Q: Determine the current through zener diode for the circuit shown in figure is: (Given: zener diode break down voltage \$V_z=5.6V\$)

For zener brak down potential difference across \$800\Omega \$ resistor will be \$5.6V\$4\$V_z=5.6V\$\$i_2=\dfrac{V_z}{800}=\dfrac{5.6}{800}=7mA\$\$\Delta V\$ across \$200\Omega =9-5.6=3.4V\$\$i_1=\dfrac{3.4}{200}=17mA\$\$i_1=1_2+i_z\$\$i_z=17mA-7mA=10mA\$

Q: In a Zener regulated power supply of a Zener diode with \$V_2 = 6V\$ is used for regulation. The load current is to be \$4 mA\$ and the unregulated input is \$10 V\$. The value of series resistor \$R_S\$ is (Assume \$I_{z}=5I_{L}\$):

Given,\$v_z=6v\$Load current\$=4mA=I_L\$unregulated input\$=10V\$To find: value of series resistance \$R_s=?\$Solution:\$I=I_Z+I_L\$\$I_Z=5I_L\$ (Given)\$I_Z=5\times 4=20mA\$\$I=20+4=24mA\$Voltage drop across \$R_s\$ is\$V_s=10-6=4V\$Therefore \$R_s=\dfrac{4}{24mA}=167r\$i.e., \$R_s\$ is greater than \$100r\$.

Question 1

How is Zener diode used as voltage regulator?

Accepted Answer

Zener Diodes are widely used as Shunt Voltage Regulators to regulate the voltage across small loads. Zener Diodes have a sharp reverse breakdown voltage and breakdown voltage will be constant for a wide range of currents. Thus we will connect the zener diode parallel to the load such that the applied voltage will reverse bias it. Thus if the reverse bias voltage across the zener diode exceeds the knee voltage, the voltage across the load will be constant.

Question 2

A Zener diode, having breakdown voltage equal to $15\ V$, is used in a voltage regulator circuit shown in figure. The current through the diode is

Accepted Answer

Hint: The voltage drop in a parallel connection is always the same. Use Ohm's law to find the current.&#10;&#10;Step1: Finding the current through the zener diode.The voltage drop in a parallel connection is always same. Therefore,&#10;&#10;$(\mathrm{i})_{1 \mathrm{k} \Omega}=\frac{15 \mathrm{~V}}{1&#10;\mathrm{k} \Omega}=15 \mathrm{~mA}$&#10;&#10;$(\mathrm{i})_{250 \Omega}=\frac{(20-15) \mathrm{V}}{250&#10;\Omega}=\frac{5 \mathrm{~V}}{250 \Omega}=\frac{20}{1000} \mathrm{~A}=20&#10;\mathrm{~mA}$&#10;&#10;$\therefore(\mathrm{i})_{\mathrm{Zener}}=(20-15)=5&#10;\mathrm{~mA} \text {. }$&#10;&#10; &#10;&#10;CORRECT ANSWER(d) 5ma

Question 3

Determine the current through zener diode for the circuit shown in figure is: (Given: zener diode break down voltage $V_z=5.6V$)

Accepted Answer

For zener brak down potential difference across $800\Omega $  resistor will be $5.6V$4$V_z=5.6V$$i_2=\dfrac{V_z}{800}=\dfrac{5.6}{800}=7mA$$\Delta V$ across $200\Omega =9-5.6=3.4V$$i_1=\dfrac{3.4}{200}=17mA$$i_1=1_2+i_z$$i_z=17mA-7mA=10mA$

Question 4

For the circuit show below, the current through the zener diode is :

Accepted Answer

Assuming zener diode does not undergo breakdown, current in circuit = $\dfrac{120}{15000} =8mA$Voltage drop across diode = 80 V > 50 V. The diode undergo breakdown. Current is $R_1 = \dfrac{70}{5000} = 14 mA$Current is $R_2 = \dfrac{50}{10000} =5A$current through diode = 9 mA

Question 5

In the given circuit, the current through zener diode is:

Accepted Answer

the current through 500 ohm resistor is 0.01 ampere as voltage across is 5V.the current through 1500 ohm resistor is 0.0067 ampere so the current through zener diode is difference in current through these resistors . so the current is $ 0.01-0.0067=0.0033 $ampere .On converting to mA we get Current through zener diode = 3.3 mAso the correct answer is B.

Question 6

In a Zener regulated power supply of a Zener diode with $V_2 = 6V$ is used for regulation. The load current is to be $4 mA$ and the unregulated input is $10 V$. The value of series resistor $R_S$ is (Assume $I_{z}=5I_{L}$):

Accepted Answer

Given,$v_z=6v$Load current$=4mA=I_L$unregulated input$=10V$To find: value of series resistance $R_s=?$Solution:$I=I_Z+I_L$$I_Z=5I_L$    (Given)$I_Z=5\times 4=20mA$$I=20+4=24mA$Voltage drop across $R_s$ is$V_s=10-6=4V$Therefore $R_s=\dfrac{4}{24mA}=167r$i.e., $R_s$ is greater than $100r$.

Question 7

Consider the following statements A and B and identify the correct answer.
Statement(A):A Zener diode is always connected in reverse bias to use it as voltage.
Statement(B):The potential barrier of a p-n junction lies between $0.1$ to $0.3 V$ , approximately.

Accepted Answer

When Zener diode is forward biased, it behaves like ordinary diode and when reverse biased, it will be a voltage stabilizer. Also, at room temperature, the voltage across the depletion layer for silicon is about $0.6-0.7V$ and for germanium is about $0.3-0.35 V$ Hence, statement A is correct but statement B is wrong.

Question 8

From the circuit shown below, the maximum and minimum value of zener diode current are :

Accepted Answer

The Zener diode acts as the voltage regulator in the circuit and the voltage supply is variable hence, excess of voltage over $50\ V$ will be across $5 K\Omega$ resistance.Thus, minimum voltage across $5 K\Omega$ resistance is $80\ V-50\ V = 30\ V$ Current through $5 K\Omega$ resistance is $I_1$(say)$I_1 = \dfrac{30}{5000} = 6\ mA $Also, maximum voltage across $5 K\Omega$ resistance is $120\ V-50\ V = 70\ V$ Current through $5 K\Omega$ resistance is $I'_1$...(say)$I'_1 = \dfrac{70}{5000} = 14\ mA $Now, voltage across load resistance of $10 K\Omega$ is $50\ V$Thus, current through $10 K\Omega$ will be$I_2 = \dfrac{50}{10000} = 5\ mA$Hence, maximum current through Zener diode is$I_z = I'_1 - I_2 = 14 - 5 = 9\ mA$   Minimum current through Zener diode is$I_z = I_1 - I_2 = 6 - 5 = 1\ mA$

Question 9

A Zener diode when used as a voltage regulator, is connected (a) in forward bias. (b) in reverse bias.(c) in parallel to the load.(d) in series to the load.

Accepted Answer

When Zener diode is biased in the forward direction it behaves like a normal signal diode passing the current increasing linearly with voltage, but as soon as a reverse voltage applied across the Zener diode attains the breakdown voltage of the device, a large current starts to flow through the diode. Thus, for a constant voltage large current will flow through it. Hence, it always connected in parallel to the load.

Question 10

Zener diode is used for

Accepted Answer

A zener diode is always operated in its reverse biased condition. A voltage regulator circuit can be designed using a zener diode to maintain a constant DC output voltage across the load in spite of variations in the input voltage or changes in the load current. The zener voltage regulator consists of a current limiting resistor $R_S$ connected in series with the input voltage $V_S$ with the zener diode connected in parallel with the load RL in this reverse biased condition.Hence we can say that ,zener diode is used for stabilization.

Zener Diode and its Working Principle

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Zener Diode