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12V zener diode has the same series as 6V2. It also has same wattage of 500 watt but VZ of 12V. Will it handle the same maximum current IZT as the 6V2? Any proof?

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    \$\begingroup\$ If the datasheet isn't proof enough then your only option is to blow a couple of them up. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 28 '14 at 17:03
  • \$\begingroup\$ @IgnacioVazquez-Abrams I'd go for blowing a few up anyway just for the fun of it... \$\endgroup\$ – Majenko Sep 28 '14 at 17:13
  • \$\begingroup\$ I would expect the data sheet to show Izt decreases at some point. For example docs-europe.electrocomponents.com/webdocs/12e2/… shows Izt is the same for a range of Vz from 2.4 to 12, then it falls sharply. However, the manufacturer is specifying that the parts in the range 2.4V to 12V have the same Izt. It may be that the 6V2 part handles more current in practice, but the manufacturer isn't making that claim for any of those parts. \$\endgroup\$ – gbulmer Sep 28 '14 at 17:48
  • \$\begingroup\$ Please post a link to the datasheet for the part. \$\endgroup\$ – gbulmer Sep 28 '14 at 18:16
  • \$\begingroup\$ Can anyone provide me datasheet of both these diodes? \$\endgroup\$ – Ali Awan Sep 29 '14 at 15:19
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Do you have a part number for that 500 watt Zener? ;)

Izt (current, Zener, test) isn't the maximum current the Zener can handle, it's the current through the Zener at which the Zener voltage drop limits are guaranteed.

Like Vf versus If for an LED, except that in the case of an LED the drop is generated by forward current through the LED, while in a Zener it's the diode's reverse current that does it.

The maximum current a normally reverse-biased Zener can handle is dictated by its power dissipation rating, and that power will be the product of the reverse current through the Zener times the voltage dropped across it at that current.

For example, since P = IE, a 6.2 volt Zener rated to dissipate 1 watt maximum continuous can sustain a Zener current of:

    Iz(max) = P/Vz = 1 watt/6.2 volts = 161 milliamperes 

Notice that if Izt is 20mA, that's well within the diode's rating.

On the other hand, Iz(max) for a 12 volt, 1 watt Zener would be about 83 milliamperes, so a 20mA Izt would still be OK.

As the Zener voltage increases, however, an Izt of 20mA will cause the one watt spec to be violated at Vz >= 50V, so Izt must be lowered, as gbulmer noticed, for the higher voltage diodes.

For example, a 100 volt 1 watt Zener with an Izt of 20mA would dissipate 2 watts, so its Izt must be lowered to 10mA just to meet the spec, and lower than that if the current into the load the Zener is working with decreases, which will offload that current into the Zener since it's a shunt regulator.

Incidentally, here's a link to an excellent, relevant resource.

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  • \$\begingroup\$ so can is it correct? P=VI,Iz=500/6.2=80.64mA. similarly,Iz=500/12=41.67mA...(for wattage of diodes 500mw) \$\endgroup\$ – Ali Awan Sep 29 '14 at 16:48
  • \$\begingroup\$ @Ali: That should be true for Iz(max) for a 500 mW diode. \$\endgroup\$ – EM Fields Sep 29 '14 at 17:38

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