An electronic circuit receives noise from switching element. If voltage on the switching device swings from 4 volt to 6 volt within 100 microsecond, during which current makes a transition from 10 microampere to 25 microampere in 10 nanoseconds, what might be the noise coupling mechanism? Identify it using suitable calculations.
Initial voltage, V₀ = 4V
Final voltage, V₁ = 6V
Change in voltage, dv = 6 - 4 = 2V
Initial current, I₀ = 10μA = 10×10⁻⁶ A
Final current, I₁ = 25μA = 25×10⁻⁶ A
Change in current, di = (25 - 10)×10⁻⁶ A = 15×10⁻⁶ATime interval, dt = 10ns = 10×10⁻⁹s = 10⁻⁸s
Final voltage, V₁ = 6V
Change in voltage, dv = 6 - 4 = 2V
Initial current, I₀ = 10μA = 10×10⁻⁶ A
Final current, I₁ = 25μA = 25×10⁻⁶ A
Change in current, di = (25 - 10)×10⁻⁶ A = 15×10⁻⁶ATime interval, dt = 10ns = 10×10⁻⁹s = 10⁻⁸s
dv
dt =
2V
10⁻⁸s = 2×10⁸ V/s
dt =
2V
10⁻⁸s = 2×10⁸ V/s
di
dt =
15×10⁻⁶A
10⁻⁸s = 15×10² A/s
dt =
15×10⁻⁶A
10⁻⁸s = 15×10² A/s
Pseudo impedance is given by,
Pseudo impedance =
dv/dt
di/dt
Pseudo impedance =
dv/dt
di/dt
=
2×10⁸ V/s
15×10² A/s = 0.0133 Ω
2×10⁸ V/s
15×10² A/s = 0.0133 Ω
Pseudo impedance = 0.01333 Ω .
As, 0.01333<<377
This reveals that the noise is coupled via inductive coupling mechanism .
As, 0.01333<<377
This reveals that the noise is coupled via inductive coupling mechanism .
0.01333 aauxa hola ni ans ta
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