A center tap full wave rectifier uses two diodes with an equivalent forward resistance 50. If the input a.C voltage to the primary side of transformer is 220v and primary to secondary turn ratio is 10:1, the load resistance of 950 . Determine: (i) peak,avg.And rms value of current (ii) efficiency (iii) ripple factor of the rectifier.
Answers
Explanation:
A center tap full wave rectifier uses two diodes with an equivalent forward resistance 50. If the input a.C voltage to the primary side of transformer is 220v and primary
Answer:
(i) The RMS current can be found using the formula:
I_rms = I_peak / √2 = 0.022 A / √2 = 0.0156 A
(ii) The efficiency is: η = P_out / P_in = 0.049 W / 0.481 W = 0.102 or 10.2%
(iii) The ripple factor is: γ = V_rms / V_avg = 6.35V / 7.0V = 0.907 or 90.7%
Explanation:
In a center tap full wave rectifier, the load current flows through both diodes in turn during alternate half-cycles of the input AC voltage.
(i) To find the peak current, we can use the formula:
I_peak = V_peak / (R_load + 2*R_forward)
Where V_peak is the peak voltage of the transformer secondary, R_load is the load resistance, and R_forward is the equivalent forward resistance of the diodes.
The peak voltage of the transformer secondary can be found using the turns ratio:
V_sec_peak = V_primary_peak / N
Where V_primary_peak is the peak voltage of the transformer primary (which is equal to the peak voltage of the AC input), and N is the turns ratio (10:1 in this case).
V_primary_peak = 220V (given)
N = 10
Therefore, V_sec_peak = 220V / 10 = 22V
Substituting the values, we get:
I_peak = 22V / (950Ω + 2*50Ω) = 0.022 A
The average current is given by:
I_avg = I_peak / π = 0.022 A / π = 0.007 A
The RMS current can be found using the formula:
I_rms = I_peak / √2 = 0.022 A / √2 = 0.0156 A
(ii) The efficiency of the rectifier is given by:
η = P_out / P_in
where P_out is the power delivered to the load, and P_in is the power supplied to the primary side of the transformer.
The output power can be found using the formula:
P_out = I_avg^2 * R_load = (0.007 A)^2 * 950Ω = 0.049 W
The input power can be found by multiplying the RMS voltage and RMS current on the primary side of the transformer:
P_in = V_primary_rms * I_rms = 220V / √2 * 0.0156 A = 0.481 W
Therefore, the efficiency is:
η = P_out / P_in = 0.049 W / 0.481 W = 0.102 or 10.2%
(iii) The ripple factor of the rectifier is given by:
γ = V_rms / V_avg
where V_rms is the RMS value of the ripple voltage, and V_avg is the average output voltage.
The RMS ripple voltage can be found by subtracting the DC component (which is equal to V_avg) from the peak voltage:
V_ripple_peak = V_sec_peak / 2 = 22V / 2 = 11V
V_rms = V_ripple_peak / √3 = 11V / √3 = 6.35V
The average output voltage can be found using the formula:
V_avg = V_sec_peak / π = 22V / π = 7.0V
Therefore, the ripple factor is:
γ = V_rms / V_avg = 6.35V / 7.0V = 0.907 or 90.7%
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