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How to calculate Zener diode power according to intrinsically safe (ATEX) standard ?

hm_fa_da

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Hi,

I'm little confused by calculation method of a Zener diode power according to IEC60079-11 (intrinsically safe devices) !
To simplify my question, I just sent a sample schematic here :

Untitled.jpg


I know in how to calculate the power of the zener in normal conditions, but in ATEX world, many things change !
and I also know that allowable wattage of a zener diode depends on it's thermal resistance and junction temperature, ambient temperature and considering safety factors ...
for example a 5W nominal zener diode maybe be only 2W after considering above calculations.

My question is how to calculate the power dissipated at the simple schematic I sent "according to ATEX".

Method 1:
Vin - Vz = 12 - 8 = 4V = Vr
Iz = Ir = 4/10 = 0.4A
Pz = Vz * Iz = 8 * 0.4 = 3.2W
considering 1.5 factor for ia, Pz will be 3.2W * 1.5 = 4.8W

Method 2:
Pz = Isc * Vz
Isc (Current at zener diode point in short circuit condition) = Vin / R = 12/10 = 1.2A
Pz = 1.2A * 8V = 9.6W
considering 1.5 safety factor for ia, Pz will be 9.6W * 1.5 = 14.4W

I know according to general electronics, Method 1 is right, but who are familiar with ATEX rules know that many things are different there and mostly the worst onerous conditions are considered.

I myself think according to the standard, Method 1 is right, but the problem is that the certification lab engineer which I'm working with says Method 2 is right !
(I know in the standard it's mentioned to consider short circuit current for semiconductors and Diodes (like Method 2) - but for Zener Diode it says to calculate power dissipation in zener mode which will be Method 1).

Thank you
 

BradtheRad

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Method 2:
...
Isc (Current at zener diode point in short circuit condition) = Vin / R = 12/10 = 1.2A
Pz = 1.2A * 8V = 9.6W

The first line assumes zener is short circuit condition and so it calculates 1.2A maximum current.
To be consistent the test should continue to assume the zener is zero Ohms, thus it has 0V and zero Watts.

Or, they might have applied a test imagining the zener knee is changed to a lesser voltage, dissipating lesser Watts.

However the test rules evidently give no consideration for a halfway situation. So it takes a strict approach and reverts to the only definite parameter available for the zener, namely 8V.

We can say Method 2 test is inconsistent in the way it calculates. Yet on the other hand it's clearly a hazard to let a damaged zener carry 1.2A. Suppose you apply some more science. What if you were to perform several calculations, based on the zener knee changing to lesser volt levels (say 1 through 7)? What is its power dissipation in each situation? Does it ever exceed your Method 1 result?
 
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hm_fa_da

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The first line assumes zener is short circuit condition and so it calculates 1.2A maximum current.
To be consistent the test should continue to assume the zener is zero Ohms, thus it has 0V and zero Watts.

Or, they might have applied a test imagining the zener knee is changed to a lesser voltage, dissipating lesser Watts.

However the test rules evidently give no consideration for a halfway situation. So it takes a strict approach and reverts to the only definite parameter available for the zener, namely 8V.

We can say Method 2 test is inconsistent in the way it calculates. Yet on the other hand it's clearly a hazard to let a damaged zener carry 1.2A. Suppose you apply some more science. What if you were to perform several calculations, based on the zener knee changing to lesser volt levels (say 1 through 7)? What is its power dissipation in each situation? Does it ever exceed your Method 1 result?
Thanks for your reply,
In fact it's not matter of only science calculations here, i.e according to ATEX rules we should use at least 2 or 3 parallel zener diodes for ib and ia levels in which the power calculations should be passed on each of them alone.
or according to the standard, for other Diodes (like schottky or rectifiers) we should calculate Isc (considering Vf = 0) and then multiply it by max Vf of the diode according to datasheet (i.e 0.5V for some schottky diodes) to calculate the maximum dissipation power ! in this case it's like method 2 in fact.
However fortunately I talked to "another Engineer" in the certification lab and he confirmed that according to the standard, for Zener Diodes, Method 1 is right (as I supposed).
 

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