[Amps] Zener Screen Regulators

[Dr. David Kirkby]

On 09/ 6/12 04:08 PM, Bill, W6WRT wrote:
> ORIGINAL MESSAGE:
> vOn Thu, 06 Sep 2012 11:55:28 +0100, David wrote:
>
>> To add a spanner into the works regarding the direction of current, I don't know
>> why these things are call zener diodes, since all the high voltage ones work on
>> avalanche breakdown.
>>
>> Dave, G8WRB
>
> REPLY:
> Zener diodes are different from regular diodes because they are designed to pass
> current in the reverse direction and survive. All silicon diodes have a zener
> point where they begin reverse conduction, but unless they are specifically
> designed for it, most of them will immediately self-destruct.
>
> More info:  http://en.wikipedia.org/wiki/Zener_diode
>
> 73, Bill W6WRT

But the regulation of high voltage so-called "zener" diodes is NOT due to the 
zener tunneling effect that occurs in low voltage zener diodes.

A 3 V zener diode will regulate the voltage by the zener tunneling effect. So 
for low-voltages devices, the name is correct.

For a 50 V "zener" diode, there will be virtually no zener tunneling, but 
avalanche breakdown will occur will occur at 50 V.

As explained in the Wikipedia article  you linked to, both effects are present. 
One has a positive temperature coefficient and the other a negative temperature 
coefficient. At around 5.6 V, the temperature coefficient of a "zener" diode is 
very low, as the two temperature coefficients cancel.

But as I said, at 50 V, there will be negligible tunneling, so no zener effect. 
Only avalanche breakdown.

Avalanche breakdown is used to good effect in a number of diode devices, in 
addition to the poorly name "zener diode".  Two that spring to mind, which I 
have both used are:

* Avalanche photo-diode.
* Noise diodes.

I suspect there are others too.

The physics behind zener and avalanche effects are very different.

To my knowledge, a modern power rectifier will not suffer any damage if you let 
it break down in the reverse direction as long as the current is limited. 
Obviously if you have a 1 kV 1 A diode, and allow it to break down in the 
reverse direction, then at 1 A it would dissipate 1 kW and die in microseconds 
or less. But I believe if the current is sufficiently limited, the breakdown 
will not harm the diode. But of course you would not use normal diodes for 
regulation, as the voltage they avalanche at will not be known.


G8WRB

-- 
Dr. David Kirkby Ph.D C.Eng MIET