Single-phase diode bridge, average and fundamental

A single-phase full bridge diode rectifier uses four diodes; two diagonal diodes conduct in each half cycle so that the load always gets a positive voltage $v_o = |v_s|$.

              P (+)
        +-------+-------+
        |               |
       D1              D3
        |               |
 a o----+               +----o b
        |               |
       D4              D2
        |               |
        +-------+-------+
              N (-)
 D1, D3: cathodes to P;  D4, D2: anodes to N

Operation

• Positive half cycle ($a$ positive): D1 and D2 conduct; $v_o = v_s$.
• Negative half cycle ($b$ positive): D3 and D4 conduct; $v_o = -v_s$.
• Each diode blocks a peak reverse voltage of $V_m$.

 vs   ___         
     /   \         /
 ---/-----\-------/---
           \_____/
 vo   ___     ___
     /   \   /   \
 ---/-----\-/-----\---
    0    180     360   (deg)
 vo = |Vm sin wt|, period 180 deg (100 Hz)

Average value

$V_m = \sqrt2 \times 220 = 311.13$ V.

$$\begin{aligned} V_{dc} &= \frac{1}{\pi}\int_0^{\pi} V_m\sin\omega t\, d(\omega t) = \frac{2V_m}{\pi} \\ &= \frac{2 \times 311.13}{\pi} = 198.07\ \text{V} \end{aligned}$$

Fundamental (lowest-frequency ac) component of output

The output has period $\pi$, so its Fourier series has only even harmonics of the supply frequency:

$$v_o(t) = \frac{2V_m}{\pi} - \frac{4V_m}{3\pi}\cos2\omega t - \frac{4V_m}{15\pi}\cos4\omega t - \frac{4V_m}{35\pi}\cos6\omega t - \cdots$$

General term: $\frac{4V_m}{\pi(n^2-1)}$ for $n = 2, 4, 6, \ldots$

The fundamental ripple component of the output is the $n = 2$ term at $2 \times 50 = 100$ Hz:

$$\begin{aligned} V_{2,peak} &= \frac{4V_m}{3\pi} = \frac{4 \times 311.13}{3\pi} = 132.05\ \text{V} \\ V_{2,rms} &= \frac{132.05}{\sqrt2} = 93.37\ \text{V} \end{aligned}$$

(The next component, at 200 Hz, has peak $\frac{4V_m}{15\pi} = 26.41$ V.)

Answer: $V_{dc} = 198.07$ V; fundamental output ripple = 132.05 V peak (93.37 V rms) at 100 Hz.