Im having trouble to understand how some of the maximum power ratings of Zener regulators are derived, I understand that Zener voltage will be dependant of current Iz, but take this for example: http://www.onsemi.com/pub_link/Collateral/1N5913B-D.PDF
If you look at datasheet, you'll notice that those diodes have a 3 Watt rating, here's what I dont get: look at diode 1N5925B on that same datasheet, its a 10V zener, and Izm is rated a 150mA. The higher the current the higher Vz will be, however with a max current of 150mA in order to achieve the 3W rating, that would mean that Vz would have to be 20V, which is twice the rated Vz of 10V. Vz will drift a bit with a change in current, but it wont drift that much!
According to my calculations a 10V zener with an Izm of 150mA will have a max power rating of around 1.5W so where exactly is the 3W rating coming from?
Best Answer
I've sampled a few rows of the table for Zeners of different voltages. It seems to be consistent that the product of the nominal voltage, and the \$I_{ZM}\$ is about 1.5W, half of the 3W.
Now, that 3W figure is something that has to be derated for ambient temperature. So there are conditions at which the maximum dissipation will in fact just be 1.5W. The derating is 24mW above 75 Celsius, so if the ambient temperature is 137 degrees, it is down to 1.5W.
So, the maximum current values can be regarded as simply being conservative. You can have that much maximum current even if the ambient temperature is quite high. Note that the table does not say anywhere that you must derate \$I_{ZM}\$ for temperature!
It looks like to achieve close to the 3W dissipation, the ambient temperature has to be 75 or less (since the 3W derates above that), and you have to violate \$I_{ZM}\$ (since those values are quoted for 1.5W dissipation).