# Zenerdiodes, and Voltage stabilizing

Discussion in 'General Electronics Chat' started by malmoit, Sep 22, 2008.

1. ### malmoitNew Member

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Hello,

would be thankful if someone could help me with the following:

In the problem there is only a text and it says (translated from Swedish):

"Calculate the components in a Zener-diode stabilizer, where the stabilizing voltage is 15V. The current through the load varies between 0 and 100mA. To one's disposal there is an unstabilized voltage, which varies between 20 and 22V."

My problem in the text above, is that I'm not sure I intepret it correctly. My interpretation can be viewed in the attached image. The answer to the problem also confuses me, i.e.:

Answer: "If P is set to 3W, we get 35 < Rs < 41 Ohm"

Would be very thankful for some help here

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2. ### ericgibbsWell-Known MemberMost Helpful Member

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hi,

http://www.woodsbas.demon.co.uk/calcs/diodes.htm
http://www.reuk.co.uk/Zener-Diode-Voltage-Regulator.htm

3. ### ecerfoglioNew Member

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In your drawing you put 0 < Iz < 100mA. That's not correct, the current that varies from 0 to 100 mA is IL (I Load).

You must calculate the R (and the zenner) at the following points:

Max ILoad = 100 mA plus the minimum zenner current (from the zenner datasheet - Let's use 10 mA for the following formulas), with Min Vsupply = 20 V. This gives the resistor's value:

R = (min Vs - Vz) / (Max IL + min Iz) = (20V - 15 V) / (100mA + 10mA) = 5V / 110mA = 45.45Ω

Min ILoad (0), Max Vs (22 V) ==> Max Iz and zenner's power:

Max Iz = 22 V / 45.45Ω = 484 mA

Pz = Vz (Max Iz) = 484 mA x 15 V = 7.26 W ==> You need at least a 10 W Zenner

PR = (Max IR)² x R = (484 mA)² x 45.45Ω = 10.6 W ==> You need at least a 15 (or better 20) W resistor.

At this power levels, perhaps a Zenner supply ("stabilizer") is not the best solution, because you need a rather big zenner diode (10W) and resistor (20W). They even keep using power (and heating up) even when the load draws no current.

It is better to amplify the zenner with a pass transistor - or even to use asingle IC solution like the 7015.

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5. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Worst case for the load is 100ma and 20v input, and Rs=50
ohms works just right for this combo.
Worst case for the zener is when the load is at 0ma and
the voltage is at 22v, and this causes only 2.1 watts
of dissipation in the zener.

On the other hand, if we also spec the zener wattage at
3 watts max, this would occur with a voltage of 22v and
load at 0ma and Rs=35 ohms. With this value we also
have no problem maintaining the load at 100ma with
20v input.

Because of these two cases, i would say the minimum
value for Rs is 35 ohms and the max value is 50 ohms.

Because this result does not agree with the stated
'answer', i would think we are missing some info
or something is completely wrong with the given
Also, the way to approach this kind of problem is by
spec'ing a min and max load, min and max voltage,
and min zener current (for proper zenering current
which makes sure the zener voltage stays within
a certain percentage accuracy).
With the given stated 'answer', it sounds like they might
have also been asking that the min zener current is
22ma in order to maintain stability within some accuracy.

Last edited: Sep 22, 2008
6. ### malmoitNew Member

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Hi guys,

first of all: Thanks a whole bunch for your help and efforts.

Well the only thing I can see missing, is the rule of thumb mentioned in my book:
Pz > 2*Uz*ILmax

Thanks again!
/Pierre

7. ### malmoitNew Member

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Ok, now I solved it, I think:

- New prerequisite and rule of thumb Iz,min = 0,1*Iz,max

Now, looking at the problem there are 4 different possibilities, or "extreme" situations. These are V, IL = {Vmin, ILmin}, {Vmin, ILmax}, {Vmax, ILmin} and finally {Vmax, ILmax}

As read on the web, and logically understood, Rs has to fullfill two conditions. 1) Rs should be designed small enough that the current through the Zenner-diode, Iz keps the diode in reverse breakdown.
2) Rs must be designed big enough so that the Zenner diode doesn't get destroyed, i.e. Iz <= Iz,max

Applied to our problem, this would mean

1) {Vmin, ILmax} = {20V, 100mA} with Rs <= (Vmin - Vz)/(Iz,min + ILmax) ~ 41 Ohm (actually 41,67 but the authors aren't that consistent regarding rounding, which I've also noticed before.)

2) With {Vmax, ILmin} and Vmax*I = Vmax*Iz,max <= Rs*(Iz,max)^2 + Pz ==> Rs = (Vmax*Iz,max - Pz)/(Iz,max)^2 >= 35 Ohm

All together: 35 <= Rs <= 41 Ohm

Last edited: Sep 23, 2008
8. ### malmoitNew Member

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One more comment: As I am new to this discipline, i.e. electric circuit theory, I had no idea if the two rules of thumb were common practice, and hence implicit to the problem solving.
Well, now I know that this isn't the case, which also explains the suggestions differing somewhat from the one presented above.

Once again: Thank you all for helping me out!

Last edited: Sep 23, 2008
9. ### Ubergeek63Well-Known Member

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1N829 with a preregulator

10. ### MrAlWell-Known MemberMost Helpful Member

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Hi E.C.,

I was wondering how you got such high numbers for the power rating,
and after looking over your post i found this:
Max Iz = 22 V / 45.45Ω = 484 mA
which isnt really correct. The Max Iz is:
Max Iz=(22v-15v)/45.45 ohms
using your value of 45.45 ohms that is.
In other words, the max current in the zener (and the series resistance)
is equal to the max voltage minus the zener voltage, divided by the
series resistance.

11. ### ecerfoglioNew Member

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Oooops. Of course you are right

It's a lot of years since I designed zenner supplies. The 78xx / LM317 are just so easy to use

12. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

I have to say it's been a very long time since i used a zener for a power supply
too (he he) as there are just too many good and cheap ic's for doing this
these days as im sure you were thinking about too. I've used them for clamps

Another thing that i realized is that the original problem seems to be implying
that the user should use a 3 watt zener (because of the given answer).
I would never use a 3 watt zener in a place where it might actually have
to dissipate 3 watts because of temperature considerations. I would probably
use a 5 watt device if it really came down to it, and that would require a
higher minimum zenering current too (last time i checked).
I think it's a little ironic that this problem provides a solution circuit that
would most likely burn up in real life unless it was used outside in

Last edited: Sep 24, 2008

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Or Sweden

Robert-Jan