Zener Diode Questions in English

(B) The voltage across the series resistor $R_s$ is given by $V_{R_s} = V_{in} - V_z = 10 \, V - 6 \, V = 4 \, V$.
For the Zener diode to act as a voltage regulator, it must maintain a minimum current $I_{z,min} = 10 \, mA$ even when the load current $I_L$ is zero.
The total current $I$ flowing through $R_s$ is $I = I_z + I_L$.
To find the maximum value of $R_s$, we consider the condition where the load current $I_L$ is minimum (i.e., $I_L = 0$).
Thus, $I = I_{z,min} = 10 \, mA$.
Using Ohm's law, $R_s = \frac{V_{R_s}}{I} = \frac{4 \, V}{10 \, mA} = \frac{4 \, V}{10 \times 10^{-3} \, A} = 400 \, \Omega$.
Therefore, the maximum value of $R_s$ is $400 \, \Omega$.

(D) To determine if the Zener diode is in breakdown, we first calculate the voltage across the parallel combination of the $1 k\Omega$ resistor and the Zener diode without the Zener diode connected.
Using the voltage divider rule, the voltage $V_{AB}$ across the $1 k\Omega$ resistor is:
$V_{AB} = V_s \times \frac{R_{parallel}}{R_s + R_{parallel}} = 10 V \times \frac{1 k\Omega}{1 k\Omega + 1 k\Omega} = 10 V \times \frac{1}{2} = 5 V$.
Since the voltage across the terminals $A$ and $B$ is $5 V$, which is less than the Zener breakdown voltage of $6 V$, the Zener diode does not conduct.
Therefore, the current through the Zener diode is $0 A$ (Zero).

(A) In the given circuit,the Zener diode is connected in parallel with the $4 \text{ k}\Omega$ resistor. Since the Zener diode is operating in its breakdown region,the voltage across it is constant at $6 \text{ V}$.
Because the Zener diode and the $4 \text{ k}\Omega$ resistor are in parallel,the voltage across the $4 \text{ k}\Omega$ resistor is also $6 \text{ V}$.
The total voltage in the circuit is $10 \text{ V}$. The voltage drop across the $6 \text{ k}\Omega$ series resistor is the difference between the source voltage and the Zener breakdown voltage:
$V_{6\text{k}\Omega} = 10 \text{ V} - 6 \text{ V} = 4 \text{ V}$.
The current flowing through the $6 \text{ k}\Omega$ resistor is given by Ohm's law:
$I = \frac{V_{6\text{k}\Omega}}{R} = \frac{4 \text{ V}}{6 \text{ k}\Omega} = \frac{4}{6} \text{ mA} = \frac{2}{3} \text{ mA}$.

(D) The Zener diode is connected in series with a resistor $R = 100 \ \Omega$ and a battery of emf $V = 10 \ V$.
When the Zener diode is in the breakdown region,it maintains a constant voltage across it equal to its breakdown voltage,$V_Z = 5.6 \ V$.
The potential difference across the resistor $R$ is given by $\Delta V = V - V_Z = 10 \ V - 5.6 \ V = 4.4 \ V$.
Using Ohm's law,the current $I$ flowing through the circuit (and thus through the Zener diode) is:
$I = \frac{\Delta V}{R} = \frac{4.4 \ V}{100 \ \Omega} = 0.044 \ A$.
Converting this to milliamperes:
$I = 0.044 \times 1000 \ mA = 44 \ mA$.

(C) The Zener diode is used for voltage regulation. The power dissipation $P_D$ of the Zener diode is given by $P_D = V_Z \times i$,where $V_Z$ is the breakdown voltage and $i$ is the current through the diode.
Given $P_D = 0.4 \ W$ and $V_Z = 10 \ V$,we have:
$0.4 = 10 \times i$
$i = \frac{0.4}{10} = 0.04 \ A$
The protective resistance $R$ is connected in series with the Zener diode. The voltage drop across the resistance $R$ is $V_R = V_{source} - V_Z = 15 \ V - 10 \ V = 5 \ V$.
Using Ohm's law,$V_R = i \times R$,we get:
$R = \frac{V_R}{i} = \frac{5 \ V}{0.04 \ A} = 125 \ \Omega$.

(D) Given: Zener voltage $V_Z = 5 \text{ V}$,load current $I_L = 5 \text{ mA}$.
Maximum Zener current $I_Z = 4 \times I_L = 4 \times 5 \text{ mA} = 20 \text{ mA}$.
Total current through the series resistor $R_S$ is $I = I_L + I_Z = 5 \text{ mA} + 20 \text{ mA} = 25 \text{ mA} = 25 \times 10^{-3} \text{ A}$.
Maximum input voltage $V_{\text{in}} = 25 \text{ V}$.
The voltage drop across the series resistor $R_S$ is $V_{R_S} = V_{\text{in}} - V_Z = 25 \text{ V} - 5 \text{ V} = 20 \text{ V}$.
Using Ohm's law,$R_S = \frac{V_{R_S}}{I} = \frac{20 \text{ V}}{25 \times 10^{-3} \text{ A}} = 800 \Omega$.

(B) The circuit in $(A)$ consists of an ideal diode and a Zener diode connected in series,which acts as a voltage clipper.
From part $(C)$,the output voltage $v_{o}(t)$ is clipped at $+5 \text{ V}$ and $-5 \text{ V}$ (since the Zener breakdown voltage $V_Z = 5 \text{ V}$).
The input voltage $v_{in}(t)$ (part $B$) is a sine wave with an amplitude of $20 \text{ V}$.
When the input is positive,the Zener diode operates in the breakdown region,clamping the output to $+5 \text{ V}$.
When the input is negative,the ideal diode is forward-biased and the Zener diode is reverse-biased,clamping the output to $-5 \text{ V}$.
Thus,the circuit between points $X$ and $Y$ consists of an ideal diode and a Zener diode connected in series. Among the given options,the configuration that represents this behavior is the series combination of two diodes (one ideal and one Zener,though the diagram in option $B$ shows two standard diodes,the question implies identifying the series combination of the two components shown in the circuit). However,based on the standard interpretation of such problems,the correct arrangement is the series connection of the two diodes.

(A) The maximum current $I_{max}$ that the Zener diode can handle is given by $P = V_z I_{max}$.
Therefore,$I_{max} = \frac{P}{V_z} = \frac{0.5 \text{ W}}{10 \text{ V}} = 0.05 \text{ A}$.
The voltage across the series resistor $R$ is $V_R = V_{supply} - V_z = 25 \text{ V} - 10 \text{ V} = 15 \text{ V}$.
For safety,the minimum resistance $R_{min}$ is calculated using Ohm's law: $R_{min} = \frac{V_R}{I_{max}} = \frac{15 \text{ V}}{0.05 \text{ A}} = 300 \text{ } \Omega$.