Question

zener shunt regulator employs a 9.1-V zener diode for which VZ = 9.1 V at IZ = 9 mA, with rz = 40 and IZK= 0.5 mA. The available supply voltage of 15 V can varyas much as ±10%. For this diode, what is the value of VZ0?For a nominal load resistance RL of 1 k and a nominal zenercurrent of 10 mA,what current must flow in the supply resistorR? For the nominal value of supply voltage, select a valuefor resistor R, specified to one significant digit, to provideat least that current. What nominal output voltage results?For a ±10% change in the supply voltage, what variationin output voltage results? If the load current is reduced by50%, what increase in VO results? What is the smallest valueof load resistance that can be tolerated while maintainingregulation when the supply voltage is low? What is the lowestpossible output voltage that results? Calculate values for theline regulation and for the load regulation for this circuit usingthe numerical results obtained in this problem.

Answer 1

Answer:

$V_z=9.1v$

$V_{zo}=8.74V$

$I=10mA$

$R=589 ohms$

Explanation:

From the question we are told that:

Zener diode Voltage $V_z=9.1-V$

Zener diode Current $I_z=9 .A$

Note

$rz = 40\\\\IZK= 0.5 mA$

Supply Voltage $V_s=15$

Reduction Percentage $P_r= 50 \%$

Generally the equation for Kirchhoff's Voltage Law is mathematically given by

$V_z=V_{zo}+I_zr_z$

$9.1=V_{z0}+9*10^{-3}(40)$

$V_{zo}=8.74V$

Therefore

$At I_z-10mA$

$V_z=V_{z0}+I_zr_z$

$V_z=8.74+(10*10^{-3}) (40)$

$V_z=9.1v$

Generally the equation for Kirchhoff's Current Law is mathematically given by

$-I+I_z+I_l=0$

$I=10mA+\frac{V_z}{R_l}$

$I=10mA+\frac{9.1}{0}$

$I=10mA$

Therefore

$R=\frac{15V-V_z}{I}$

$R=\frac{15-9.1}{10*10^{-3}}$

$R=589 ohms$