When a pn-junction diode is biased by the voltage v_D = v_AC = 0.2 V, the total

Question

When a pn-junction diode is biased by the voltage v_D = v_AC = 0.2 V, the total current flow through the diode is the consequence of the cumulative effect of: yes no not applicable the transition of electrons through the depletion region from the n-side to the p- side of the pn-junction, the transition of holes through the depiction region from the n-side to the p-side of the pn-junction. the transition of electrons through the depletion region from the p-side to the n- side of the pn-junction. the transition of holes through the depletion region from the p-side to the n-side of the pn-junction. The magnitude of the depletion region capacitance of a pn-junction diode: yes no not applicable depends on the concentration of donor atoms in the n-region, depends on the magnitude of the reverse bias voltage applied to the diode. is always greater than the diffusion capacitance of the diode. depends on the concentration of acceptor atoms in the p-region. The height of the potential barrier V_bo of the pn-junction in thermal equilibrium, at room temperature, yes no not applicable depends on the temperature of the pn-junction. depends on the atmospheric pressure at the surface of the piece of semiconductor material which contains the pn-junction, depends on the magnitude of the external electric field, depends on the concentration of acceptor atoms on the p-side of the junction. In a pn-junction diode: yes no not applicable the terminal called the anode is connected to the semiconductor region which is doped by the loins of a three-valent element. the terminal called the cathode is connected to the semiconductor region which is doped by the atoms of a five-valent element. the terminal called the cathode is connected to the semiconductor region which is doped by the atoms of a three-valent element. the terminal called the anode is connected to the semiconductor region which is doped by the atoms of a five-valent element.

Explanation / Answer

Explanation for 1.1:

When pn junction is forward biased, junction voltage and depletion region width decreases. In forward bias, junction voltage is less than the applied voltage. When the diode is forward biased junction voltage opposes the flow of minority carriers from one side of the junction to other side but the majority carriers move towards the junction. This happens because the junction voltage is less than the applied forward voltage.

On p-side holes are the majority carriers and electrons are the minority carriers whereas on n-side holes are the minority carriers and electrons are the majority carriers. So, holes from p side move through the depletion region from p-side to n-side and electrons move from n-side to p-side through the depletion region. The answers are yes,no,no,yes.

Expalnation for 1.2:

Transistion capacitance comes into picture when the diode is reverse biased. It is also called space charge capacitance (or) depletion capacitance. When the junction is reverse biased, the depletion region acts like dielectric material while the p and n regions on the either sides acts as the plates. Thus pn junction may be considered a parallel plate capacitor. A reverse bias causes majority carriers to move away from the junction , thereby uncovering more immobile charges. So the thickness(area) of the depletion region increases with the increase in reverse bias voltage. But, when the diode is forward biased the thickness(area) of the depletion region decreases and so is the diffusion capcitance. Hence the diffusion capacitance is less than the depletion region capacitance (Since the value of a capacitance is directly proportional to area i.e., C=A0/d. Here as the area of the depletion region increases the capcitance also increase) So the answers are yes, yes, no, yes.

Expalnation for 1.3:

The potential barrier developed across the junction due to ions opposes the flow of majority carriers from one side to other side and it allows minority carriers to move from one side of the junction to other side. The potential is given by KTln(NAND / ni2) volts. where NA = concerntration of acceptor impurities, ND = concerntration of donor impurities, ni = intrinsic carrier concerntration, T is the temperature and K is the Boltzmann's constant. So the answers are yes, no, not applicable, yes.

Explanation for 1.4:

When the atoms of a five valent element (donor impurities) are introduced into one side and atoms of a three valent element (acceptor impurities) are introduced into other side of a single crystal of silicon then its called a pn junction. The terminal where the acceptor impurities are doped is called the anode while the terminal where donor impurities are doped is called the cathode. So the answers are yes,yes,no,no.

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