Question

In a Zener-regulated power supply a Zener diode with ${V_z} = 6$ volt is used for regulation. The load current is to be $4mA$ and the unregulated input is $10$volt. What should be the value of the series resistor ${R_s}$, if the current through the diode is five-time the load current?

Answer 1

Hint:The total current ($I$) passing through the circuit is divided into two parts i.e. the current through Zener diode (${I_z}$) and the current through the load (${I_L}$ ).
$\therefore I = {I_z} + {I_L}$ .
Also, we have to find the voltage drop across the series resistance
Finally we get the series resistance by using the formula.

Formula used:
Total current, $I = {I_z} + {I_L}$ where ${I_z}$= the current through Zener diode, and ${I_L}$= the current through the load.
The voltage drop across the series resistance ${R_S}$, =Input voltage – Zener voltage.
Also, ${R_S} = \dfrac{{voltage - drop}}{I}$

Complete step by step answer:
In the given circuit system, an unregulated input voltage is connected in a side and a current is passing through the series resistance.

The total current ($I$) passing through the circuit is divided into two part i.e. the current through Zener diode (${I_z}$) and the current through the load (${I_L}$) is as follows,

Here we can write it as $\therefore I = {I_z} + {I_L}....\left( 1 \right)$
Given the current through Zener diode (${I_z}$) is 5 times the load current (${I_L}$) i.e. ${I_z} = 5{I_L}$.
It is given that the question , ${I_L}$= $4mA$ and ${I_z} = 5{I_L}....\left( 2 \right)$
Putting the value ${I_L}$ in equation $\left( 2 \right)$we get,
$ \Rightarrow {I_z} = 5 \times 4mA$
Let us multiply the term we get
$ \Rightarrow {I_z} = 20mA$
So, we have to find the total current on putting these values we get,
$I = {I_z} + {I_L}$
$ \Rightarrow I = 20 + 4$
On adding the terms we get,
$ \Rightarrow I = 24mA$
$ \Rightarrow I = 24 \times {10^{ - 3}}mA$
Now the voltage drop across the series resistance ${R_S}$,
Input voltage – Zener voltage=$10 - 6 = 4$ volt.
Therefore the series resistance,
${{\text{R}}_{\text{S}}}{\text{ = }}\dfrac{{{\text{voltage - drop}}}}{{\text{I}}}$
Putting the values and we get
$ \Rightarrow {R_S} = \dfrac{4}{{24}}$
$ \Rightarrow {R_S} = \dfrac{4}{{24 \times {{10}^{ - 3}}}}\Omega $
On simplifying we get
$ \Rightarrow {R_S} = 166.66\Omega $
$\therefore $The value of the series resistor should be approx $167\Omega $ .

Additional information:
-For the above circuit system, if the input unregulated voltage is changed increased or decreased, the current through the Zener diode is also changed in a direct manner (increased or decreased).
-Hence, the increased or decreased voltage drop across the series resistor does not affect the potential difference.
-This is because the break-down voltage is not increased due to an increase of the current through the Zener diode.
-This characteristic is used to make a circuit work as a voltage regulator.

Note: If the Rating of a Zener Diode is $({V_z} - {P_z})$, the maximum safe current flow should be ${\operatorname{I} _{\max }} = \dfrac{{{P_z}}}{{{V_z}}}$ , where ${P_Z}$ is called watt rating, and ${V_Z}$ is called voltage rating of a diode.
In the mentioned circuit system the series resistor is chosen such as a value for which the Zener Current can not be greater than the maximum current (${I_{\max }}$ ).