Assignment #3
1-33. Design an inverting amplifier with voltage gain ≈ -5.6 and an input resistance ≈ 10 kΩ
using an OP27 OpAmp. The OP27 has the following typical performance parameters:
Input resistance -2 GΩ
Voltage gain -1.5 MV/V
Output resistance -- 70 Ω
Determine the variation from the ideal design goals due to the non-ideal properties of the
OP27 OpAmp.
Solution: MathCAD
#1-33
9
6
Rf := 56000 RS := 10000
Ri := 2⋅ 10
Ro := 70
Guess
v1_vi := 1
vo_vi := −1
Given
1
1
1
1 ⎞
1 ⎞
v1_vi ⋅ ⎛⎜
+
+
− vo_vi ⋅ ⎛⎜
RS
⎝ RS Rf Ri ⎠
⎝ Rf ⎠
0
v1_vi ⋅ ⎛⎜
A
⎝ Ro
−
⎞ − vo_vi ⋅ ⎛ 1 + 1 ⎞
⎜
Rf ⎠
⎝ Rf Ro ⎠
1
a := Find( v1_vi , vo_vi )
v1_vi := a
0
vo_vi := a
−6
v1_vi = −3.73802× 10
variation_percent :=
A := 1.5⋅ 10
−5.6 − ( vo_vi )
−5.6
⋅ 100
1
vo_vi = −5.60002
variation_percent = −0.00044
%
1-40. For the amplifier shown below, determine the values of the load resistor, RL, that will
lead to gain that deviates from the ideal value by -0.01%. Assume the OpAmp has the
following properties:
Av = 500,000
Ri = 1 MΩ
Ro = 75Ω
R
R
1
1k Ω
2
22kΩ
+15 V
R
v
−
v
3
10kΩ
i
+
RL
-15 V
Solution: MathCAD
o
#1-40
The following are circuit and OpAmp parameters:
3
6
A := 500⋅ 10
Ri := 10
Ro := 75
3
3
9
R3 := 10⋅ 10
R1 := 1000
R2 := 22⋅ 10
Confirm the gain of the amplifier using the solve block
Guess
v1i := 1.001
RL := 10
Av := 25
The two simultaneous equations are:
Given
v1i⋅ ⎛⎜
1
v1i⋅ ⎡⎢
−1
⎝ R1
⎣ R2
+
−
1
R2
A
Ro
+
1
⋅ ⎛⎜ −
⎝
⎞ − Av ⋅ ⎛ 1 ⎞
⎜ R2
⎝ ⎠
Ri + R3 ⎠
Ri
1
Ri⋅ R3
⎞ ⎤ + Av ⋅ ⎛ 1 + 1 + 1 ⎞
⎥
⎜ R2 RL Ro
⎝
⎠
Ri + R3 ⎠ ⎦
A ⋅ Ri
Ro⋅ ( Ri + R3)
Solve for the gain, Av = ov/vi , and the value of 1v/vi :
Find( Av , v1i) =
⎛ 23.021⎞
⎜
⎝ 1 ⎠
So the gain of the amplifier using the simplified OpAmp mod
is 23.021 which is in good agreement with the results
calculated from the ideal model of the OpAmp.
Now find value of RL such that gain -0.01% of ideal. That gain is:
Av_new := 0.999923
⋅
Av_new = 22.998
Av := Av_new
Let
Solve for RL for Av_new = 22.998 :
Guess
v1i := 1
RL := 75
Given
v1i⋅ ⎛⎜
1
⎞ − Av ⋅ ⎛ 1 ⎞
⎜ R2
Ri
⋅ R3
⎝ ⎠
Ri ⎞ ⎤
1
1
A ⎛
1
1 ⎞
v1i⋅ ⎡⎢ −
−
⋅⎜ −
+ Av ⋅ ⎛⎜
+
+
⎥
⎣ R2 Ro ⎝ Ri + R3 ⎠ ⎦
⎝ R2 RL Ro ⎠
1
⎝ R1
+
1
R2
Find( v1i , RL) =
+
1
Ri + R3 ⎠
A ⋅ Ri
Ro⋅ ( Ri + R3)
So RL = 65.8 Ω for a -0.01% change in gain.