A half-wave rectifier is a device that takes an oscillating voltage as input and

Question

A half-wave rectifier is a device that takes an oscillating voltage as input and outputs only the positive portion of that input. In one particularly simple type of half-wave rectifier (see the figure (Figure 1) ), the alternating voltage source is changed by using a diode to essentially eliminate current flow in one direction. Doing this forces the output voltage to take only positive values (or a value of zero), even though the input voltage will alternate between positive and negative.

The key to the working of a half-wave rectifier is the diode. The standard circuit symbol for a diode is a triangle with a line at the point. The point of the triangle corresponds to the n region, so that current flowing in the direction in which the triangle points corresponds to positive current (i.e., current arising from positive potential difference Vd across the diode). A diode with positive Vd is said to be forward biased. A diode with negative Vd is said to be reverse biased. The current-voltage relationship for a diode is given by

I=Is(eeVd/kBT1),

where Is is called the saturation current, Vd is the potential difference across the diode (positive if the potential is highest in the p region), kB is Boltzmann's constant, T is the temperature of the diode, and e in the exponent is the magnitude of the charge on the electron. Use 1.602×1019C for the magnitude of the charge on the electron. The e in the base of the exponential is the familiar constant e=2.718281828….

Assume that the circuit is at room temperature (300K) and that the saturation current of the diode is Is=1.0×108A. Take Boltzmann's constant to be 1.38 x 1023J/K.

Explanation / Answer

Vd is not mentioned in the question and the figure is also missing. Let us assume it to be 0.5 V (Generally set at this value for diodes)

then I = Is*exp(eVd/KBT - 1)
The formula written in the question is wrong.
So, now to obtain the value of current just substitute the values
I = (10^-8)*exp((0.5*1.6*10^-19)/(300*1.38*10^-23)-1)
I = 0.90754 Ampere

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