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?
Zener diodes comparison 12V vs 6V2
diodeszener
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It's just the basic operating principle of a diode. An ideal diode would allow current to flow in one direction but block current flow in the other direction. This is based on how it's made, with a p-type region, an n-type region, and a depletion zone in between. Like the bottom diode in this picture:
When you apply a some voltage, in your case 0.9V then the p-type holes and the n-type electrons move into the depletion region because they are repelled by their respective battery terminal. With enough voltage (0.9V in your case) the free electrons in the depletion region get moving and current begins to flow like this:
Now in the ideal case if you were to reverse that battery the opposite will happen and you'll get no current flowing:
In the real world though you can only apply so much reverse voltage or push before you hit the breakdown voltage and current begins to flow freely in the reverse direction. Zener diodes take advantage of this fact and are constructed to breakdown at lower voltages such as your 3.3V.
Sources:
You can read more about how zeners are made here
Or see the article I got all the pictures from here
I think you've got it mixed up. The body diode is a part of the structure of the MOSFET and does not help when switching an inductive load with a single MOSFET.
The zener in the symbol represents the avalanche rating of the MOSFET and is of use when switching an unclamped inductive load. You won't see it conduct unless you exceed the voltage rating \$V_{DS}\$ and get to \$V_{(BR)DSS}\$.
If you are actually getting +/-12V out of the pulse transformer, I don't see any obvious reason why that part would not work. Your negative swing out of the pulse transformer does have to be sufficiently large to turn the control MOSFET on, when loaded with both gate charges (-10V would be good).
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Best Answer
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:
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.