A full-wave bridge rectifier with a \(120-\mathrm{V} \mathrm{rms}\) sinusoidal input has a load resistor of \(1 \mathrm{k} \Omega\). a. If silicon diodes are employed, what is the dc voltage available at the load? b. Determine the required PIV rating of each diode. c. Find the maximum current through each diode during conduction. d. What is the required power rating of each diode?

The DC voltage calculation in a full-wave bridge rectifier is a fundamental concept when working with AC to DC conversion. It represents the average output voltage provided to the load and is critical to the functioning of the rectified circuit. For a bridge rectifier that uses silicon diodes, the DC voltage (\(V_{DC}\)) at the output can be computed by subtracting the voltage drop across two diodes from the peak input voltage (\(V_m\)). Silicon diodes typically have a voltage drop (\(V_d\)) of approximately 0.7V.

In our example, the input RMS voltage of 120V gets converted to a peak voltage (\(V_m\)) using the formula \(V_m = \sqrt2 \times V_{rms}\) which equates to 169.7V. Therefore, the DC voltage available at the load resistor comes out to \(V_{DC} = \sqrt2 \times 120V - 2 \times 0.7V = 168.3V\). This value represents the steady voltage that the load will experience from the rectifier circuit.

Peak Inverse Voltage (PIV) is the maximum voltage a diode can withstand when it is reverse biased in a circuit without conducting. In the realm of rectifiers, it's a key parameter for ensuring the reliability and longevity of diodes under reverse voltage conditions. During the non-conductive phase of the diode's operation, the PIV must be higher than the maximum reverse voltage that the diode will encounter.

For a full-wave bridge rectifier, the PIV rating of each diode should match or exceed the peak voltage of the input AC signal. In our exercise, since the peak voltage (\(V_m\)) is 169.7V, each diode should have a PIV rating of at least 169.7V to prevent breakdown during the reverse bias portion of the cycle.

Understanding diode current conduction is vital for calculating the current flow through the diode when it is forward biased. In a bridge rectifier circuit, each diode conducts in turns, allowing the full utilization of both halves of the AC input cycle. The maximum current each diode carries is ideally the same as the current flowing through the load.

The maximum diode current can be determined using Ohm's Law, stating that current \(I = V_{DC} / R\), where \(V_{DC}\) is the DC voltage across the load and \(R\) is the resistance of the load. In our example, with a load resistance of 1k\(\Omega\), the maximum current through each diode during conduction is calculated to be \(0.1683 A\) or 168.3 mA.

The power rating of a diode is a critical specification that determines the amount of power it can safely dissipate. This rating ensures the diode operates within its thermal limits, preventing damage due to overheating. The power dissipated by a diode (\(P\)) is given by the product of the current passing through it (\(I\)) and the voltage drop across it when forward biased (\(V_d\)).

When calculating the power rating for each diode in a bridge rectifier, the formula \(P = I \times V_d\) is used. As per the provided solution, for a current of 0.1683 A and a voltage drop of 0.7V, the power rating required for each diode would be 0.118W or 118mW. It is advisable to choose diodes with a power rating higher than this calculated value to account for possible surges and to enhance the reliability of the circuit.