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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
