Half-Wave Rectifier Behavior

An ideal half-wave rectifier allows current to flow in only one direction. It behaves as a closed switch when forward-biased and an open switch when reverse-biased.

The circuit consists of a sinusoidal voltage source $V_S(t)$, an ideal diode, and a purely resistive load $R$. Let the source voltage be represented as $V_S(t) = V_p \sin(\omega t)$.

Rectifier Voltage Analysis

We examine the voltages across the load ($V_L(t)$) and the diode ($V_D(t)$) during the positive and negative half-cycles of the source voltage.

• Positive Half-Cycle ($V_S(t) > 0$):
  • The diode is forward-biased and acts as a short circuit ($V_D(t) = 0$).
  • The entire source voltage appears across the load: $V_L(t) = V_S(t)$.
• Negative Half-Cycle ($V_S(t) < 0$):
  • The diode is reverse-biased and acts as an open circuit (infinite resistance).
  • No current flows through the load, so the load voltage is zero: $V_L(t) = 0$.
  • The voltage across the diode becomes equal to the source voltage: $V_D(t) = V_S(t)$.

Average Voltage Calculations

The average value of a function $f(t)$ over one period $T$ is given by $f_{avg} = \frac{1}{T} \int_0^T f(t) dt$. The period for a sinusoidal source is $T = 2\pi/\omega$. Let $V_p$ be the peak source voltage.

• Average Load Voltage ($V_{L, avg}$):

  During the positive half-cycle, $V_L(t) = V_p \sin(\omega t)$. During the negative half-cycle, $V_L(t) = 0$.

  $ V_{L, avg} = \frac{1}{T} \int_0^{T/2} V_p \sin(\omega t) dt + \frac{1}{T} \int_{T/2}^T 0 dt $

  $ V_{L, avg} = \frac{V_p}{T} \left[ -\frac{\cos(\omega t)}{\omega} \right]_0^{T/2} = \frac{V_p}{\omega T} [-\cos(\pi) - (-\cos(0))] $

  Since $\omega T = 2\pi$, $V_{L, avg} = \frac{V_p}{2\pi} [-(-1) - (-1)] = \frac{V_p}{2\pi} (1+1) = \frac{V_p}{\pi}$.

• Average Diode Voltage ($V_{D, avg}$):

  During the positive half-cycle, $V_D(t) = 0$. During the negative half-cycle, $V_D(t) = V_p \sin(\omega t)$.

  $ V_{D, avg} = \frac{1}{T} \int_0^{T/2} 0 dt + \frac{1}{T} \int_{T/2}^T V_p \sin(\omega t) dt $

  $ V_{D, avg} = \frac{V_p}{T} \left[ -\frac{\cos(\omega t)}{\omega} \right]_{T/2}^T = \frac{V_p}{\omega T} [-\cos(2\pi) - (-\cos(\pi))] $

  Since $\omega T = 2\pi$, $V_{D, avg} = \frac{V_p}{2\pi} [-1 - (-(-1))] = \frac{V_p}{2\pi} [-1 - 1] = -\frac{V_p}{\pi}$.

Average Value Comparison

We found $V_{L, avg} = \frac{V_p}{\pi}$ and $V_{D, avg} = -\frac{V_p}{\pi}$.

These values are equal in magnitude ($|\frac{V_p}{\pi}|$) and opposite in sign.

Evaluating the options:

• Option 1 is incorrect.
• Option 2 is correct as it matches our findings.
• Option 3 is incorrect.
• Option 4 is incorrect.

The statement that is always true is: The average values of $V_L(t)$ and $V_D(t)$ are equal in magnitude and opposite in sign.