Question

a semiconductor diode,the forward and reverse characteristics of which can considered as ideal, is used in a half wave rectifier circuit supplying a resistive load of 1100 ohms, if the rms value of the sinusoidal supply voltage is 240v. Determine the peak diode current

Answer 1

Answer:

approximately 307.9 milliamps.

Explanation:

In a half-wave rectifier circuit, only one half of the input AC waveform is used to produce the output DC waveform. The diode allows current to flow in one direction, and blocks it in the other direction.

The peak voltage of the sinusoidal supply voltage can be found by multiplying the RMS value by the square root of 2, which gives:

Vp = Vrms * √2

Vp = 240 * √2

Vp = 339.4 volts

When the diode is forward biased, it conducts current and the voltage drop across the diode is typically around 0.7 volts. Therefore, the peak voltage across the load resistor is:

Vload = Vp - Vdiode

Vload = 339.4 - 0.7

Vload = 338.7 volts

The peak current through the diode can be found using Ohm's Law, which states that:

I = V / R

where I is the current, V is the voltage, and R is the resistance. In this case, the resistance is the load resistor, which is 1100 ohms. Therefore, the peak current through the diode is:

I = Vload / R

I = 338.7 / 1100

I = 0.3079 amps or 307.9 milliamps (rounded to 3 significant figures)

Therefore, the peak diode current in the half-wave rectifier circuit is approximately 307.9 milliamps.