1 Answer
In the context of a **half-wave rectifier**, **VDC** refers to the **average or DC (direct current) output voltage** of the rectified signal. A half-wave rectifier converts an AC (alternating current) input signal into a pulsating DC signal by allowing only one half (positive or negative) of the AC waveform to pass through, blocking the other half.
---
### **Understanding the Half-Wave Rectifier**
* **Input**: An AC voltage (typically sinusoidal).
* **Working**: A single diode is used to allow only the **positive half-cycles** of the input AC to pass, while the negative half-cycles are blocked.
* **Output**: A pulsating signal that is zero during the negative half-cycles and follows the AC waveform during the positive half-cycles.
---
### **Definition of VDC**
The **VDC** (DC output voltage) is the **average value of the rectified output voltage** over one complete cycle of the input AC.
Mathematically, it is calculated as:
$$
V_{\text{DC}} = \frac{1}{T} \int_{0}^{T} v_{\text{out}}(t) \, dt
$$
For a **pure sinusoidal AC input** voltage of the form:
$$
v(t) = V_m \sin(\omega t)
$$
where:
* $V_m$ = peak voltage of the AC signal
* $\omega$ = angular frequency of the input
* $T$ = time period of one full AC cycle
Since the diode in a half-wave rectifier conducts only for half the cycle (0 to $\pi$), the average DC value becomes:
$$
V_{\text{DC}} = \frac{1}{2\pi} \int_{0}^{\pi} V_m \sin(\theta) \, d\theta
$$
Solving the integral:
$$
V_{\text{DC}} = \frac{V_m}{2\pi} \left[ -\cos(\theta) \right]_0^\pi = \frac{V_m}{2\pi} \left( -\cos(\pi) + \cos(0) \right)
$$
$$
V_{\text{DC}} = \frac{V_m}{2\pi} (1 + 1) = \frac{V_m}{\pi}
$$
---
### ✅ Final Formula:
$$
\boxed{V_{\text{DC}} = \frac{V_m}{\pi}} \approx 0.318 \, V_m
$$
---
### **Key Points**
* This result is **ideal**, assuming no diode drop or load resistance effects.
* In practical circuits, if you consider the **diode’s forward voltage drop** (usually 0.7V for silicon), the peak output voltage and hence VDC will be slightly less.
* This output is **not pure DC**, but a **pulsating DC**. A **capacitor filter** is usually added to smooth it out.
---
Would you like a visual graph or waveform diagram to help illustrate this?
---
### **Understanding the Half-Wave Rectifier**
* **Input**: An AC voltage (typically sinusoidal).
* **Working**: A single diode is used to allow only the **positive half-cycles** of the input AC to pass, while the negative half-cycles are blocked.
* **Output**: A pulsating signal that is zero during the negative half-cycles and follows the AC waveform during the positive half-cycles.
---
### **Definition of VDC**
The **VDC** (DC output voltage) is the **average value of the rectified output voltage** over one complete cycle of the input AC.
Mathematically, it is calculated as:
$$
V_{\text{DC}} = \frac{1}{T} \int_{0}^{T} v_{\text{out}}(t) \, dt
$$
For a **pure sinusoidal AC input** voltage of the form:
$$
v(t) = V_m \sin(\omega t)
$$
where:
* $V_m$ = peak voltage of the AC signal
* $\omega$ = angular frequency of the input
* $T$ = time period of one full AC cycle
Since the diode in a half-wave rectifier conducts only for half the cycle (0 to $\pi$), the average DC value becomes:
$$
V_{\text{DC}} = \frac{1}{2\pi} \int_{0}^{\pi} V_m \sin(\theta) \, d\theta
$$
Solving the integral:
$$
V_{\text{DC}} = \frac{V_m}{2\pi} \left[ -\cos(\theta) \right]_0^\pi = \frac{V_m}{2\pi} \left( -\cos(\pi) + \cos(0) \right)
$$
$$
V_{\text{DC}} = \frac{V_m}{2\pi} (1 + 1) = \frac{V_m}{\pi}
$$
---
### ✅ Final Formula:
$$
\boxed{V_{\text{DC}} = \frac{V_m}{\pi}} \approx 0.318 \, V_m
$$
---
### **Key Points**
* This result is **ideal**, assuming no diode drop or load resistance effects.
* In practical circuits, if you consider the **diode’s forward voltage drop** (usually 0.7V for silicon), the peak output voltage and hence VDC will be slightly less.
* This output is **not pure DC**, but a **pulsating DC**. A **capacitor filter** is usually added to smooth it out.
---
Would you like a visual graph or waveform diagram to help illustrate this?