Half-Wave Rectification

Half-wave rectification converts AC to pulsed DC by blocking one half of the AC cycle using a diode.

Circuit

An AC source $V_{\text{in}} = V_m \sin(\omega t)$ in series with a diode and load resistor $R_L$
During the positive half-cycle: diode is forward-biased → conducts → $V_O = V_m - V_D \approx V_m$
During the negative half-cycle: diode is reverse-biased → blocks → $V_O = 0$

    +ve half:  ┌─── V_in ──▶|─── R_L ───┐
               │                           │
               └───────────────────────────┘
    -ve half:  No current

Where Does 0.318 Come From?

TL;DR: It's $1/\pi$.

The average (DC) value of a periodic signal is:

$$V_{DC} = \frac{1}{T} \int_0^T V_O(t) , dt$$

For a half-wave rectified sine wave (only positive half conducts):

$$V_O(t) = \begin{cases} V_m \sin(\omega t), & 0 \le t \le T/2 \ 0, & T/2 \le t \le T \end{cases}$$

Integrate over one period:

$$V_{DC} = \frac{1}{T} \int_0^{T/2} V_m \sin(\omega t) , dt$$

Substitute $\omega = 2\pi/T$:

$$V_{DC} = \frac{V_m}{T} \int_0^{T/2} \sin\left(\frac{2\pi t}{T}\right) dt$$

$$V_{DC} = \frac{V_m}{T} \left[-\frac{T}{2\pi} \cos\left(\frac{2\pi t}{T}\right)\right]_0^{T/2}$$

$$V_{DC} = \frac{V_m}{T} \cdot \frac{T}{2\pi} \left[-\cos\pi + \cos 0\right]$$

$$V_{DC} = \frac{V_m}{2\pi} \left[-(-1) + 1\right] = \frac{V_m}{2\pi} \cdot 2$$

$$V_{DC} = \frac{V_m}{\pi} \approx 0.318 V_m$$

Accounting for the diode voltage drop $V_D$ (≈0.7 V for silicon):

$$V_{DC} = \frac{V_m - V_D}{\pi} = 0.318(V_m - V_D)$$

Key Takeaways

0.318 is $1/\pi$, not arbitrary — it comes from integrating half a sine wave
The diode drop ($V_D$) reduces the peak, which reduces the average proportionally
Full-wave rectification gives $V_{DC} = 2V_m/\pi \approx 0.636 V_m$ (twice the average)

Half-Wave Rectification

Circuit

Where Does 0.318 Come From?

Key Takeaways

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