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### Network # Zener diode shunt regulator

#### Ledwardz

Dec 21, 2010
43
hey again,.....

ive attached the question and heres my working, can anyone tell me if im on the right lines?

i then know zener voltage is equal to load voltage
which means the voltage through Rs =
(20 - 14.2)
= 5.8V

i then said that if the maximim current through the zener is 60ma then there will be no current through RL so Rs must be equal to 5.8 / 0.06 = 96.66 ohms? i dunno its a wild guess tbh.

actually thats wrong because

load current + zener current flows through Rs

i think knee current is the amount the zener can handle when the voltage is reversed, not to sure on this tho n if it is how does this help in any way?

thanks again.

#### Resqueline

Jul 31, 2009
2,848
Ok, here's a load of questions in return for you Where did you get 14.2V from?
And you mention a voltage through a resistor, is that a sensible statement?
If you Google zener knee current what do you come up with?
And what is the normal operation mode of a zener, forward or reverse?
Approximately how much voltage will you have across a zener if operated in the opposite of normal direction?
And approximately how much current is it likely to handle then before it gets as hot as when it's operated in normal mode?

#### Ledwardz

Dec 21, 2010
43
Where did you get 14.2V from?
And you mention a voltage through a resistor, is that a sensible statement?
If you Google zener knee current what do you come up with?
And what is the normal operation mode of a zener, forward or reverse?
Approximately how much voltage will you have across a zener if operated in the opposite of normal direction?
And approximately how much current is it likely to handle then before it gets as hot as when it's operated in normal mode?

okay erm....

1.) my notes sheet says that Vl = Vd and maximum v through zener = 7.1
so i presume that Voltage drop over Rs = (vcc - vd -vl)
= 20-7.1-7.1
= 5.8v

I then know mazimum zener current = 60ma hence Rs = 96.66 ohms.

allows me to work out power dissipation Vrs * I= 348 mW

2.)voltage drop across resistor i knew someone would mention this but had already logged out.

3.)reverse current break down point but that is from wiki.....

4.)normal operation = forward biased but can operate in reverse when in normal operation

5.) reverse biase cut in is 1.8 v

6.) if 3 is right 1ma?

#### Resqueline

Jul 31, 2009
2,848
Ok, I'm guessing here; Vl = V load, Vd = V diode. Your notes are correct in stating that they are the same value. What else could they be? But why do you add them?
Are they in series or in parallell? You may want to correct the wording in those notes by the way; voltage is never through anything, it's across - or over something.
It seems to me that you are confusing/mixing how to treat voltages with how to treat currents in this first presumption/calculation; it is not correct, and not only verbally.

2) Yes, that sounds better.

3) Yes, that's correct, but what does it actually point to? What part of the curve is it, what lies above, and what lies below, in terms of usefulness?

4) You contradict yourself in the same sentence.. I thought Wiki stated something about this. Normal zener operation is a controlled reverse (avalanche) breakdown state.

5) Maybe it gets that high at high currents, but the forward voltage drop of zeners usually starts at 0.7V or so (a little higher than normal diodes which are at 0.5V or so).

6) 1mA * 1.8V = 1.8mW. I don't see how this can get as hot as 60mA * 7.1V = 426mW.. Zener knee current is part of normal (reverse bias) operation, not forward bias.

So to sum it up; rethink (that is reread & use what I believe is called Thevenins theorem) and recalculate voltage drops and currents.
You also need to look more closely at zener curves and the definitions of their different parts, understanding what is their significance.

Last edited:

#### Ledwardz

Dec 21, 2010
43
ok so usin thevenins i get voltage drop to be (20 -7.1)= 12.9 v
ur right parallel dunno wot the hell i was thinking

this divided by max current (60ma) = 215 ohms for Rs

i read somewhere that you assume Iz (min)
= 10 * I knee
=10ma

providin tht is right
IL max = 60-10 = 50ma
so min RL = 7.1 / 50ma = 142 ohms?

if thts wrong ill ask someone to show me an example.

#### lexroxas

Apr 15, 2011
37
This problem was posted Dec 29, 2010 and too old already.

1. 20V - 7.1V = 12.9V
2. 12.9V divided by 60mA = 215 ohms ---> this is the value of Rs
3. 60mA minus 1mA = 59mA
4. 7.1V divided by 59mA = 120.34 ohms ---> this is minimum value of RL

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
lexroxas gives the answer the examiner probably wanted, while Ledwardz gave the answer an engineer would possibly prefer The difference is that between the theoretical and the practical.

#### lexroxas

Apr 15, 2011
37
Steve is correct. Ledwardz suggested that it's a good idea to multiply the knee current by a factor of 10. Voltage breakdown starts at the knee of the curve so it's important that the zener diode operates above the knee because it's linear above the knee current. The problem with using the factor of 10 is that it's subjective. One can argue using a factor of 8 or a factor of 15. The original problem was a textbook problem and will most probably require textbook approach when solving the unknown values. Finally, Ledwardz may submit his answer to his professor using a factor of 10 and see what will be his professor's comment.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
I agree 99% with lexroaxas.

A zener's knee is always soft to some extent. The further away from the knee that you can operate, the better if relative current through the zener will vary (in fact the more it will vary, the further from the knee you want to be)

Choosing a factor of 10 (or 8, or 15) is just rule of thumb stuff. If you want a voltage regulated by a zener that has certain limits to regulation then you need to look at the voltage vs current graphs of the zener and determine exactly at which point you're going to operate the circuit so that the defined variations in load current produce changes in the regulated voltage within limits.

However, if Ledwardz explains his reasoning clearly when submitting his results (and we hope he has already done so) then I would hope that any good professor would award him at least as many bonus marks for this as he might remove for getting an answer that was not exactly as expected. If he did get marked down then it says something about any institution that would mark down answers showing a deeper understanding than is obtained from models of perfect devices.

#### lexroxas

Apr 15, 2011
37
Very very clear indeed. Thanks Steve!

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