The Junction Rectifier — Question 5

Question 5

Full-wave rectification doubles the ripple frequency

An ideal bridge and resistive load give $v_o=V_m|\sin\omega t|$. The source frequency is f, and there is no smoothing capacitor.

1. Derive average and rms output voltage.

2. Find the ripple frequency and the rms ripple factor $r=\sqrt{V_{\mathrm{rms}}^2-V_{\mathrm{dc}}^2}/V_{\mathrm{dc}}$.

Question 5 - Solution

Full-wave rectification doubles the ripple frequency

(a) Method and calculation

$$\boxed{V_{\mathrm{dc}}=\frac1\pi\int_0^\pi V_m\sin\theta\,d\theta=2V_m/\pi}.$$$$\boxed{V_{\mathrm{rms}}=V_m/\sqrt2}.$$

(b) Interpretation and further calculation

$$v_o(t+\pi/\omega)=v_o(t)\Rightarrow\boxed{f_r=2f}.$$$$\boxed{r=\sqrt{\pi^2/8-1}=.4834}.$$

Physical check

Both half cycles feed the load with the same polarity. The rms voltage is unchanged by taking the absolute value because its square is unchanged.