Zener Diode Questions in English

(B) Zener breakdown occurs in a $p-n$ junction diode when it is under a high reverse bias condition.
As the reverse bias voltage increases,the electric field across the depletion region becomes very strong.
This strong electric field exerts a force on the electrons in the valence band,which is sufficient to break the covalent bonds.
As a result,a large number of electron-hole pairs are generated,leading to a sudden increase in the reverse current.
This phenomenon is known as Zener breakdown.

(B) Zener breakdown occurs in heavily doped $P-N$ junction diodes.
In heavily doped diodes,the depletion layer is very thin.
When a reverse bias voltage is applied,the electric field across the thin depletion layer becomes very strong,which pulls electrons out of their covalent bonds.
This process is known as Zener breakdown.
In contrast,lightly doped diodes have a wider depletion layer,and avalanche multiplication is the primary mechanism for breakdown in those cases.

(D) Statement $A$ is wrong because the potential barrier for a silicon $PN$ junction is approximately $0.7 \, V$,and for a germanium $PN$ junction,it is approximately $0.3 \, V$. The range $0.1 \, V$ to $0.3 \, V$ is not universally applicable to all $PN$ junctions.
Statement $B$ is correct because a Zener diode is specifically designed to operate in the reverse breakdown region to act as a voltage regulator. When forward-biased,it behaves like an ordinary $PN$ junction diode.
Therefore,statement $A$ is wrong and statement $B$ is correct.

(A) The symbol for a Zener diode is represented by a standard p-n junction diode symbol,but the cathode line is bent at both ends,resembling the letter '$Z$'.
Option $A$ correctly depicts this standard symbol for a Zener diode.

(D) Zener diode is a special type of diode designed to operate in the reverse breakdown region.
When the input voltage varies,the Zener diode maintains a constant voltage across its terminals,provided the current through it remains within its rated limits.
This property allows it to act as a $dc$ voltage regulator or stabilizer.
Therefore,the correct option is $D$.

(B) Zener diode is specifically designed to operate in the reverse breakdown region. It is primarily used as a voltage regulator to maintain a constant output voltage across a load,which is known as voltage stabilisation. In contrast,a standard $p-n$ junction diode is typically used for rectification.

(D) The maximum power dissipation $(P)$ of the Zener diode is given as $364 \;mW = 364 \times 10^{-3} \;W$.
The breakdown voltage $(V_Z)$ is given as $9.1 \;V$.
The maximum current $(I_{Z_{\max}})$ that the diode can handle is calculated using the formula:
$I_{Z_{\max}} = \frac{P}{V_Z}$
Substituting the given values:
$I_{Z_{\max}} = \frac{364 \times 10^{-3} \;W}{9.1 \;V}$
$I_{Z_{\max}} = 40 \times 10^{-3} \;A$
Converting to milliamperes $(mA)$:
$I_{Z_{\max}} = 40 \;mA$.

(D) The Zener diode is in parallel with the load resistor of $1 \, k\Omega$. Since the Zener voltage is $5 \, V$,the voltage across the load resistor is also $5 \, V$.
The current through the load resistor is $I_L = \frac{V_Z}{R_L} = \frac{5 \, V}{1 \, k\Omega} = 5 \, mA$.
The total current $I$ flowing from the source through the $500 \, \Omega$ resistor is $I = \frac{V_{source} - V_Z}{R} = \frac{10 \, V - 5 \, V}{500 \, \Omega} = \frac{5 \, V}{500 \, \Omega} = 0.01 \, A = 10 \, mA$.
Applying Kirchhoff's Current Law at the node,the current through the Zener diode $I_Z$ (or $I_1$) is $I_Z = I - I_L = 10 \, mA - 5 \, mA = 5 \, mA$.

(A) Given: $E = 9 \ V$,$V_z = 6 \ V$,$R_L = 1000 \ \Omega$,and $R_s = 100 \ \Omega$.
The potential drop across the series resistor $R_s$ is $V_R = E - V_z = 9 \ V - 6 \ V = 3 \ V$.
The total current flowing through the series resistor $R_s$ is $I = \frac{V_R}{R_s} = \frac{3 \ V}{100 \ \Omega} = 0.03 \ A$.
The current flowing through the load resistor $R_L$ is $I_L = \frac{V_z}{R_L} = \frac{6 \ V}{1000 \ \Omega} = 0.006 \ A$.
The current through the Zener diode is $I_z = I - I_L = 0.03 \ A - 0.006 \ A = 0.024 \ A$.
The power dissipated in the Zener diode is $P_z = V_z \times I_z = 6 \ V \times 0.024 \ A = 0.144 \ W$.

(A) The current flowing through the load resistor $R_L$ is given by:
${I_L} = \frac{{{V_Z}}}{{{R_L}}} = \frac{{15}}{{1 \times {{10}^3}}} = 15 \, mA$
Applying Kirchhoff's voltage law to the input loop:
${V_i} = I \cdot R + {V_Z}$
Since the total current $I = {I_Z} + {I_L}$,we have:
${V_i} = ({I_Z} + {I_L})R + {V_Z}$
Rearranging to solve for the Zener current ${I_Z}$:
${I_Z} = \frac{{{V_i} - {V_Z}}}{R} - {I_L}$
Substituting the given values (${V_i} = 20 \, V$,${V_Z} = 15 \, V$,$R = 250 \, \Omega$,${I_L} = 15 \, mA$):
${I_Z} = \frac{{20 - 15}}{{250}} - 15 \times {10^{ - 3}}$
${I_Z} = \frac{5}{{250}} - 0.015$
${I_Z} = 0.02 - 0.015 = 0.005 \, A = 5 \, mA$

(A) The voltage drop across the $1\,k\Omega$ resistor is equal to the Zener breakdown voltage,$V_Z = 15\,V$.
The current through the $1\,k\Omega$ resistor $(I')$ is calculated as:
$I' = \frac{V_Z}{R_L} = \frac{15\,V}{1 \times 10^3\,\Omega} = 15 \times 10^{-3}\,A = 15\,mA$.
The voltage drop across the $250\,\Omega$ series resistor is the difference between the input voltage and the Zener voltage:
$V_R = 20\,V - 15\,V = 5\,V$.
The total current $(I)$ flowing through the $250\,\Omega$ resistor is:
$I = \frac{V_R}{R} = \frac{5\,V}{250\,\Omega} = 0.02\,A = 20\,mA$.
Applying Kirchhoff's Current Law at the junction,the current through the Zener diode $(I_Z)$ is:
$I_Z = I - I' = 20\,mA - 15\,mA = 5\,mA$.

(D) $Zener$ diode is specifically designed to operate in the reverse breakdown region. To achieve a sharp breakdown voltage,the depletion layer must be very thin,which is accomplished by heavy doping of both the $p$-type and $n$-type regions. In contrast,photodiodes,$LEDs$,and solar cells are typically moderately doped to optimize their light-sensitive or light-emitting characteristics. Therefore,the correct answer is $(c)$ only.

(B) The Zener diode is connected in parallel with the resistor $2R$.
Since the Zener diode is in the breakdown region,the voltage across it is constant at $9\ V$.
Because the resistor $2R$ is in parallel with the Zener diode,the voltage across $2R$ is also $9\ V$.
The total voltage applied across terminals $A$ and $B$ is $17\ V$.
Let $V_R$ be the voltage across resistor $R$.
According to Kirchhoff's voltage law,the total voltage is the sum of the voltage across $R$ and the voltage across the parallel combination of the Zener diode and $2R$.
$V_{total} = V_R + V_{Zener}$
$17\ V = V_R + 9\ V$
$V_R = 17\ V - 9\ V = 8\ V$.
Therefore,the potential voltage across $R$ is $8\ V$.

(A) In the given circuit,the first Zener diode (top) is connected in forward bias,so it acts as a simple conducting wire with negligible voltage drop.
The second Zener diode (bottom) is connected in reverse bias. Since the source voltage is $20 \, V$ and the Zener breakdown voltage of the second diode is $V_{z2} = 15 \, V$,the diode will operate in the breakdown region.
The voltage drop across the $10 \, k\Omega$ resistor is $V_R = V_{source} - V_{z2} = 20 \, V - 15 \, V = 5 \, V$.
The current $I$ through the circuit is given by Ohm's law: $I = \frac{V_R}{R} = \frac{5 \, V}{10 \, k\Omega} = 0.5 \, mA$.

(D) The Zener diode is connected in parallel with resistor $R_{2} = 1500 \, \Omega$. Since the Zener breakdown voltage is $V_{z} = 10 \, V$,the voltage across $R_{2}$ is fixed at $V_{R_{2}} = 10 \, V$.
The current through $R_{2}$ is:
$I_{R_{2}} = \frac{V_{R_{2}}}{R_{2}} = \frac{10 \, V}{1500 \, \Omega} = \frac{1}{150} \, A \approx 6.67 \times 10^{-3} \, A = 6.67 \, mA$.
The voltage drop across resistor $R_{1} = 500 \, \Omega$ is:
$V_{R_{1}} = V_{source} - V_{z} = 15 \, V - 10 \, V = 5 \, V$.
The total current supplied by the source through $R_{1}$ is:
$I_{R_{1}} = \frac{V_{R_{1}}}{R_{1}} = \frac{5 \, V}{500 \, \Omega} = 10^{-2} \, A = 10 \, mA$.
Applying Kirchhoff's Current Law at the node,the current through the Zener diode $(I_{z})$ is:
$I_{z} = I_{R_{1}} - I_{R_{2}} = 10 \, mA - 6.67 \, mA = 3.33 \, mA$.

(B) In the given circuit,the Zener diode is connected in parallel with the $1\,k\Omega$ resistor. The Zener diode acts as a voltage regulator,maintaining a constant voltage of $5\,V$ across it.
Since the $1\,k\Omega$ resistor is in parallel with the Zener diode,the voltage across the $1\,k\Omega$ resistor is also $5\,V$.
Using Ohm's law,the current $I$ flowing through the $1\,k\Omega$ resistor is given by:
$I = \frac{V}{R} = \frac{5\,V}{1\,k\Omega} = \frac{5}{1 \times 10^3\,\Omega} = 5 \times 10^{-3}\,A = 5\,mA$.

(D) The input voltage is $V_{in} = 30 \, V$ and the Zener breakdown voltage is $V_Z = 6 \, V$. The load resistance is $R_L = 1 \, k\Omega$.
For the Zener diode to regulate the voltage,it must operate in the breakdown region. The current through the load resistance is $I_L = \frac{V_Z}{R_L} = \frac{6 \, V}{1 \, k\Omega} = 6 \, mA$.
The voltage drop across the series resistor $R_S$ is $V_{R_S} = V_{in} - V_Z = 30 \, V - 6 \, V = 24 \, V$.
For the Zener diode to just maintain regulation,the current through the Zener diode $I_Z$ should be at least $0$ (the limiting case for maximum $R_S$).
Thus,the total current through $R_S$ is $I_S = I_L + I_Z = 6 \, mA + 0 = 6 \, mA$.
Using Ohm's law for $R_S$,we have $R_S = \frac{V_{R_S}}{I_S} = \frac{24 \, V}{6 \, mA} = 4 \, k\Omega$.
Therefore,the maximum value of $R_S$ is $4 \, k\Omega$.

(C) The total current $I$ supplied by the source is given by Ohm's law applied to the series resistor:
$I = \frac{V_{in} - V_z}{R} = \frac{9\,V - 6\,V}{50\,\Omega} = \frac{3\,V}{50\,\Omega} = 0.06\,A = 60\,mA$.
The Zener diode must operate between its knee current $I_{z,min} = 5\,mA$ and its maximum current $I_{z,max}$ determined by the power dissipation limit $P_{max} = 300\,mW$.
$I_{z,max} = \frac{P_{max}}{V_z} = \frac{300\,mW}{6\,V} = 50\,mA$.
The load current $I_L$ is given by $I_L = I - I_z$.
To find the maximum load current $I_{L,max}$, we use the minimum Zener current:
$I_{L,max} = I - I_{z,min} = 60\,mA - 5\,mA = 55\,mA$.
To find the minimum load current $I_{L,min}$, we use the maximum Zener current:
$I_{L,min} = I - I_{z,max} = 60\,mA - 50\,mA = 10\,mA$.
Thus, the minimum and maximum load currents are $10\,mA$ and $55\,mA$ respectively.

$(A)$ The Zener diode maintains a constant voltage $V_z = 9\,V$ across the load resistance $R_L = 450\,\Omega$.
The current through the load resistance is $I_L = \frac{V_z}{R_L} = \frac{9\,V}{450\,\Omega} = 0.02\,A$.
The maximum current through the Zener diode is $I_{Z,max} = \frac{P_{max}}{V_z} = \frac{0.27\,W}{9\,V} = 0.03\,A$.
The total current flowing through the series resistor $R_s$ is $I = I_L + I_{Z,max} = 0.02\,A + 0.03\,A = 0.05\,A$.
The voltage drop across the series resistor $R_s$ is $V_{R_s} = V_{in} - V_z = 12\,V - 9\,V = 3\,V$.
Therefore,the value of the series resistor is $R_s = \frac{V_{R_s}}{I} = \frac{3\,V}{0.05\,A} = 60\,\Omega$.

(A) Zener diode is specifically designed to operate in the reverse breakdown region. To function as a voltage regulator,it must be connected in parallel with the load resistor $R_L$ so that the voltage across the load remains constant at the Zener breakdown voltage $V_Z$. Furthermore,the Zener diode must be connected in reverse bias,meaning its cathode (the side with the bar) is connected to the positive terminal of the input voltage source through a series current-limiting resistor $R_S$. This configuration ensures that if the input voltage or load current changes,the Zener diode adjusts its current to maintain a constant output voltage across $R_L$. Looking at the provided options,the first diagram (represented by image $828-$a134) shows the Zener diode in reverse bias parallel to the load,which is the correct configuration.

(B) The voltage across the load resistor $(R_L = 1 \, k\Omega)$ is equal to the Zener breakdown voltage,$V_Z = 5 \, V$.
The current through the load resistor is:
$I_L = \frac{V_Z}{R_L} = \frac{5 \, V}{1 \, k\Omega} = 5 \, mA$.
The total current supplied by the $20 \, V$ source through the series resistor $(R = 150 \, \Omega)$ is:
$I = \frac{V_{in} - V_Z}{R} = \frac{20 \, V - 5 \, V}{150 \, \Omega} = \frac{15 \, V}{150 \, \Omega} = 0.1 \, A = 100 \, mA$.
The current through the Zener diode $(I_Z)$ is the difference between the total current and the load current:
$I_Z = I - I_L = 100 \, mA - 5 \, mA = 95 \, mA$.

(D) During the regulation action of a Zener diode,the input voltage or load current may vary.
To maintain a constant output voltage across the load,the Zener diode adjusts its internal resistance.
As the Zener resistance changes,the current flowing through the Zener diode also changes.
Consequently,the voltage drop across the series resistance $(R_s)$ changes to compensate for the variations,ensuring the voltage across the Zener remains constant.
Therefore,both the current through the series resistance and the resistance offered by the Zener change.

(D) The Zener diode is connected in parallel with resistor $R_{2} = 1500 \, \Omega$. Since the Zener breakdown voltage is $V_{z} = 10 \, V$,the voltage across $R_{2}$ is fixed at $V_{R_{2}} = V_{z} = 10 \, V$.
The current through $R_{2}$ is:
$I_{R_{2}} = \frac{V_{R_{2}}}{R_{2}} = \frac{10 \, V}{1500 \, \Omega} = \frac{1}{150} \, A \approx 6.67 \times 10^{-3} \, A = 6.67 \, mA$.
The voltage drop across $R_{1} = 500 \, \Omega$ is:
$V_{R_{1}} = V_{source} - V_{z} = 15 \, V - 10 \, V = 5 \, V$.
The total current supplied by the source through $R_{1}$ is:
$I = \frac{V_{R_{1}}}{R_{1}} = \frac{5 \, V}{500 \, \Omega} = 10^{-2} \, A = 10 \, mA$.
Applying Kirchhoff's Current Law at the junction,the current through the Zener diode $(I_{z})$ is:
$I_{z} = I - I_{R_{2}} = 10 \, mA - 6.67 \, mA = 3.33 \, mA$.

(B) The voltage across the zener diode is $V_{Z} = 10 \ V$. Since the zener diode is in parallel with resistor $R_{2} = 1500 \ \Omega$,the voltage across $R_{2}$ is $V_{R_{2}} = V_{Z} = 10 \ V$.
The current through $R_{2}$ is:
$I_{R_{2}} = \frac{V_{R_{2}}}{R_{2}} = \frac{10 \ V}{1500 \ \Omega} = \frac{1}{150} \ A \approx 6.67 \times 10^{-3} \ A = 6.67 \ mA$.
The voltage across $R_{1} = 500 \ \Omega$ is:
$V_{R_{1}} = V_{source} - V_{Z} = 15 \ V - 10 \ V = 5 \ V$.
The total current through $R_{1}$ is:
$I_{R_{1}} = \frac{V_{R_{1}}}{R_{1}} = \frac{5 \ V}{500 \ \Omega} = 0.01 \ A = 10 \ mA$.
Applying Kirchhoff's Current Law at the junction,the current through the zener diode $I_{Z}$ is:
$I_{Z} = I_{R_{1}} - I_{R_{2}} = 10 \ mA - 6.67 \ mA = 3.33 \ mA \approx 3.3 \ mA$.

(C) The power dissipation $P$ in the protective resistance $R$ is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage across the resistor.
To determine the minimum voltage range of the $DC$ source required to test the diode up to its maximum power rating,we use the power formula $P = \frac{V^2}{R}$.
Given $P = 1 \,W$ and $R = 100 \,\Omega$,we have $V^2 = P \times R$.
$V^2 = 1 \,W \times 100 \,\Omega = 100 \,V^2$.
Taking the square root,we get $V = 10 \,V$.
Therefore,the minimum voltage range of the $DC$ source must be at least $0-10 \,V$. Among the given options,$0-12 \,V$ is the appropriate range that covers the required $10 \,V$ limit.

(A) In the given $dc$ voltage regulator circuit,the total current flowing through the series resistor $R_S$ is the sum of the load current $I_L$ and the Zener diode current $I_Z$.
From the circuit diagram,the Zener diode current is given as $I_Z = nI_L$.
Therefore,the total current $I$ flowing through $R_S$ is $I = I_L + I_Z = I_L + nI_L = (n + 1)I_L$.
The voltage drop across the resistor $R_S$ is $V_i - V_L$.
Using Ohm's law,$V_i - V_L = I \times R_S$.
Substituting the value of $I$,we get $V_i - V_L = (n + 1)I_L \times R_S$.
Thus,the value of the resistor $R_S$ is $R_S = \frac{V_i - V_L}{(n + 1)I_L}$.

(D) Given: Series resistance $R = 4 \, k\Omega = 4 \times 10^3 \, \Omega$,Input voltage $V_i = 60 \, V$,Zener voltage $V_Z = 10 \, V$,Load resistance $R_L = 2 \, k\Omega = 2 \times 10^3 \, \Omega$.
$1$. The load current $I_L$ is determined by the Zener voltage across the load resistance:
$I_L = \frac{V_Z}{R_L} = \frac{10 \, V}{2 \times 10^3 \, \Omega} = 5 \times 10^{-3} \, A = 5 \, mA$.
$2$. The total current $I$ flowing through the series resistor $R$ is determined by the voltage drop across it:
$I = \frac{V_i - V_Z}{R} = \frac{60 \, V - 10 \, V}{4 \times 10^3 \, \Omega} = \frac{50 \, V}{4000 \, \Omega} = 12.5 \times 10^{-3} \, A = 12.5 \, mA$.
$3$. Applying Kirchhoff's Current Law at node $A$:
$I = I_Z + I_L$
$I_Z = I - I_L = 12.5 \, mA - 5 \, mA = 7.5 \, mA$.
Thus,the currents are $I = 12.5 \, mA$,$I_Z = 7.5 \, mA$,and $I_L = 5 \, mA$.

(A) The Zener diode is connected in parallel with the $10 \, k\Omega$ resistor. Since the Zener breakdown voltage is $50 \, V$,the voltage across the $10 \, k\Omega$ resistor is also $50 \, V$.
The current through the $10 \, k\Omega$ resistor is $I_L = \frac{V_Z}{R_L} = \frac{50 \, V}{10 \, k\Omega} = 5 \, mA$.
The total current supplied by the $120 \, V$ source passes through the $5 \, k\Omega$ series resistor. The voltage drop across this resistor is $120 \, V - 50 \, V = 70 \, V$.
The total current $I_S$ is $I_S = \frac{70 \, V}{5 \, k\Omega} = 14 \, mA$.
Applying Kirchhoff's Current Law at the node,the current through the Zener diode $I_Z$ is $I_Z = I_S - I_L = 14 \, mA - 5 \, mA = 9 \, mA$.

(A) First,we check if the Zener diode is in the breakdown region. Assume the Zener diode is not conducting $(i_Z = 0)$.
The circuit becomes a series combination of $R_1 = 500 \ \Omega$ and $R_2 = 1500 \ \Omega$ in parallel with another $R_2 = 1500 \ \Omega$ (as shown in the diagram,the load resistor is $1500 \ \Omega$).
The voltage across the parallel combination is $V_p = 12 \times \frac{750}{500 + 750} = 12 \times \frac{750}{1250} = 7.2 \ V$.
Since $7.2 \ V < 10 \ V$ (the Zener breakdown voltage),the Zener diode does not conduct.
Therefore,the current through the Zener diode is $0 \ mA$.

(A) The total voltage supplied is $V = 9\, V$. The Zener diode is in parallel with the load resistor $R_L = 800\, \Omega$,so the voltage across the load is $V_z = 5.6\, V$.
The voltage drop across the series resistor $R_s = 200\, \Omega$ is $V_{R_s} = V - V_z = 9 - 5.6 = 3.4\, V$.
The total current flowing through the series resistor $R_s$ is $I = \frac{V_{R_s}}{R_s} = \frac{3.4}{200} = 0.017\, A = 17\, mA$.
The current flowing through the load resistor $R_L$ is $I_L = \frac{V_z}{R_L} = \frac{5.6}{800} = 0.007\, A = 7\, mA$.
Applying Kirchhoff's current law at the junction,the current through the Zener diode is $I_z = I - I_L = 17\, mA - 7\, mA = 10\, mA$.

(D) Given: Breakdown voltage of Zener diode $V_Z = 6\,V,$ load resistance $R_L = 4\,k\Omega,$ series resistance $R_i = 1\,k\Omega.$
The current through the load resistance is constant because the Zener diode maintains a constant voltage $V_Z$ across it:
$I_L = \frac{V_Z}{R_L} = \frac{6\,V}{4\,k\Omega} = 1.5\,mA.$
The total current $I_i$ supplied by the battery through the series resistor $R_i$ is given by $I_i = \frac{V_B - V_Z}{R_i}.$
For minimum battery voltage $V_B = 8\,V$:
$I_{i,min} = \frac{8\,V - 6\,V}{1\,k\Omega} = \frac{2\,V}{1\,k\Omega} = 2\,mA.$
The Zener current $I_Z = I_i - I_L = 2\,mA - 1.5\,mA = 0.5\,mA.$
For maximum battery voltage $V_B = 16\,V$:
$I_{i,max} = \frac{16\,V - 6\,V}{1\,k\Omega} = \frac{10\,V}{1\,k\Omega} = 10\,mA.$
The Zener current $I_Z = I_i - I_L = 10\,mA - 1.5\,mA = 8.5\,mA.$
Thus,the minimum and maximum currents through the Zener diode are $0.5\,mA$ and $8.5\,mA$ respectively.

(B) The Zener diode acts as a voltage regulator, maintaining a constant voltage of $V_Z = 6\,V$ across the load resistance $R_L = 4\,k\Omega$ when it is in the breakdown region.
The load current $I_L$ is constant and is given by:
$I_L = \frac{V_Z}{R_L} = \frac{6\,V}{4\,k\Omega} = 1.5\,mA$
The total source current $I_S$ is given by:
$I_S = \frac{V_{in} - V_Z}{R_S}$
The Zener current $I_Z$ is given by:
$I_Z = I_S - I_L = \frac{V_{in} - 6}{2\,k\Omega} - 1.5\,mA$
To maximize $I_Z$, we must use the maximum input voltage $V_{in} = 16\,V$:
$I_{S,max} = \frac{16\,V - 6\,V}{2\,k\Omega} = \frac{10\,V}{2\,k\Omega} = 5\,mA$
Therefore, the maximum Zener current is:
$I_{Z,max} = I_{S,max} - I_L = 5\,mA - 1.5\,mA = 3.5\,mA$

(B) The circuit consists of a $16 \ V$ $DC$ source, a series resistor $R_s = 1 \ k\Omega$, a Zener diode with breakdown voltage $V_z = 6 \ V$, and a load resistor $R_L = 1 \ k\Omega$.
First, we calculate the total current $I_s$ flowing from the source through the series resistor:
$I_s = \frac{V_{in} - V_z}{R_s} = \frac{16 \ V - 6 \ V}{1 \ k\Omega} = \frac{10 \ V}{1 \ k\Omega} = 10 \ mA$.
Next, we calculate the current $I_L$ flowing through the load resistor $R_L$:
$I_L = \frac{V_z}{R_L} = \frac{6 \ V}{1 \ k\Omega} = 6 \ mA$.
Applying Kirchhoff's Current Law at the junction, the current $I$ through the Zener diode is:
$I = I_s - I_L = 10 \ mA - 6 \ mA = 4 \ mA$.

(A) When a Zener diode is connected in reverse bias,as the applied voltage reaches the Zener breakdown voltage,the strong electric field causes the rupture of covalent bonds. This releases a large number of charge carriers,leading to a sudden increase in current in the Zener diode.

(B) Zener diode is specifically designed to operate in the reverse breakdown region. When connected in a circuit,it maintains a constant voltage across its terminals regardless of variations in the input voltage or load current. Therefore,it is primarily used as a voltage regulator. Thus,option $(b)$ is correct.

(D) The total resistance in the circuit is $R = 200 \; \Omega + 200 \; \Omega = 400 \; \Omega$.
The voltage drop across the resistors is $V_R = V_{in} - V_{out} = 12 \; V - 8 \; V = 4 \; V$.
The current flowing through the circuit is $I = \frac{V_R}{R} = \frac{4 \; V}{400 \; \Omega} = 0.01 \; A = 10 \; mA$.
Since the two identical Zener diodes are connected in series, the voltage across each diode is $V_d = \frac{8 \; V}{2} = 4 \; V$.
The power dissipated in each diode is $P = V_d \times I = 4 \; V \times 10 \; mA = 40 \; mW$.

(C) The value of $R_{S}$ should be such that the current through the Zener diode is significantly larger than the load current to ensure good load regulation.
Typically, we choose the Zener current $(I_{z})$ to be five times the load current $(I_{L})$, i.e., $I_{z} = 5 \times 4.0 \; mA = 20 \; mA$.
The total current through $R_{S}$ is $I = I_{z} + I_{L} = 20 \; mA + 4.0 \; mA = 24 \; mA$.
The voltage drop across $R_{S}$ is $V_{in} - V_{z} = 10.0 \; V - 6.0 \; V = 4.0 \; V$.
Using Ohm's law, $R_{S} = \frac{V_{in} - V_{z}}{I} = \frac{4.0 \; V}{24 \times 10^{-3} \; A} \approx 167 \; \Omega$.
The nearest standard value for a carbon resistor is $150 \; \Omega$. Therefore, a series resistor of $150 \; \Omega$ is appropriate.

(N/A) Zener diode is a special-purpose semiconductor diode,named after its inventor,$C$. Zener.
The symbol for a Zener diode is shown in figure $(a)$.
$A$ Zener diode is fabricated by heavily doping both the $p$ and $n$-sides of the junction. Due to this,the depletion region formed is very thin $\left(< 10^{-6} \ m\right)$ and the electric field at the junction is extremely high $\left(\approx 5 \times 10^{6} \ V/m\right)$ even for a small reverse bias voltage of about $5 \ V$.
The $I-V$ characteristics of a Zener diode are shown in figure $(b)$.
It is observed that when the applied reverse bias voltage $(V)$ reaches the breakdown voltage $\left(V_{Z}\right)$ of the Zener diode,there is a large change in the current.
Note that after the breakdown voltage $V_{Z}$,a large change in the current can be produced by an almost insignificant change in the reverse bias voltage. Thus,the Zener voltage remains constant.
This property of the Zener diode is used for regulating supply voltages so that they remain constant.
We know that the reverse current is due to the flow of electrons (minority carriers) from $p \rightarrow n$ and holes from $n \rightarrow p$.

(N/A) When the $AC$ input voltage of a rectifier fluctuates,its rectified output also fluctuates.
To obtain a constant $DC$ voltage from the unregulated $DC$ output of a rectifier,a Zener diode can be used.
The circuit diagram of a $DC$ voltage regulator using a Zener diode is shown in the figure.
The unregulated $DC$ voltage (filtered output of a rectifier) is connected to the Zener diode through a series resistance $R_S$ such that the Zener diode is reverse-biased.
If the input voltage increases,the current through $R_S$ and the Zener diode also increases. This increases the voltage drop across $R_S$ without any change in the voltage across the Zener diode. This is because,in the breakdown region,the Zener voltage remains constant even though the current through the Zener diode changes.
Similarly,if the input voltage decreases,the current through $R_S$ and the Zener diode also decreases. The voltage drop across $R_S$ 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 $R_S$ without any change in the voltage across the Zener diode.
Thus,the Zener diode acts as a voltage regulator. To maintain a constant voltage across the Zener diode,we must select the Zener diode according to the required output voltage and choose the series resistance $R_S$ accordingly.

(N/A) Zener diode is primarily used as a voltage regulator.
It is designed to operate in the reverse breakdown region without being damaged.
When connected in parallel with a load,it maintains a constant output voltage across the load,even if the input voltage or the load current varies,provided the input voltage remains above the Zener breakdown voltage.

(B) The power rating of the Zener diode is $P = 1 \, W$ and its breakdown voltage is $V_Z = 5 \, V$.
The maximum current that can flow through the Zener diode is given by $I_{Z, \text{max}} = \frac{P}{V_Z} = \frac{1 \, W}{5 \, V} = 0.2 \, A$.
For the circuit to operate safely,the Zener diode must handle the maximum input voltage without exceeding its power rating. The voltage across the series resistor $R_s$ is $V_{R_s} = V_i - V_Z$.
To ensure the Zener diode is not damaged,we consider the maximum input voltage $V_{i, \text{max}} = 7 \, V$.
The current through the resistor $R_s$ is $I = \frac{V_{i, \text{max}} - V_Z}{R_s}$.
For safe operation,this current must be at least the maximum current the Zener can handle: $I \ge I_{Z, \text{max}}$.
Thus,$R_s \le \frac{V_{i, \text{max}} - V_Z}{I_{Z, \text{max}}} = \frac{7 \, V - 5 \, V}{0.2 \, A} = \frac{2 \, V}{0.2 \, A} = 10 \, \Omega$.
Therefore,the value of $R_s$ should be $10 \, \Omega$.

(A) The circuit consists of two Zener diodes connected in series in reverse polarity.
When the input voltage $V_{in}$ is positive, the first Zener diode is in reverse bias and the second is in forward bias.
When $V_{in}$ exceeds the breakdown voltage of $6\, V$, the first Zener diode operates in the breakdown region, maintaining a constant output voltage of $6\, V$.
When $V_{in}$ is between $0\, V$ and $6\, V$, the output voltage follows the input voltage.
Similarly, when the input voltage $V_{in}$ is negative, the second Zener diode is in reverse bias and the first is in forward bias.
When $V_{in}$ becomes more negative than $-6\, V$, the second Zener diode operates in the breakdown region, maintaining a constant output voltage of $-6\, V$.
Thus, the output voltage is clipped at $+6\, V$ and $-6\, V$, resulting in a waveform that follows the input sine wave between $-6\, V$ and $+6\, V$ and remains constant at $\pm 6\, V$ outside this range.
Comparing this with the given options, the correct graph is represented by option $A$.

(A) $1$. For $V_{\text{in}} < 4\, V$: Neither Zener diode $A$ $(6\, V)$ nor $B$ $(4\, V)$ is in the breakdown region. Thus,the output voltage $V_{0}$ follows the input voltage $V_{\text{in}}$,i.e.,$V_{0} = V_{\text{in}}$.
$2$. For $4\, V < V_{\text{in}} < 6\, V$: Zener diode $B$ reaches its breakdown voltage of $4\, V$. Since it is in parallel with the load,the output voltage $V_{0}$ is clamped at $4\, V$. The excess voltage $(V_{\text{in}} - 4\, V)$ drops across the series resistance associated with diode $B$.
$3$. For $V_{\text{in}} > 6\, V$: Zener diode $A$ also reaches its breakdown voltage of $6\, V$. Since $A$ is in parallel with the load,the output voltage $V_{0}$ is now clamped at $6\, V$. The excess voltage drops across the series resistance of the source or the circuit.
$4$. Therefore,the graph of $V_{0}$ versus time will show a linear increase up to $4\, V$,a constant plateau at $4\, V$,and then a step increase to a constant plateau at $6\, V$.

(D) The total voltage of the source is $V_{in} = 30\, V$.
The voltage across the Zener diode is $V_z = 6\, V$.
Since the Zener diode is in parallel with the resistor $R$, the voltage across the resistor $R$ is $V_R = V_z = 6\, V$.
The voltage across the $1\, k\Omega$ series resistor is $V_s = V_{in} - V_z = 30\, V - 6\, V = 24\, V$.
The current flowing through the series resistor is $I_s = \frac{V_s}{1\, k\Omega} = \frac{24\, V}{1\, k\Omega} = 24\, mA$.
Let $I_z$ be the current through the Zener diode and $I_R$ be the current through the resistor $R$. By Kirchhoff's Current Law, $I_s = I_z + I_R$.
For the resistance $R$ to be maximum, the current $I_R$ must be minimum. The minimum current through the Zener diode is $I_{z,min} = 0$.
Thus, $I_R = I_s = 24\, mA$.
Using Ohm's law for resistor $R$, $R = \frac{V_R}{I_R} = \frac{6\, V}{24\, mA} = 0.25\, k\Omega$.
Wait, re-evaluating the circuit: The Zener diode is in parallel with $R$. The voltage across the series resistor $1\, k\Omega$ is $V_{series} = 30 - 6 = 24\, V$. The current through the series resistor is $I = \frac{24\, V}{1\, k\Omega} = 24\, mA$. This current splits into the Zener diode and the resistor $R$. To maximize $R$, we minimize the current through it. If we assume the Zener diode takes all excess current, the minimum current through $R$ is not well-defined without a minimum Zener current. However, if the question implies the Zener diode is just at the breakdown point, then $I_R = 24\, mA$, giving $R = 0.25\, k\Omega$. Given the options, let's re-read the circuit. If $R$ is in series with the Zener, $R = (30-6)/I_z$. If $I_z = 6\, mA$ (as per diagram label), then $R = 24/6 = 4\, k\Omega$.

(A) $1$. The voltage across the series resistor $R_{S}$ is $V_{R_{S}} = V_{in} - V_{Z} = 22 \text{ V} - 15 \text{ V} = 7 \text{ V}$.
$2$. The total current flowing through the series resistor $R_{S}$ is $I = \frac{V_{R_{S}}}{R_{S}} = \frac{7 \text{ V}}{35 \Omega} = 0.2 \text{ A} = \frac{1}{5} \text{ A}$.
$3$. The current flowing through the load resistor $R_{L} = 90 \Omega$ is $I_{L} = \frac{V_{Z}}{R_{L}} = \frac{15 \text{ V}}{90 \Omega} = \frac{1}{6} \text{ A}$.
$4$. The current flowing through the Zener diode is $I_{Z} = I - I_{L} = \frac{1}{5} \text{ A} - \frac{1}{6} \text{ A} = \frac{6 - 5}{30} \text{ A} = \frac{1}{30} \text{ A}$.
$5$. The power dissipated across the Zener diode is $P = V_{Z} \times I_{Z} = 15 \text{ V} \times \frac{1}{30} \text{ A} = 0.5 \text{ W}$.
$6$. Expressing this in the form $x \times 10^{-1} \text{ W}$, we get $0.5 \text{ W} = 5 \times 10^{-1} \text{ W}$.
$7$. Therefore, the value of $x$ is $5$.

(C) The Zener diode is connected in parallel with the $2\, k\Omega$ resistor. Since the Zener breakdown voltage is $5\, V$,the voltage across the $2\, k\Omega$ resistor will be constant at $5\, V$ as long as the Zener diode is in the breakdown region.
Using Ohm's law,the current $I$ flowing through the $2\, k\Omega$ resistor is given by:
$I = \frac{V}{R} = \frac{5\, V}{2 \times 10^3\, \Omega}$
$I = 2.5 \times 10^{-3}\, A$
To express this in terms of $10^{-4}\, A$:
$I = 25 \times 10^{-4}\, A$
Thus,the value is $25$.

(B) Zener breakdown is a phenomenon that occurs in $p-n$ junction diodes that are heavily doped.
Due to heavy doping,the depletion layer becomes very narrow (typically less than $10^{-6} \ m$).
When a reverse bias voltage is applied,the electric field across this narrow depletion layer becomes extremely high.
This high electric field is sufficient to pull electrons out of their covalent bonds,leading to a large increase in current,which is known as Zener breakdown.

(B) The total voltage supplied is $V_{in} = 90\, V$ and the series resistance is $R_{s} = 4\, k\Omega = 4000\, \Omega$.
The Zener diode maintains a constant voltage of $V_{z} = 30\, V$ across it.
The voltage drop across the series resistor $R_{s}$ is $V_{R} = V_{in} - V_{z} = 90\, V - 30\, V = 60\, V$.
The total current $i$ flowing from the source through the series resistor is $i = \frac{V_{R}}{R_{s}} = \frac{60\, V}{4000\, \Omega} = 0.015\, A = 15\, mA$.
The load resistor is $R_{L} = 5\, k\Omega = 5000\, \Omega$. The current $i_{1}$ flowing through the load resistor is $i_{1} = \frac{V_{z}}{R_{L}} = \frac{30\, V}{5000\, \Omega} = 0.006\, A = 6\, mA$.
Applying Kirchhoff's Current Law at the junction,the current through the Zener diode $i_{z}$ is given by $i_{z} = i - i_{1} = 15\, mA - 6\, mA = 9\, mA$.

(B) The current through the load resistor $R_L$ is given by:
$i = \frac{V_z}{R_L} = \frac{10 \, V}{5 \, k\Omega} = 2 \, mA$
The total current $I$ flowing through the series resistor is:
$I = \frac{V_{in} - V_z}{R_s} = \frac{24 \, V - 10 \, V}{1 \, k\Omega} = \frac{14 \, V}{1 \, k\Omega} = 14 \, mA$
The current through the zener diode $I_z$ is:
$I_z = I - i = 14 \, mA - 2 \, mA = 12 \, mA$
The power dissipated across the zener diode is:
$P = I_z \times V_z = 12 \, mA \times 10 \, V = 120 \, mW$

(B) The power rating of the Zener diode is $P_z = 2 \, W$ and its breakdown voltage is $V_z = 10 \, V$.
The maximum current that can flow through the Zener diode is $I_{z,max} = \frac{P_z}{V_z} = \frac{2 \, W}{10 \, V} = 0.2 \, A$.
To ensure safe operation, the Zener diode must handle the maximum input voltage of $V_{in,max} = 14 \, V$ without exceeding its power rating.
When $V_{in} = 14 \, V$, the voltage drop across the series resistor $R_s$ is $\Delta V_{Rs} = V_{in,max} - V_z = 14 \, V - 10 \, V = 4 \, V$.
Assuming the load current is negligible for the worst-case calculation (or that the Zener must handle the full current when no load is connected), the current through $R_s$ is $I = I_{z,max} = 0.2 \, A$.
Using Ohm's law, $R_s = \frac{\Delta V_{Rs}}{I} = \frac{4 \, V}{0.2 \, A} = 20 \, \Omega$.