May 20, 2022
X ZIKHALI 21905225
Introduction
The intention of this practical simulation is to reaffirm some of the more essential differential amplifier
features and applications. The attributes of the Differential Amplifier should be familiarized by the
readers. Under specific scenarios, the Differential Amplifier's behaviour will also be considered.
2. Differential Amplifier Circuit Analysis
Figure 1
2.1. (a) At Ve
π£=
π
4
× π£2
π
3 + π
4
Using KCL at V
π£ − π£1 π − ππ
+
=0
π
1
π
2
π
2
π
2
( × π) − ( × π1) + π = ππ
π
1
π
1
π
2
π
2
( + 1) × π −
× π1 = ππ
π
1
π
1
π
2
π
4
π
2
( + 1) (
) × π2 − ( × π1) = ππ
π
1
π
3 + π
4
π
1
∴ ππ = (
π
2
π
3
π
2
+ 1) [1 ÷ ( + 1)] × π2 − ( × π1)
π
1
π
4
π
1
(b)Letting V1=0
ππ = (
π
2
π
3
π
2
+ 1) [1 ÷ ( + 1)] × π2 − ( × π1)
π
1
π
4
π
1
ππ = (
π
2
π
3
π
2
+ 1) [1 ÷ ( + 1)] × π2 − ( × 0)
π
1
π
4
π
1
ππ = (
π
2
π
3
+ 1) [1 ÷ ( + 1)] × π2
π
1
π
4
(c)Letting V2=0
ππ = (
π
2
π
3
π
2
+ 1) [1 ÷ ( + 1)] × π2 − ( × π1)
π
1
π
4
π
1
ππ = (
π
2
π
3
π
2
+ 1) [1 ÷ ( + 1)] × 0 − ( × π1)
π
1
π
4
π
1
π
2
ππ = − ( × π1)
π
1
(d) If
π
1
π
2
=
π
3
π
4
then :
ππ = (
π
2
π
3
π
2
+ 1) [1 ÷ ( + 1)] × π2 − ( × π1)
π
1
π
4
π
1
ππ = (
π
2
π
1
π
2
+ 1) [1 ÷ ( + 1)] × π2 − ( × π1)
π
1
π
2
π
1
ππ =
π
2
π
2
× π2 −
π1
π
1
π
1
∴ ππ =
π
2
(π2 − π1)
π
1
(e) It is noticeable that Vout value is made up by the combination of the inputs of v1 and v2.
When V1 was 0, the output depended on V2 multiplied by the value of the resistors. When V2
was = 0, the output depended on the V1 with the combination of resistors of R1 and R2.
Figure 2
2.2(a)KCL at node A
π1 π1 + ππ1
+
=0
π
1
π
2
π1π
2 + (π1 + ππ1 )π
1
=0
π
1π
2
π1(π
1 + π
1) − ππ1 π
1 = 0
ππ1 π
1 = π1(π
2 + π
1)
π01 =
π1(π
2 + π
1)
π
1
Now apply KCL at node B
π2 − ππ1 π2 − ππ
+
=0
π
3
π
4
(π2 − ππ1 )π
4 − (π2 − ππ)π
3
π
3π
4
π2(π
4 + π
2) − ππ1 π
4 − πππ
1 = 0
π2(π
4 + π
3) −
ππ =
π1(π
2 − π
1)
× π
4 = πππ
3
π
1
π2(π
3 + π
4) π1(π
2 − π
1)
−
π
4
π
3
π
1π
3
ππ = (
π
4
π
4
π
2
(1 + ) π1
+ 1) π2 −
π
3
π
3
π
1
(b)Let V1 =0
ππ = (
π
4
π
4
π
2
(1 + ) π1
+ 1) π2 −
π
3
π
3
π
1
ππ = (
π
4
π
4
π
2
(1 + ) 0
+ 1) π2 −
π
3
π
3
π
1
ππ = (
π
4
+ 1) π2
π
3
© Let V2=0
ππ = (
π
4
π
4
π
2
(1 + ) π1
+ 1) π2 −
π
3
π
3
π
1
ππ = (
π
4
π
4
π
2
(1 + ) π1
+ 1) 0 −
π
3
π
3
π
1
ππ = −
π
4
π
2
(1 + ) π1
π
3
π
1
(d)Expression of Vout
π
2
π
3
if π
1 = π
4
π
4
π
4
π
2
(1 + ) π1
ππ = ( + 1) π2 −
π
3
π
3
π
1
π
2
π
3
Substitute π
1 = π
4
ππ = (
ππ =
π
4
π
4
π
3
(1 + ) π1
+ 1) π2 −
π
3
π
3
π
4
π
3 + π
4
× (π2 − π1)
π
3
ππ = (1 +
π
4
) × (π2 − π1)
π
3
∴ ππ = (1 +
π
2
) × (π2 − π2)
π
1
Question 3
Figure 3
3.1 Differential Amplifier Circuit Applications with one-Amp(A)
−π
2
)ππ = ( π
1 ) × π1
ππ = (
−40
) × π1
10
ππ = −4π1
Given: Vcc=12V
Given: Vcc=12V
V1= 2V
π2
(a)π = (π
3+π
4) × π
4
ππ2 = (1 +
π
2
)×π
π
1
π
2
π2π
4
(b) ππ = (1 + π
1) × π
3+π
4
ππ = (1 +
40π πβππ
π2 × 10π πβππ
)×(
)
10π πβππ
10π πβππ + 15π πβππ
ππ2 = 5 × 0.6π2
ππ2 = 3π2
Total πππ’π‘ = 3π2 − 4π1
πππ’π‘ = 3π2 − 4(2)
πππ’π‘ = 3π2 − 8
Range of V2
Vout =12
πππ’π‘ = 3π2 − 8
12 = 3π2 − 8
π2 = 6.67 π (Upper limit)
Vout= -12
πππ’π‘ = 3π2 − 8
−12 = 3π2 − 8
π2 = −1.33π (Lower limit)
(c) Circuit in Multism
Figure 4
Figure 5
3.2. Differential Amplifier circuit with two Op-Amps
Figure 6
π
π
(a)ππ₯ = [1 + π
ππ] π1
ππ₯ = [1 +
10
] π1
10
ππ₯ = 2π1
π01 = [1 +
π
π
] π2
π
ππ
ππ1 = [1 +
20
] π2
10
ππ1 = 3π2
Vout using gain formula
ππ2 = [
π
π
]π
π
1 π₯
ππ2 = [1 +
20
] × 2π1
10
ππ2 = −4π1
Total Output=ππ1 + ππ2 = 3π2 − 4π1
Vout at V1=2
ππ = 3π2 − 8
(b)Range of linear operation
•
Upper limit:ππ = 12
12 = 3π2 − 8
3π2 = 12 + 8
3π2 = 20
π2 = 6.67 π
•
Lower limit: ππ = −12
ππ = 3π2 − 8
−12 = 3π2 − 8
π2 = −1.33π
(c)Op-Amp Circuit in multism
Figure 7
Figure 8
(3.3)
Question 4
(a)πππ’π‘ = 2ππ − 4π
πππ’π‘ = 2ππ + (−4π)……….1
π
π
π
π
πππ’π‘ = (π
1 + 1) ππ + (− π
1 ππ)………2
π
′ (
π
π
+ 1) = 2
π
π
π
π
′ ( π
+ 1) = 2
4π
1
π
′ ( π
1 + 1) = 2
2
π
′ = 5
π
2
π
2+π
3
2
=5
and
−
π
π
π
1
= −4
and π
π = 4π
1
π
3
5
1 + π
2 = 2
π
3
π
2
3
=2
2
π
3 = 3 × π
2
Figure 9
(b)First op-amp
π
π
(π
1 + 1) = 2
π
π
π
1
=1
π
π = π
1
Figure 10
Second op-amps
π
π
(π
1 + 1) = 4
π
π
π
1
=3
π
π = 3π
1
Figure 11
Question 5
Gain of Instrumentation amp = 2
Gain of inverting amplifiers = -1
At 25° R15 is equal to 12πβ¦
π
15
π
15
) π1 + (
πππ’π‘ = 2 [− (
+ 1) π2]
π
7
π
7
πππ’π‘ = 2 [− (
0 = 2 [− (
12000
12000
) π1 + (
+ 1) π2]
1200
1200
12000
12000
) π1 + (
+ 1) π2]
1200
1200
0 = −(20)π1 + 22π2
(20)V1= (22)V2
∴ π1 = 1.1π2
At 85% R7 is unknown
π
15
π
15
) π1 + (
πππ’π‘ = 2 [− (
+ 1) π2]
π
7
π
7
−10 = 2 [− (
12000
12000
) π1 + (
+ 1) π2]
π
7
π
7
12000
12000
) π1 + (
−5 = [− (
+ 1) π2]
π
7
π
7
12000
) (π2 − π1)] + π2
−5 = [(
π
7
−5 = [(
12000
) (0.5 − (1.1 × 0.5)] + 0.5
π
7
−12000 × 0.5
)
π
7 = (
−5.5
π
7 = 109.0909 πβππ
Therefore, the thermistor is equal:
π=
π
π
πΌπ (π
π1 )
2
1
1
−
π1 π2
π=
1200
πΌπ (109.091)
1
1
−
298.15 358.15
π = 4267.55
Figure 13
Figure 14
Conclusion
In the instrumentation system, the temperature is independent of the voltage. The range of the
one-Amp and that of the two-Amp is the same even though the voltage is different.
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