I post some schematics, some are mine, mostly aren't.... I do understand some basics on electronics but as year's pass somethings are getting forgotten so here's an exercise to remember....

Zeners are tiny little funny devices.... they sit quiet in their room until someone drops them a voltage over their rated value, basically like diodes but on a bigger scale (voltage wise), and then they tend to conduct some electrons.... I never thought it was so difficult to explain what components do..... let's say a zener is the equivalent of a series connection of diodes but connected the other way around, polarity wise.

So let's see a tipical circuit with a zener:

There's a power supply, Vs a resistor R, a zener (that weird diode) and the famous (or not) NE602.

So let's see from an engineering point of view.... the potencial diferencial sum (basicaly the sum of voltages) in a circuit should be "0" (zero)... if not you are creating an energy monster :) but this doesn't matter for the zener subject...just for puting arrows in the circuit case some old teacher of mine sees this...

On we go...

Let's supose we need to stabilize the voltage in the NE602 at 6.2V (just because it eases some math and circuit simplicity). It's also nice to know that the NE602 needs about 2.5mA (0.0025A) of current to work and we have a power supply of

13.8V (most of the time) and our zener can witstand 41mA (normaly more but this is a tested current value from the datasheet and we whant things cool)

Now, we can do this in two ways....

The worst case scenario and the low power scenario...

* The worst case scenario:

Let's say we can have a problem in the power supply and voltage can raise to 24V

We would have to drop some 17.8V in the resistor to maintain 6.2V in the NE602 so our resistor should be (U/I) 17.8 / (Iz + Ine602) = 17.8 / (41mA + 2.5mA) = 409 Ohm...

Power dissipated in the resistor is: 0.77W

Power dissipated in the zener is: 0.25W

So what about when we are runing the nominal 13.8V:

We would have to drop 7.6V in the resistor (13.8V less the zener 6.2V), if our resistance is 409 Ohm then the current is: U/R = 7.6/409 = 18.5mA that is 2.5mA for the NE602 and 16mA in the zener diode.

Power dissipated in the resistor is: 0.14W

Power dissipated in the zener is: 0.099W

* The low power scenario:

Let's save the planet and make low CO2 emissions:

We have a good 13.8 power supply the NE602 consuption is still 2.5mA and we will give just 5mA for the zener (he must do something and still have some margin for power supply flutuation)

Our resistor is now: U/I = (13.8-6.2) / 7.5mA = 7.6 / 7.5mA = 1013 Ohm

Power dissipated in the resistor is: 0.057W

Power dissipated in the zener is: 0.031W

Much less power than on the worst case scenario...

...let's just see what max voltage the circuit can withstand...

We use the same 41mA max on the zener and 2.5mA on the NE602 so voltage on the resistor is R*I = 1013 * 43.5mA = 44V plus the 6.2 on the zener that's: 50.2V

And what if the voltage drops: let's acount only for the NE602 current so the voltage in the resistor is 1013*2.5mA = 2.53V plus the minimum for the NE602 that we consider is 6.2 then the voltage in the circuit should be no less than: 8.73V.

Conclusion:

The worst case scenario let's you drop more on the power supply voltage (409*2.5mA is less than 1013*2.5mA).

The low power scenario covers you more on over voltage and has less power consumption at normal operating points...you choose!

Have fun!

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