In the given circuit, the breakdown voltage of the Zener diode is 3.0 V. What is the value of Iz ?
- 7 mA
- 1.5 mA
- 5.5 mA
- 10 mA
The Correct Option is C
Solution and Explanation
Step 1: Identify the Zener Breakdown Voltage
Given \(V_Z = 3 \, \text{V}\).
Step 2: Determine the Potentials in the Circuit
Let the potential at point \(B = 0 \, \text{V}\). The potential at \(E(V_E) = 10 \, \text{V}\). Therefore, \(V_C = V_A = 3 \, \text{V}\) across the Zener diode.
Step 3: Calculate the Total Current \(I\)
Using Ohm’s law, the total current \(I\) through the \(1 \, \text{k}\Omega\) resistor:
\[ I = \frac{10 - 3}{1000} = \frac{7}{1000} = 7 \, \text{mA} \]
Step 4: Find \(I_1\), the Current through the \(2 \, \text{k}\Omega\) Resistor
Voltage across the \(2 \, \text{k}\Omega\) resistor is \(3 \, \text{V}\):
\[ I_1 = \frac{3}{2000} = 1.5 \, \text{mA} \]
Step 5: Calculate \(I_Z\), the Zener Current
By Kirchhoff’s Current Law (KCL), \(I = I_Z + I_1\):
\[ I_Z = I - I_1 = 7 \, \text{mA} - 1.5 \, \text{mA} = 5.5 \, \text{mA} \]
So, the correct answer is: 5.5 mA
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Concepts Used:
Applications of diode
Diodes are used in many applications. By removing the signal component, several of them convert AC current to DC current also are known as rectifiers when used in this position. Diodes act as electric switches and can block voltage spikes, making them ideal for surgical protectors. They are hired to do digital logic. Some things are made from diodes such as power supply and voltage doubles. The sensors, as well as light on lighting devices and lasers, depend on LEDs. For electronic tuning, varactors are used, and for compressing transients in AC lines varistors are used. Zener diodes are used as voltage regulators where varactors are used for electronic tuning, and varistors are used for compressing transients in AC lines.
On the base of diodes, transistors and op-amps are built. The most common diode mode is the p-n junction. One (n) object with electrons as the charging conductor closes the second object (p) with holes (well-charged particles acted as depleted areas of electrons) as the charge carriers on this type of diode. Where they connect, the narrowing space forms, where electrons are dispersed to fill holes in the p-side. This effectively stops the flow of electrons. When a positive voltage is given to the p-side of this junction, electrons may easily move from it and the current flows into the diode to fill the holes. The shrinkage area expands as the negative bias interacts (i.e., negative voltage is given to the p-side), making the electrons harder to move across.