Answer to Question #148894 in Electric Circuits for Ali Haider

Question

Searches related to Consider the full-wave bridge rectifier circuit. The input signal is 120 V (rms) at 60 Hz. The load resistance is RL = 250 . The peak output voltage is to be 9 V and the ripple voltage is to be no more than 5 percent. Determine the required turns ratio and the load current.

Answer 1

Determine the peak value $v_{s(max)}$

$v_{s(max)} = v_{0(max)} + 2V_γ$

$V_γ$ is the diode cut-in voltage

$v_{0(max)}$ is the desired peak value of the output voltage

$V_γ = 0.7 \;V \\ v_{0(max)} = 9 \;V \\ v_{s(max)} = 9 + 2 \times 0.7 = 10.4 \;V$

Determine the root mean square of the output sinusoid signal:

$v_{s,rms} = \frac{v_{s(max)}}{\sqrt{2}} \\ = \frac{10.4}{\sqrt{2}} = 7.35 \;V$

Determine the required turn ratio:

$k = \frac{N_1}{N_2} = \frac{v_{I,rms}}{v_{s,rms}}$

$N_1$ is the number of winding on the primary side of the transformer

$N_2$ is the number of winding on the secondary side of the transformer

$v_{I,rms}$ is the root mean square of the input voltage

$v_{s,rms}$ is the root mean square of the sinusoid output signal

$v_{I,rms} = 120 \;V(rms) \\ k = \frac{120}{7.35} = 16.3$

The required turns ratio is 16.3