Respuesta :
Given Information:
Frequency = f = 60 Hz
Transformer output voltage = Vrms = 6.3 V
Diode voltage drop = Vd = 1 V
Ripper voltage = Vr = 0.25 V
Load resistance = R = 0.5 Ω
Required Information:
a) dc output voltage = V₀ = ?
b) Capacitane = C = ?
Answer:
a) dc output voltage = V₀ = 2.52 V
b) Capacitane = C = 0.336 F
Explanation:
a) The average or dc output voltage of a half-wave rectifier is given by
[tex]V_0 = V_p/\pi[/tex]
Where Vp is given by
[tex]V_p = (V_{rms} \times \sqrt{2}) - V_d \\\\V_p = (6.3 \times \sqrt{2}) - 1 \\\\V_p = 8.91 - 1 \\\\V_p = 7.91 \: V \\\\[/tex]
So, the dc output voltage is
[tex]V_0 = 7.91/\pi \\\\V_0 = 2.52 \: V[/tex]
b) The minimum value of C required to maintain the ripple voltage to less than 0.25 V is given by
[tex]$ C = \frac{I}{Vr \cdot f} $[/tex]
Where I is current, Vr is the ripple voltage and f is the frequency
[tex]$ I = \frac{V_0}{R} $[/tex]
[tex]$ I = \frac{2.52}{0.5} $[/tex]
[tex]I = 5.04 \: A[/tex]
[tex]$ C = \frac{5.04}{0.25 \cdot 60} $[/tex]
[tex]C = 0.336 \: F[/tex]
Therefore, 0.336 F is the minimum value of capacitance required to maintain the ripple voltage to less than 0.25 V
