maximum value of 160 kΩ and α is restricted to the range of 0.22 ≤ α ≤ 0.7.

(a) Calculate the range of v_o if v_g = 40 mV

(b) If α is not restricted, at what value of α will the op amp saturate?

Homework Equations

for a difference amplifier: v_o = -(R_f/R_i)v_in

The Attempt at a Solution

Firstly, I don't understand how the adjustable resistor functions in this circuit.
My interpretation is that the resistance ranges from a minimum of 0.22(160 kΩ / 0.7)
to a maximum of 160 kΩ (given).

PART A:

for maximum gain, R_f is maximized
so R_f = 20k + 160k = 180k
then v_o = -(180/4)(40mV) = -1.8 V, which is not the right answer

for minimum gain, R_f = 20k + 50.29k = 70.29k
so v_o = -(70.29/4)(40mV) = -0.703 V, which is again not the right answer

PART B:

for saturation, we would require a gain: A = 12V/40mV = 300
so R_f should be 300*4 kΩ = 1200 kΩ
then the variable resistor should provide a resistance of 1180 kΩ
how is this even possible, if the maximum resistance is way smaller than this value?

I must be misunderstanding how the resistor works, so I would appreciate any help
Thanks in advance.

 

[berkeman]

Homework Statement

The op-amp in the circuit shown is ideal (please see attachment). The adjustable resistor RΔ has a maximum value of 160 kΩ and α is restricted to the range of 0.22 ≤ α ≤ 0.7.

(a) Calculate the range of v_o if v_g = 40 mV

(b) If α is not restricted, at what value of α will the op amp saturate?

Homework Equations

for a difference amplifier: v_o = -(R_f/R_i)v_in

The Attempt at a Solution

Firstly, I don't understand how the adjustable resistor functions in this circuit.
My interpretation is that the resistance ranges from a minimum of 0.22(160 kΩ / 0.7)
to a maximum of 160 kΩ (given).

PART A:

for maximum gain, R_f is maximized
so R_f = 20k + 160k = 180k
then v_o = -(180/4)(40mV) = -1.8 V, which is not the right answer

for minimum gain, R_f = 20k + 50.29k = 70.29k
so v_o = -(70.29/4)(40mV) = -0.703 V, which is again not the right answer

PART B:

for saturation, we would require a gain: A = 12V/40mV = 300
so R_f should be 300*4 kΩ = 1200 kΩ
then the variable resistor should provide a resistance of 1180 kΩ
how is this even possible, if the maximum resistance is way smaller than this value?

I must be misunderstanding how the resistor works, so I would appreciate any help
Thanks in advance.

I believe they are saying that the pot can be adjusted from 0.22 to 0.7 of its full scale setting. And I don't think you can just use the standard inverting amplifier equation, not with that grounded pot sitting in there. I'd write out the new KCL equations, with a new node at the wiper of the pot. What equation do you get now for the transfer function?

And when they remove the restrictions on the pot, that just means that it can go all the way to the ends, with the wiper at 0 Ohms or up at 160 kOhms.

 

[Sunny16]

Thanks for the explanation. I think I understand it better now, even though I still can't get the correct answer.
The attachment is a quick sketch of a section of the circuit with my variables labelled.

Here's my KCL equations:

1)

V_g/4000 = -V_x/20000
so V_x = -5V_g

2)

V_x/(aR) = (V_o-V_x)/(1-a)R
(1-a)V_x = a(V_o-V_x)
V_o = V_x/a = -5V_g/a

PART A:

I used a = 0.22, then a = 0.7 to get the range of V_o

PART B:

I set V_o = -12 = -5V_g/a
and got a = 0.0167