Barrier potential of a p-n junction diode does not depend on:
A. Diode design
B. Doping density
C. Temperature
D. Forward bias
Answer
638.4k+ views
Hint: The barrier potential depends on the width of the depletion region at the junction of the diode. The width of the depletion region depends on temperature and the doping density of the diode. It also depends on forward biased state.
Complete answer:
Let us first understand what a p-n junction diode is.
When we join a p-type semiconductor with an n-type semiconductor, we get a p-n junction diode. The p-type of the diode consists of excess holes whereas the n-type part of the diode consists of excess electrons. At the junction, the electrons from n-type diffuse into p-type and combine with the holes. This creates a small region with separation of positive and negative charges with high resistivity at the junction of the diode called the depletion region.
The depletion region creates an electric field at the junction. Therefore, there is some potential difference across the depletion region. This potential difference is called barrier potential.
The barrier potential is directly proportional to width of the depletion region. This means that larger the width of the depletion region, greater will be the barrier potential.
The width of the depletion region depends on the way the diode is connected to the voltage source (whether forward biased or reverse biased). It also depends on temperature and the doping concentration of the diode.
Since the barrier potential depends on the width of the depletion region, the barrier potential depends on temperature, forward bias and the doping density of the diode.
Therefore, the barrier potential does not depend on the design of the diode. Hence, the correct option is A.
Note: Note that the width of depletion region increases with temperature.It decreases if the doping density of the diode is increased. When the diode is in forward bias, the width of the depletion region reduces. When the diode is in reverse bias, the width of the depletion region increases.
Complete answer:
Let us first understand what a p-n junction diode is.
When we join a p-type semiconductor with an n-type semiconductor, we get a p-n junction diode. The p-type of the diode consists of excess holes whereas the n-type part of the diode consists of excess electrons. At the junction, the electrons from n-type diffuse into p-type and combine with the holes. This creates a small region with separation of positive and negative charges with high resistivity at the junction of the diode called the depletion region.
The depletion region creates an electric field at the junction. Therefore, there is some potential difference across the depletion region. This potential difference is called barrier potential.
The barrier potential is directly proportional to width of the depletion region. This means that larger the width of the depletion region, greater will be the barrier potential.
The width of the depletion region depends on the way the diode is connected to the voltage source (whether forward biased or reverse biased). It also depends on temperature and the doping concentration of the diode.
Since the barrier potential depends on the width of the depletion region, the barrier potential depends on temperature, forward bias and the doping density of the diode.
Therefore, the barrier potential does not depend on the design of the diode. Hence, the correct option is A.
Note: Note that the width of depletion region increases with temperature.It decreases if the doping density of the diode is increased. When the diode is in forward bias, the width of the depletion region reduces. When the diode is in reverse bias, the width of the depletion region increases.
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