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Solving for Resistor Values in an Analog Electronics Circuit to Achieve Desired Gain, Assignments of Electrical and Electronics Engineering

The solution to homework #2 in ee 321 analog electronics, fall 2008. The problem involves determining resistor values for a circuit to achieve an input resistance of 100 kω and a gain that varies from -1 to -10 as the potentiometer is adjusted. The document derives the relationship between input and output voltages as a function of the potentiometer position and then calculates the resistor values for the minimum and maximum gain settings.

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Uploaded on 08/08/2009

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EE 321 Analog Electronics, Fallw 2008
Homework #2 Solution
2.35
Choose resistor values for the following circuit such that it has an input resistance of 100 kΩ,
and such that as the potentiometer, R4, is turned from one end to the other the gain varies
from 1 to 10.
The requirement on the input resistance means that R1= 100 kΩ. I will define the variable
x[0; 1] such that the resistance b etween voand R2is xR4, and the resistance between
R2and R3is (1 x)R4. Also, the voltage at the center pin of the potentiometer (the pin
connected to R2), I will call vx. Note these are all labeled on the figure. Then we write down
the node equations.
vI
R1
+vx
R2
= 0
vovx
xR4
+vx
R3+ (1 x)R4
+vx
R2
= 0
From these we want to find the relationship betweeen vIand vOas a function of x. Thus we
need to eliminate vx. From the first node equation we get
vx=vI
R2
R1
From the second node equation we get
vO
xR4
=vx1
xR4
+1
R3+ (1 x)R4
+1
R2
Eliminating vxby inserting we get
vO
xR4
=vI
R2
R11
xR4
+1
R3+ (1 x)R4
+1
R2
Rearranging in the form of a transfer function we get
1
pf3
pf4
pf5

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EE 321 Analog Electronics, Fallw 2008

Homework #2 Solution

Choose resistor values for the following circuit such that it has an input resistance of 100 kΩ, and such that as the potentiometer, R 4 , is turned from one end to the other the gain varies from −1 to −10.

The requirement on the input resistance means that R 1 = 100 kΩ. I will define the variable x ∈ [0; 1] such that the resistance between vo and R 2 is xR 4 , and the resistance between R 2 and R 3 is (1 − x)R 4. Also, the voltage at the center pin of the potentiometer (the pin connected to R 2 ), I will call vx. Note these are all labeled on the figure. Then we write down the node equations.

vI R 1

vx R 2

vo − vx xR 4

−vx R 3 + (1 − x)R 4

−vx R 2

From these we want to find the relationship betweeen vI and vO as a function of x. Thus we need to eliminate vx. From the first node equation we get

vx = −vI

R 2

R 1

From the second node equation we get

vO xR 4

= vx

xR 4

R 3 + (1 − x)R 4

R 2

Eliminating vx by inserting we get

vO xR 4

= −vI

R 2

R 1

xR 4

R 3 + (1 − x)R 4

R 2

Rearranging in the form of a transfer function we get

vO vI

R 2

R 1

xR 4 R 3 + (1 − x)R 4

xR 4 R 2

R 2

R 1

xR 4 R 2 R 3 R 1 + (1 − x)R 4 R 1

xR 4 R 1

In the two cases, x = 0 and x = 1 we have a gain or either −1 or −10. First for x = 0

vO vI

(x = 0) = −

R 2

R 1

Next for x = 1

vO vI

(x = 1) = −

R 2

R 1

R 4 R 2

R 3 R 1

R 4

R 1

It looks like the x = 1 case has the larger gain, so let’s assume the gain is −1 for x = 0 which gives us R 2 = R 1. Now we only need to determined R 3 such that the gain is −10 for x = 1.

R 4

R 3

R 4

R 1

From which we get

R 4

R 3

and

R 4

R 3

R 3 =

R 4

= 1.1 kΩ

When the potentiometer is set in the middle, x = 0.5, then the voltage gain is

vO vI

(x = 0.5) = −

R 2

R 1

0. 5 R 4

R 3 + 0. 5 R 4

0. 5 R 4

R 2

I am going to assume that it is acceptable to invert the sine-wave signal. In that case we are just to construct a inverting summing coupling with the two inputs being the sine wave and and the 2 V reference. In that case we want the input-output relation to be

vO = −vI −

vref

The transfer function for this circuit is

Vx = −V 1

RB

RA

Vo = − RF

Vx R 1

V 2

R 2

= − RF

V 1 RB

RAR 1

V 2

R 2

If we choose RF = 100 kΩ, RA = RB = R 1 = R 2 = 10 kΩ, then we get the transfer equation

Vo = −10(V 2 − V 1 ) = 10(V 1 − V 2 )

For this circuit we have the node equation

vO Rf

5 s 3 10

5 s 2 20

5 s 1 40

5 s 0 80

or

vO = − Rf

( (^) s 0 16

s 1 8

s 2 4

s 3 2

Rf 16

(s 0 + 2s 1 + 4s 2 + 8s 3 )

Rf 16

20 s 0 + 2^1 s 1 + 2^2 s 2 + 2^3 s 3

When s 0 = s 1 = s 2 = s 3 = 0 the output voltage is vO = 0. When s 0 = s 1 = s 2 = s 3 = 1, the output voltage is

vO = −Rf

In order to make vO = −12 V under these circumstances, we must choose

Rf = 12 ×

kΩ = 12.8 kΩ