Explanation
The following logic gate circuit is equivalent to
In p-type materials holes are majority carriers and electrons are majority carriers in n-type materials. When the two types of semiconductor materials are joined together, the electrons from the n-type material diffuse into p-type material and combines with holes as their concentration is higher in n-type layer. This creates a layer of negative ions near the junction in p-type material. Negative ions are formed because the trivalent impurities (e.g., Aluminum) now has an extra electron from the n-type material. Similarly, the holes from the p-type material diffuse into n-type material resulting in a layer of positive ions in the n-type material.These negative ions creates an electric field in the direction from n-type to p-type. As more electrons diffuse into p-type material, the electric field strength goes on increasing. The electrons from n-type material now diffusing into p-type material will have to overcome the electric field due to negative ions. At one point, the electric field becomes sufficiently strong to stop further diffusion of electrons.
Hint: The voltage drop in a parallel connection is always the same. Use Ohm's law to find the current.
Step1: Finding the current through the zener diode.
The voltage drop in a parallel connection is always same. Therefore,
$$(\mathrm{i})_{1 \mathrm{k} \Omega}=\frac{15 \mathrm{~V}}{1 \mathrm{k} \Omega}=15 \mathrm{~mA}$$
$$(\mathrm{i})_{250 \Omega}=\frac{(20-15) \mathrm{V}}{250 \Omega}=\frac{5 \mathrm{~V}}{250 \Omega}=\frac{20}{1000} \mathrm{~A}=20 \mathrm{~mA}$$
$$\therefore(\mathrm{i})_{\mathrm{Zener}}=(20-15)=5 \mathrm{~mA} \text {. }$$
CORRECT ANSWER(d) 5ma
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