Ooops, that was my bad sketching the circuit quickly to put here.1) A zener diode should be reverse biased. Yours is forward biased.
No, it is a 6.8V zener diode.The Zener is not dropping 6.8V, so it's 5.2V x 133mA Iz Max = ~0.7W. That leaves a lot of overhead or derating for temp.
The resistor does not have 12V across it, it has 12V - 6.8V= 5.2V across it.Where I am confused is in determining the value of R with a load attached. Without the load, R should limit the Zener to less than Iz Max. One book gives a figure of using a value of 5mA for Iz and another 100mA Iz for the circuit under load. So how much current is left to use for load resistance and for the Resistor? Apparently calculating Iz Max from the 1W rating isn't the answer since the PDF gives a different value.
So if I use 133mA Max for R without a load, that gives me V/I = R = 12V/0.133A = ~90Ω. So, depending on what current I allow for the Zener that leaves me ~33-128mA for the load? Am I going on the right path here?
The Zener has a ±5% tolerance, so the worst-case current should be 1W/(6.8*1.05) = 140mA.Which tells me Max Iz is 133mA
No. You are looking at it backwards.Apparently, I have some thinking and reassessing to do... I need to start looking at Vz as a voltage divider with R1.
No.R1 = (Vs - Vz) / Imax That's simple enough. And the Imax value comes from the PDF instead of the Watt rating. K, got it, thanks.
It depends upon the Zener, as shown in the data sheet (Izt).I read that the Zener should have 5-100mA of current under load. 5 sounds a bit light
by Jake Hertz
by Jake Hertz