Question

for the shown circuit, assume v_in,rms = 3.96v with frequency of 50 hz, v_d,on = 0.8v, c_1 = 100f and the average current through r_l = 5ma. what is the value of the ripple voltage? select one: a. 1v b. 2v c. 0.25v d. 0.5v

Explanation:

Step1: Calculate peak - input voltage

For an AC input with $V_{in,rms}=3.96V$, the peak - input voltage $V_{in,peak}=\sqrt{2}V_{in,rms}=\sqrt{2}\times3.96V\approx5.6V$.

Step2: Consider diode - drop

After passing through the diodes in the bridge - rectifier, the voltage available for charging the capacitor (after accounting for the diode - on voltage) is $V = V_{in,peak}-V_{D,on}$. Since $V_{D,on} = 0.8V$, $V=5.6 - 0.8=4.8V$.

Step3: Use ripple - voltage formula

The formula for the ripple voltage $\Delta V$ in a capacitor - filtered rectifier is $\Delta V=\frac{I_{L}}{fC}$, where $I_{L}$ is the load current, $f$ is the frequency of the input AC, and $C$ is the capacitance of the filter capacitor. Given $I_{L}=5mA = 5\times10^{- 3}A$, $f = 50Hz$, and $C = 100F$. Then $\Delta V=\frac{5\times10^{-3}}{50\times100}=1\times10^{-6}V$. However, there is a mistake above. The capacitance value should be $C = 100\mu F=100\times10^{-6}F$.
Recalculate: $\Delta V=\frac{I_{L}}{fC}=\frac{5\times10^{-3}}{50\times100\times10^{-6}} = 1V$.

Answer:

A. 1 V