In a Zener diode,breakdown occurs in reverse bias due to

$A$ Zener diode in the circuit shown has a knee current of $5\,mA$ and a maximum allowed power dissipation of $300\,mW$. What are the minimum and maximum load currents that can be drawn safely from the circuit, keeping the output voltage $V_0$ constant at $6\,V$?

The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at $V_Z = 5 \text{ V}$ and the desired current in the load is $5 \text{ mA}$. The unregulated voltage source can supply up to $25 \text{ V}$. Considering the Zener diode can withstand four times the load current, the value of resistor $R_S$ (shown in the circuit) should be . . . . . . $\Omega$. (in $\text{ } \Omega$)

$A$ Zener diode,having a breakdown voltage equal to $15\,V$,is used in a voltage regulator circuit as shown in the figure. The current through the diode is......$mA$.

The value of the resistor,$R_S$,needed in the $dc$ voltage regulator circuit shown here,equals

If the voltage between the terminals $A$ and $B$ is $17\ V$ and the Zener breakdown voltage is $9\ V$,then the potential voltage across $R$ is.......$V$.