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Question

# How is a Zener diode fabricated? What causes the setting up of high electric field even for small reverse bias voltage across the diode? Describe, with the help of a circuit diagram, the working of Zener diode as a voltage regulator OR (a) Explain with the help of a diagram, how depletion region and potential barrier are formed in a junction diode. (b) If a small voltage is applied to a p-n junction diode, how will the barrier potential be affected when it is (i) forward biased, and (ii) reveres biased?

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Solution

## A Zener diode is fabricated by heavily doping both p and n sides of the junction Because of heavy doping, a very thin (<10−6m) depletion region is formed between the p and n sides, and hence, the electric field of the junction is extremely high( 5×106V/m) even for a small reverse bias voltage of about 5V.Zener diode as a voltage regulator:To get a constant d.c. voltage from the d.c. unregulated output of a rectifier, we use a Zener diode. The circuit diagram of a voltage regulator using a Zener diode is shown in the figure below The unregulated d.c. voltage is connected to the Zener diode through a series resistance Rs such that the Zener diode is reverse biased. If the input voltage increases, the current through Rs and the Zener diode also increases. This increases the voltage drop across Rs without any change in the voltage across the Zener diode. This is because the Zener voltage remains constant in the breakdown region even though the current through it changes. Similarly, if the input voltage decreases, the current through Rs and the Zener diode also decreases. The voltage drop across Rs decreases without any change in the voltage across the Zener diode. Thus, any increase or decrease in the input voltage results in an increase or decrease of the voltage drop across Rs without any change in the voltage across the Zener diode. Hence, the Zener diode acts as a voltage regulator. OR (a) We know that in an n-type semiconductor, the concentration of electrons is more compared to the concentration of holes. Similarly, in a p-type semiconductor, the concentration of holes is more than the concentration of electrons. During the formation of a p–n junction and because of the concentration gradient across the p and n sides, holes diffuse from the p-side to the n-side (p→n) and electrons diffuse from the n-side to the p-side (n→p). This motion of charge gives rise to a diffusion current across the junction. When an electron diffuses from n→p, it leaves behind an ionised donor on the n-side. This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from n→p, a layer of positive charge (or positive space–charge region) on n-side of the junction is developed. Similarly, when a hole diffuses from p→n due to the concentration gradient, it leaves behind an ionised acceptor (negative charge) which is immobile. As the holes continue to diffuse, a layer of negative charge (or negative space–charge region) on the p-side of the junction is developed. This space–charge region on either side of the junction together is known as the depletion region. Because of the positive space–charge region on the n-side of the junction and negative space charge region on the p-side of the junction, an electric field directed from the positive charge towards the negative charge develops. Due to this field, an electron on the p-side of the junction moves to the n-side and a hole on the n-side of the junction moves to the p-side. The loss of electrons from the n-region and the gain of electrons by the p-region cause a difference of potential across the junction of the two regions. This is how the barrier potential is formed. (b) (i) If a small voltage is applied to a p–n junction diode in the forward bias, then the barrier potential decreases. (ii)If a small voltage is applied to a p–n junction diode in the reverse bias, then the barrier potential increases.

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