#### hm_fa_da

##### Full Member level 5

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 :

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