A full-wave rectifier circuit is driven by a sinusoidal input voltage with Vrms=15V and frequency 50Hz. If the load resistance is 100Ω, with a filter capacitance of 1.5mF? (Assume the diodes to be ideal, with Von=0V.)
what is the maximum instantaneous reverse voltage (approximately) across each diode?
Assuming a center-tapped transformer is used for the secondary winding,
Vp = Vrms * sqrt(2) = (15/2) * 1.414 = 10.61 Volts = Peak inverse voltage of each diode.
Note: Each diode sees 1/2 of the 15-volt rms supply.
Correction: The output voltage should be added to Vp. The max. possible voltage across the capacitor is 10.61 volts. Therefore, Vpmax = 2*(-10.61) = -21.2 Volts.
To find the maximum instantaneous reverse voltage across each diode in a full-wave rectifier circuit, we need to consider how the circuit works.
In a full-wave rectifier circuit, the input sinusoidal voltage is converted into a rectified output voltage. The diodes in the rectifier circuit play a crucial role in this conversion. When the input voltage is positive, one diode conducts and allows current to flow through the load resistance. When the input voltage is negative, the other diode conducts and allows current to flow through the load resistance in the opposite direction.
To calculate the maximum instantaneous reverse voltage across each diode, we need to determine when the diode starts to conduct. In this case, since the diodes are ideal with a Von (threshold voltage) of 0V, they start conducting when the reverse voltage across them reaches 0V.
The reverse voltage across each diode can be calculated as follows:
Reverse Voltage = |Peak Input Voltage - Forward Voltage Drop|
To find the peak input voltage, we can use the formula:
Peak Input Voltage = √2 × Vrms
Given that Vrms = 15V, we can calculate the peak input voltage as follows:
Peak Input Voltage = √2 × 15V ≈ 21.21V
Since the diodes are ideal with a Von of 0V, the forward voltage drop is also 0V.
Therefore, the maximum instantaneous reverse voltage across each diode is approximately:
Reverse Voltage = |21.21V - 0V| = 21.21V
Hence, the maximum instantaneous reverse voltage across each diode is approximately 21.21V.