Fullwave Rectifier Fundamentals

A fullwave rectifier is an electronic circuit designed to convert alternating current (AC) into pulsating direct current (DC) by rectifying both the positive and negative half-cycles of the input AC signal. This process is more efficient than half-wave rectification because it utilizes the entire input waveform, resulting in a smoother DC output. Typically, fullwave rectifiers can be implemented using either a center-tapped transformer with two diodes or a bridge rectifier configuration with four diodes.

To accurately determine the current flow in such a circuit, it's crucial to understand the roles of the various components, including the load resistance and the inherent resistance of the diodes.

Peak Voltage Determination

The input voltage applied to each diode in the fullwave rectifier is given by the expression: \(V = 240 \sin 50 t \text{ V}\). This is a standard sinusoidal waveform representation, \(V = V_m \sin(\omega t)\), where \(V_m\) represents the peak voltage of the AC signal and \(\omega\) is the angular frequency.

• From the provided voltage equation, we can directly identify the peak voltage (\(V_m\)). In this case, \(V_m = 240 \text{ V}\).

Total Resistance Calculation

When a diode in the fullwave rectifier conducts, it acts as a very low resistance path for the current. The total resistance that limits the peak current in the circuit includes the external load resistance and the internal forward resistance of the conducting diode. The given resistance values are:

• Load resistance (\(R_L\)) = \(2 \text{ k}\Omega\). Converting this to ohms: \(R_L = 2 \times 1000 \Omega = 2000 \Omega\).
• Diode forward resistance (\(R_f\)), which corresponds to the slope characteristic, is given as \(400 \Omega\).

The total effective resistance (\(R_{total}\)) in the path of the current flow during either the positive or negative half-cycle (when one diode conducts) is the sum of the load resistance and the diode's forward resistance:

\[R_{total} = R_L + R_f\]

\[R_{total} = 2000 \Omega + 400 \Omega\]

\[R_{total} = 2400 \Omega\]

Peak Current (\(I_m\)) Through the Load

Using Ohm's Law, the peak current (\(I_m\)) that flows through the load during the peak of each half-cycle can be calculated by dividing the peak voltage (\(V_m\)) by the total resistance (\(R_{total}\)):

\[I_m = \frac{V_m}{R_{total}}\]

Substituting the values we determined:

\[I_m = \frac{240 \text{ V}}{2400 \Omega}\]

\[I_m = 0.1 \text{ A}\]

To express this peak current in milliamperes (mA), we multiply by 1000:

\[I_m = 0.1 \times 1000 \text{ mA}\]

\[I_m = 100 \text{ mA}\]

Average DC Current (\(I_{dc}\)) Calculation

The question asks for "the peak value of current, \(I_{dc}\)". In the context of rectifiers, \(I_{dc}\) most commonly refers to the average DC current produced by the rectifier. For a fullwave rectifier, the relationship between the average DC current (\(I_{dc}\)) and the peak current (\(I_m\)) is defined by the following formula:

\[I_{dc} = \frac{2 I_m}{\pi}\]

Now, we substitute the calculated peak current (\(I_m = 100 \text{ mA}\)) into this formula:

\[I_{dc} = \frac{2 \times 100 \text{ mA}}{\pi}\]

\[I_{dc} = \frac{200}{\pi} \text{ mA}\]

Using the approximate value of \(\pi \approx 3.14159\), we calculate the average DC current:

\[I_{dc} \approx \frac{200}{3.14159} \text{ mA}\]

\[I_{dc} \approx 63.6619 \text{ mA}\]

Rounding this value to two decimal places, we get approximately \(63.66 \text{ mA}\). This calculated value is very close to one of the provided options, \(63.84 \text{ mA}\). The minor discrepancy might be due to slight differences in the value of \(\pi\) used or internal rounding in the problem's original values.

Comparing with Options

Let's compare our calculated average DC current with the given options to find the best match:

Based on our detailed calculations, the average DC current (\(I_{dc}\)) is approximately \(63.66 \text{ mA}\), which aligns most closely with the option stating \(63.84 \text{ mA}\). Therefore, option 3 is the correct answer.