zener shunt stabiliser

[Nigel Goodwin]

Current goes up (diode turns on hard) - voltage goes down. I'm paraphrasing, but that's what I understand this statement to mean. This is negative resistance. Zeners don't have negative resistance. In a zener, dv/di is always positive, on any portion of the V-I curve.

The voltage across the resistor is an accurate measure of the current, by ohms law (as I'm sure you will agree).

As the zener draws current from the resistor the voltage across the resistor will increase (as will the current through it). By the same reasoning (as they are in series) the voltage across the zener will decrease as it's current increases.

As soon as the voltage across the zener decreases, the current it draws will decrease as well (as it turns off), and the voltage across the zener will increase again.

This action will stabilise at a specific voltage and current, dependent on the zener voltage, the value of the resistor, the supply voltage, and the current drawn off the top of the zener.

 

[Roff]

Nigel wrote:

the voltage across the zener will decrease as it's current increases

This is incorrect. It implies negative resistance. Look at the VI curve of a zener. As the current increases, the voltage increases.

As soon as the voltage across the zener decreases, the current it draws will decrease as well

This is correct. The two quotes are contradictory.

 

[Nigel Goodwin]

Nigel wrote:

the voltage across the zener will decrease as it's current increases

This is incorrect. It implies negative resistance. Look at the VI curve of a zener. As the current increases, the voltage increases.

I'm not talking about VI curves, I'm talking about how it works.

 

[Roff]

Nigel wrote:

the voltage across the zener will decrease as it's current increases

This is incorrect. It implies negative resistance. Look at the VI curve of a zener. As the current increases, the voltage increases.

I'm not talking about VI curves, I'm talking about how it works.

I guess that means you don't see any inconsistency between your two quotes that I listed in my previous post. Amazing.
And VI curves are a graphical representation of "how it works".

 

[Nigel Goodwin]

I guess that means you don't see any inconsistency between your two quotes that I listed in my previous post. Amazing.

That's because there is none, you still seem to be thinking of it as a resistance, it's nothing like a resistance.

Would you agree with the following two statements? (ignoring the fact that it's not perfect, the conduction actually following a 'knee' curve).

1) A zener diode conducts at a specific voltage, the specified zener voltage of the component.

2) If the voltage across the zener is below the zener voltage, it doesn't conduct.

And VI curves are a graphical representation of "how it works".

No, they are a graphical representation of "how it performs", a totally different thing!.

 

[Roff]

Are you thinking of a zener which has a load in parallel with it? In this case, when the load current increases, the zener current (and voltage) decrease. That's not what you said, and the original circuit had no load, but that's the only thing I can think of that makes sense.
Please think about this statement again:

Nigel wrote:

the voltage across the zener will decrease as it's current increases.

Really think about it. If you come up with the same answer, I challenge you to make this happen in the lab. Copy the OP's circuit, and vary either the input voltage or the resistance. Tell us what happens to the voltage across the zener when the current through it increases.

 

[Russlk]

I agree with Ron, the zener resistance is positive. Some times very small, and usually noisy. A 6.8 volt zener makes a good RF noise generator.

 

[Nigel Goodwin]

Really think about it. If you come up with the same answer, I challenge you to make this happen in the lab. Copy the OP's circuit, and vary either the input voltage or the resistance. Tell us what happens to the voltage across the zener when the current through it increases.

You ignored the question, do you agree with the two statements I made?, if so I'll go further from there.

And I'll repeat again, I'm not discussing how it performs, but how it works, a totally different thing!.

 

[Roff]

Really think about it. If you come up with the same answer, I challenge you to make this happen in the lab. Copy the OP's circuit, and vary either the input voltage or the resistance. Tell us what happens to the voltage across the zener when the current through it increases.

You ignored the question, do you agree with the two statements I made?, if so I'll go further from there.

And I'll repeat again, I'm not discussing how it performs, but how it works, a totally different thing!.

OK, I'll address your questions:
Nigel wrote:

1) A zener diode conducts at a specific voltage, the specified zener voltage of the component.

The voltage is specified at a current. If the current goes up, the voltage goes up. If the current goes down, the voltage goes down. The change in voltage divided by the change in current is the dynamic resistance, and it varies with current, and with zener type, but is always positive. See the dreaded (typical) VI curve below for a graphical view of this. This**broken link removed** shows zener impedance for various zener voltages at various currents. The relevant specs are Zzt and Zzk.

2) If the voltage across the zener is below the zener voltage, it doesn't conduct.

There will always be a measurable current except at zero volts. This current is generally insignificant until the voltage approaches the zener voltage. The point at which the current becomes significant depends on the application and the type of zener.
Nigel, my purpose here isn't to crucify you or back you into a corner. You provide great answers to questions that I know nothing about. This forum is primarily for helping hobbyists, and my concern is here is they don't walk away with an erroneous view of how zeners work - pardon me - perform (a totally different thing). :wink:

 

[Nigel Goodwin]

[quote="Ron H]
Nigel, my purpose here isn't to crucify you or back you into a corner. [/quote]

Hi Ron,

This isn't an argument, it's a discussion 8) and the advantage of an open forum is that anyone can contribute - and all can learn, or not, from different points of view.

You certainly hedged the answers to my two questions :lol: are you perhaps a lawyer? - or is that a serious insult?.

I'll try and explain what I'm getting at, which (as I've said) is how I perceive the circuit to work - which isn't explained in any of the graphs.

Going back to the original circuit with one resistor, one capacitor, and one zener.

As you apply power to the top of the resistor, the capacitor starts to charge, so the voltage on the top of the zener diode increases (the capacitor doesn't need to be there, but it probably makes this idea more understandable as it will happen slower).

At a certain point the zener diode will start to conduct significantly, so current through the zener increases.

As the only source of current is the feed resistor, drawing more current through the resistor will increase the voltage drop across the resistor, thus decreasing the voltage across the zener diode.

As the voltage across the zener diode has decreased, the current through it will now decrease as well - as the current through it decreases, the voltage drop across the resistor will decrease as well - causing the voltage across the zener to increase again.

This continues on, round and round, causing the voltage on the top of the zener diode to apparently stabilise - in actual fact it's continually correcting itself - which is why it's often used as a white noise source.

Does that longer explanation make any more sense?.

Would using a Vbe multiplier as a variable zener help to explain it any better?.

 

[Russlk]

Ah! Nigel, I finally understand what you are saying, but the noisy zener does not represent a negative resistance. I don't know what the mechanism is that causes the noise but it is probably more like the frying sound that comes from an arc.

 

[ljcox]

hey nigel
thats exactly what i was trying to explain ... and then the capacitor is simply used to "soften" this noise of the quick switching?

but then ljcox said something about the capacitor being charged by the zener ...which totally threw me off!

Thanks alot
Kane

I did not say it was "charged by the Zener". I said

"When the supply voltage is above the Zener voltage, the Zener is on and current flows through it. The capacitor is charged to the Zener voltage"

RonH is correct, the current is 390 uA.

Len

 

[Roff]

Nigel, I think we could discuss this endlessly and never agree. Your original explanation implied negative dynamic resistance to me. Your most recent posting sounded to me like you envision a feedback mechanism in a zener. Neither is an explanation that I can accept. Your way of looking at zeners obviously satisfies you, and I think I have adequately made my point. I think we have to agree to disagree.
At the bottom of **broken link removed** is a discussion of the mechanism for zener noise. I didn't see any feedback mechanism mentioned or implied.

 

[Nigel Goodwin]

Nigel, I think we could discuss this endlessly and never agree.

I think we'll have to agree to disagree :lol:

 

[pike]

noting both explanations, what is positve and negative resistance??? I have never heard of such a term used ??

 

[Roff]

noting both explanations, what is positve and negative resistance??? I have never heard of such a term used ??

Resistance, in the most general sense, is change in voltage divided by change in current, expressed mathematically as dv/di. In any circuit where DC current flows, this can be measured.
I cobbled together a simulation, and am posting the schematic and resulting voltage vs current curve below. In the graph, Vin of the schematic was swept from zero to two volts, and the current from Vin is plotted on the vertical axis. The regions with positive slope (0 to 0.9v and above 1.2v) have positive resistance. The region in between has negative resistance. The dynamic resistance at any point is the slope of the curve (dv/di) at that point.
For comparison, if we had plotted the curve of a 1kohm resistor, we would see a straight line from the origin (0 volts, 0 current) to the point (2v, 2ma).

 

[ljcox]

RonH's explaination is very good. But I would like to clarify one point.

Resistance is the constant of proportionality between voltage and current.

V = R I. Where R could be called the "static" resistance.

If R was negative, then a positive voltage would cause a negative current.

The derative of voltage with respect to current dV/dI is (as RonH said) the dynamic resistance, sometimes called the slope resistance. Devices such as Tunnel diodes, Unijunction transistors, SCRs etc. all have regions of negative dynamic resistance.

With a normal resistor its "static" and dynamic restances are equal because if you differentiate V = R I then dV/dI = R.

Len

 

[pike]

ahh..I remember seeing this circuit in a PIC programmer. It limits the amount of current and (in my programmer, it allows 100ma) stops the PIC from frying if you put it in backwards.

Thanks guys

 

[Roff]

ahh..I remember seeing this circuit in a PIC programmer. It limits the amount of current and (in my programmer, it allows 100ma) stops the PIC from frying if you put it in backwards.

Thanks guys

I doubt that. I conjured this circuit up just to illustrate the concept of negative resistance. I can't imagine how it would be used for reverse polaity protection. You might have seen something similar. Can you post a schematic?

 

[laroche73]

negative resistance

Anyone remember tunnel diodes? They were a little before my time. This guy has some interesting experiments on his page using homemade negative resistance materials:

**broken link removed**