<P> The value of R must satisfy two conditions: </P> <Ol> <Li> R must be small enough that the current through D keeps D in reverse breakdown . The value of this current is given in the data sheet for D. For example, the common BZX79C5V6 device, a 5.6 V 0.5 W Zener diode, has a recommended reverse current of 5 mA . If insufficient current exists through D, then U is unregulated and less than the nominal breakdown voltage (this differs to voltage - regulator tubes where the output voltage will be higher than nominal and could rise as high as U). When calculating R, allowance must be made for any current through the external load, not shown in this diagram, connected across U . </Li> <Li> R must be large enough that the current through D does not destroy the device . If the current through D is I, its breakdown voltage V and its maximum power dissipation P correlate as such: I D V B <P max (\ displaystyle I_ (D) V_ (B) <P_ (\ text (max))). </Li> </Ol> <Li> R must be small enough that the current through D keeps D in reverse breakdown . The value of this current is given in the data sheet for D. For example, the common BZX79C5V6 device, a 5.6 V 0.5 W Zener diode, has a recommended reverse current of 5 mA . If insufficient current exists through D, then U is unregulated and less than the nominal breakdown voltage (this differs to voltage - regulator tubes where the output voltage will be higher than nominal and could rise as high as U). When calculating R, allowance must be made for any current through the external load, not shown in this diagram, connected across U . </Li> <Li> R must be large enough that the current through D does not destroy the device . If the current through D is I, its breakdown voltage V and its maximum power dissipation P correlate as such: I D V B <P max (\ displaystyle I_ (D) V_ (B) <P_ (\ text (max))). </Li>

To determine the reverse breakdown voltage of zener diode