# Simple Zener Voltage Regulator

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#### ckck20

##### Member level 1 Hi, it's a simple problem for most of you guys. The problem is as follows:
We have Vin=23-25V , Vz=12.4, Izmin=10mA, Izmax=250mA and the load current IL=20-400mA. (Suppose that instead the RL we have a current source).
For simplicity we don't care about RL or about the zener resistance.
What we want to find is the limits (min,max) for the resistance R(the min is my problem).

My thought:
The Rmax should allow with Vinmin the max IL.
Therefore Rmax=(Vinmin-Vz)/(Izmin+ILmax)=25.8 Ohms
Similarly with Rmin
Rmin=(Vinmax-Vz)/(Izmax+ILmin)=46.7 Ohms
Obviously something is wrong. Something that you hopefully point out.

#### aryajur Calculating the range of Rmax is like calculating the extreme values of Rmax, for which the regulation will still work, even though for just 1 operating point.
By this meaning Rmax and Rmin should be calculated as follows:

Rmax = (Vinmax-Vz)/(Izmin+Ilmin) = 420 ohms
Rmin = (Vinmin - Vz)/(Izmax+Ilmax) = 16.3 ohms

#### ckck20

##### Member level 1 I see what you mean: Rmax=max/min and Rmin=min/max.
But then the Iz wouldn't go outside it's limits? For example for Vin=25V, Il=400mA, R=31 Ohms(in the range) then the Iz=6mA<Izmin and if we increase more the resistance R then Iz would become negative.
Don't we want for the given range of Vin and IL to find the range of R
so that Izmin<=Iz<=Izmax?

#### Borber As I see the situation, calculation of Rmax or Rmin has no sense because zener diode max current is les than max load current. When load current is max you calculate R for min diode current, but when load current is min your zener diode will burn (400mA>250mA).

### ckck20

Points: 2

#### aryajur Like I said, this range is the extreme. If you set R at a endpoint then u are assuming that it will work only for one point of operation and that won't be regulation. The range calculated signifies that R will never be any value outside this range in any case.
Now to evaluate R to make a regulator would be a different thing. Say you want Il to vary from 20mA to 260mA (i.e. ur load resistance may vary from 48 ohms to 620 ohms) in a worst case input voltage i.e. Vin = 23 V. Now you can get your R as:

R = (23-12.4)/270mA = 39.2 ohms
This value of R will give you load regulation if Rl varies from 48 to 620 ohms. If however Vin goes to 25 volts then Il can vary from 311mA to 71mA for regulation, that means then ur regulation will be there for a load resistance variation of 40ohms to 175 ohms.
So in this case when R is 39.2 ohms ur regulation will work perfectly for a Rl variation from 48 ohms to 175 ohms.
I hope this makes things clear. Note R is in the range. Your Iz can go outside its limit always since the margin in Il is so huge.
Had your load current range been say 20mA to 260mA then things would work without Iz going out of range and then you would be able to calculate Rmax and Rmin using your method. I hope this helped.

### ckck20

Points: 2

#### ckck20

##### Member level 1 Thank you guys, i really appreciate your help.

##### Member level 1 If you want to find R min, then RL must be opened,
the current throgh Vz must never exceed Iz max =250mA,
and the input voltage is maximum = 25 V

So,
R min = (Vin(max) - Vz) / Iz(max)
= (25 - 12.4) / 0.25 Ohm
= 90.4 Ohm

If you get a smaller R min (e.g. 46.7 Ohm), since the RL is open
circuit, and Vin(max) happens, then the current throgh zenner
diode will be

Iz = (Vin(max) - Vz) / R min
= (25 - 12.4) / 46.7
= 269.8 mA
It is dangerious for the zenner diode.

#### hill

##### Full Member level 5 Hi,
You can always boost the current rating of your design by using a transistor!

#### VVV First of all, the problem does not have a solution, for the input data you considered.

The load current for the Zener regulator cannot be higher than the maximum permissible Zener current. That is because when there is no load, the entire current must be drawn by the Zener. In your case, the maximum Zener current is 250mA, but the maximum load current is 400mA! That does not work.
This is a quick first-check. As soon as I saw the requirement, I knew it was wrong.

At the maximum input voltage, with the load drawing its minimum current (it should be considered zero, but we'll go for 20mA), you get:
Rmin=(Vinmax-Vz)/(Izmax+Ilmin)=46.66Ω.

Now, at minimum input voltage you need to still maintain a minimum Zener current to ensure regulation. In your case, Izmin=10mA.
The maximum available current, when using Rmin calculated above will be:
Iavail=(Vinmin-Vz)/Rmin=227mA
Since your Zener requires 10mA to maintain regulation, the maximum current available for the load is: Ilmax=Iavail-Izmin=217mA. If the load draws more than 217mA, you lose regulation at the lower limit of the input voltage.

So, assuming that the maximum load current had been specified at 200mA, then you would have obtained:
Rmax=(Vinmin-Vz)/(Izmin+Ilmax)=50.47Ω.
The closer the maximum load current gets to the calculated 217mA limit, the closer Rmax will get to Rmin, limiting your choices.

Note that if the load current varies only slightly, you can obtain a valid solution, even for maximum load currents that exceed the maximum Zener current, but this is not generally the case in the real world. Usually, the load current varies widely, or should be considered to vary within large limits.

So, your equations were written correctly. The problem was formulated incorrectly.

I hope this helps.
VVV

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