Answer to Question #124907 in Electric Circuits for LLLL

Question

Design and draw a zener voltage regulator to provide 12V output if load resistance varies between 100 ohm and 200 ohm. Assume that the input unregulated dc voltage is 18V and minimum zener operating current is 10 mA. Calculate the wattage of the Zener diode, value and wattage of RS

Answer 1

Explanations & Calculations

• Refer to the sketch.

• Safety resistor is to be always fed with (18 V-12 V =) 6 V, due to the Zener diode in operation.
• So the current through it is a constant, part of which is conducted through the load as it's requirement while the remaining part is conducted through the zener diode.
• Therefore, Zener is there to supply load with the constant voltage needed while facilitating the conduction of the excess current ($\bold{i_s -i_{r1}}$).
• Therefore, minimum zener current is possible when the load conducts the maximum @ the minimum resistance.
• Therefore,

$\qquad\qquad \begin{aligned} \small i_{R1-max} = \small \frac{12V}{100\Omega} = \bold{0.12A} \end{aligned}$ and $\begin{aligned} \small i_{R1-min} = \small \frac{12V}{200\Omega} = 0.06A \end{aligned}$

• Therefore,

$\qquad\qquad \begin{aligned} \small i_s = \small i_z + i_{r1} = 10mA + 120mA = \bold{130mA} \end{aligned}$

• Therefore, the value & the wattage of the R_s safety resistor is,

$\qquad\qquad \begin{aligned} \small R_s = \small \frac{V_s}{i_s} = \frac{6V}{130mA} =\bold{ 46.154\Omega} \end{aligned}$

$\qquad\qquad \begin{aligned} \small P_s = \small V_s\times i_s = 6V\times 0.13A = \bold{0.78 W} \end{aligned}$

• Maximum zener current is possible when the load conducts minimum. Therefore,

$\qquad\qquad \begin{aligned} \small i_{z-max} &= i_s - i_{R1-min}\\ &= \small 130mA - 60mA = \bold{70mA} \end{aligned}$

• Therefore, power is,

$\qquad\qquad \begin{aligned} \small P_z =\small V\times i_{z-max} = 12V\times 0.07A = \bold{0.84W} \end{aligned}$