When you balance a redox equation using the half-reaction method, what do you do after you write the oxidation and reduction hal
2 answers:
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Answer:
a) Differential mode gain = 48
b) Common mode gain = 0.4
c) CMRR = 120
Explanation:
The output of a difference amplifier is related to the input by the equation:
When V₁ = 6.4 V, V₂ = 5.1 V and V₀ = 64.7 V, the equation becomes
6.4 A₁ + 5.1 A₂ = 64.7.....................(1)
When V₁ = 5.6 V, V₂ = 4.9 V and V₀ = 35.7 V, the equation becomes
5.6 A₁ + 4.9 A₂ = 35.7.....................(2)
Multiply equation (1) by 5.6 and (2) by 6.4
35.84 A₁ + 28.56A₂ = 362.32.....................(3)
35.84 A₁ + 31.36 A₂ = 228.48....................................(4)
Subtract equation (3) from (4)
2.8 A₂ = -133.84
A₂ = -133.84/2.8
A₂ = -47.8
Put the value of A₂ into equation (1)
6.4 A₁ + 5.1 (-47.8) = 64.7
6.4 A₁ = 64.7 + 243.78
A₁ = 308.48/6.4
A₁ = 48.2
a) Common mode gain = A₁ + A₂ = 48.2 + (-47.8)
Common mode gain = 0.4
b) Differential mode gain = (A₁ -A₂)/2
Differential mode gain = (48.2 - (-47.8))/2
Differential mode gain = 96/2
Differential mode gain = 48
c) Common Mode Rejection Ratio (CMRR)