Zener question

[tnadys]

I have a piezo connected to a zener and am noticing that it seems to be scaling instead of clipping the signal. I prefer it to scale but this doesn't seem right, am I missing something? I am just using a piezo to try to learn about rectifiers, opamps, filters, comparators etc.

[Tac Eht Xilef]

I have a piezo connected to a zener ...

How? 

If you provide a schematic or circuit diagram we might be able to explain what's happening.

[Hideki]

Maybe you should add "How to explain something clearly in a forum post." to the list of things to learn? 

Don't worry. You're not the only one here with this particular problem. It's a widespread phenomenon. Don't assume that other people will be able to read your mind all the time. We don't know what's in there, so you actually have to explain what you're doing in more detail than in your first post.

[AG6QR]

I have a piezo connected to a zener ...

If you provide a schematic or circuit diagram we might be able to explain what's happening.

Also, describe what kind of piezo device you're working with.  Buzzer?  Speaker?  Microphone?  Transducer?   While you're at it, the part number of the zener, or at least its zener voltage, might be useful.  Finally, what kind of signal are you using, and how are you determining whether it's "clipping" or "scaling"?  Where and how are you injecting the signal, and where and how are you measuring it?

[tnadys]

the zener effectively closes the circuit with the scope being the load. It was part of a larger piece but right now I am just trying to understand the effect of the zenor. I am getting voltage regulation just not in the way I expected. Am I right to assume that the zener should clip and it is just some aspect of my setup that is causing this? Piezo's are really messy so I assumed that trying to get a predictable signal would be a good way to learn this stuff. I am one of those programmer geeks Dave talks about and I am having trouble coming to terms with the difference between ideal and real life.

[tnadys]

FYI. I am not asking you to debug my circuit I am asking for a clarification of the behavior of a zener diode. I mentioned the piezo for background only. The question cannot possibly be more simple. Does a zener clip or scale? Shouldn't require a circuit diagram.

[sleemanj]

. Does a zener clip or scale? Shouldn't require a circuit diagram.

A voltage applied such that the diode is forward biased, will act like any other diode, a small forward voltage drop and current will pass.

A voltage applied such that the diode is reverse biased, will act like any other diode when the voltage is below a certain threshold, it will block the current, above the threshold current will pass with a voltage drop of the zener voltage.
 

[AG6QR]

FYI. I am not asking you to debug my circuit I am asking for a clarification of the behavior of a zener diode. I mentioned the piezo for background only. The question cannot possibly be more simple. Does a zener clip or scale? Shouldn't require a circuit diagram.

The zener's behavior is described on its data sheet.  The current versus voltage curve(s) will be there.  What does the data sheet say?

The data sheet doesn't directly say whether it clips or scales, because it depends on what you're doing with it, and how you define "clip" versus "scale".

We don't know the context of your circuit, and you have ignored our pleas for information.  We certainly don't know more about your zener than the manufacturer who wrote the data sheet.  So, unless you want to provide some context for your question, I think the data sheet is where you'll have to find your answer.

[Tac Eht Xilef]

FYI. I am not asking you to debug my circuit I am asking for a clarification of the behavior of a zener diode. I mentioned the piezo for background only. The question cannot possibly be more simple. Does a zener clip or scale? Shouldn't require a circuit diagram.

Piezo strain gauge, speaker element, or buzzer (with integral driver)? Zener in series or parallel? Value of zener?

[Hideki]

A zener alone does not clip, nor does it scale. No, it works like a diode in both directions, with the reverse direction having a larger voltage drop than ~0.7V.
To get the clipping behavior you assume it has, you have to connect it as a voltage divider with the zener pointing up from ground and a resistor as the upper part of the divider.
When you increase the voltage, the zener will at some point abruptly start conducting and keep the output near the zener voltage, the rest of the input voltage will be found across the resistor.

I guess you have a round 'golden'-looking disk that has no driver circuitry that you're tapping with your finger to get an output (see, you never said what you did?). A rough approximation of a piezo is a voltage source (that spikes when you tap it) in series with a capacitor.

So in your case you have no resistor, there's only the capacitance of the piezo that briefly passes some current.

[tnadys]

A zener alone does not clip, nor does it scale. No, it works like a diode in both directions, with the reverse direction having a larger voltage drop than ~0.7V.
To get the clipping behavior you assume it has, you have to connect it as a voltage divider with the zener pointing up from ground and a resistor as the upper part of the divider.
When you increase the voltage, the zener will at some point abruptly start conducting and keep the output near the zener voltage, the rest of the input voltage will be found across the resistor.

I guess you have a round 'golden'-looking disk that has no driver circuitry that you're tapping with your finger to get an output (see, you never said what you did?). A rough approximation of a piezo is a voltage source (that spikes when you tap it) in series with a capacitor.

So in your case you have no resistor, there's only the capacitance of the piezo that briefly passes some current.

you are correct about tapping the golden disk . My question though, is more general because I thought I understood zeners until that stupid yellow disk. If you have a 0-10v sine wave and a 5v zener what will the wave look like on the other side. I assumed the top 5 volts would be cut off and you would see flat tops on the scope, but I'm not really seeing that. What I see looks more like a voltage divider. Could it have something to do with passing in and out of the breakdown voltage (very fast with a piezo)? FYI I originally rectified the signal first and had a load resistor but I pulled everything out to try to figure this out and it didn't seem to matter.

[tnadys]

FYI. I am not asking you to debug my circuit I am asking for a clarification of the behavior of a zener diode. I mentioned the piezo for background only. The question cannot possibly be more simple. Does a zener clip or scale? Shouldn't require a circuit diagram.

Piezo strain gauge, speaker element, or buzzer (with integral driver)? Zener in series or parallel? Value of zener?

Given my limited experience  . I have never seen a zener used in series, they always seem to be in parallel. Under what circumstances would you use a zener in series? Just curious.

[AG6QR]

If you have a 0-10v sine wave and a 5v zener what will the wave look like on the other side. I assumed the top 5 volts would be cut off and you would see flat tops on the scope, but I'm not really seeing that.

On the other side of what?  A zener diode is just a two terminal device.  What sides are you talking about?  What's the source impedance of your signal?  How have you connected the zener to the signal source?  What are you measuring across? Are you measuring voltage, current, or both?


If you have a low enough impedance 0-10V sine wave, you can connect anything across it, including a 5V zener, and you'll still see a 0-10V sine wave.  The current won't be sinusoidal, though, and you might burn up your zener, depending on its ratings.

OTOH, if you have a high impedance signal (perhaps created by connecting a low impedance signal in series with a suitable resistor) you can use a 5V zener, oriented properly, to clip the voltage range to about 5V, according to the voltage/current curve in the data sheet.

[AG6QR]

Given my limited experience  . I have never seen a zener used in series, they always seem to be in parallel. Under what circumstances would you use a zener in series? Just curious.

Check out http://en.wikipedia.org/wiki/Zener_diode

All of the example circuits use a zener in series with something, either a resistor or another zener.

[tnadys]

Given my limited experience  . I have never seen a zener used in series, they always seem to be in parallel. Under what circumstances would you use a zener in series? Just curious.

Check out http://en.wikipedia.org/wiki/Zener_diode

All of the example circuits use a zener in series with something, either a resistor or another zener.

ah yeah, I actually read about the Waveform clipper an hour ago. Must be getting old.

[David_AVD]

Just for interest:

I've used a zener diode to drop a voltage down by a (relatively) fixed amount.  In my case the load was not constant (20-40mA) so using a resistor instead would not have worked.

As long as you do the calculation for power dissipation and ensure the zener can handle the current, it can be a cheap alternative to a voltage regulator.

[tnadys]

If you have a 0-10v sine wave and a 5v zener what will the wave look like on the other side. I assumed the top 5 volts would be cut off and you would see flat tops on the scope, but I'm not really seeing that.

On the other side of what?  A zener diode is just a two terminal device.  What sides are you talking about?  What's the source impedance of your signal?  How have you connected the zener to the signal source?  What are you measuring across? Are you measuring voltage, current, or both?


If you have a low enough impedance 0-10V sine wave, you can connect anything across it, including a 5V zener, and you'll still see a 0-10V sine wave.  The current won't be sinusoidal, though, and you might burn up your zener, depending on its ratings.

OTOH, if you have a high impedance signal (perhaps created by connecting a low impedance signal in series with a suitable resistor) you can use a 5V zener, oriented properly, to clip the voltage range to about 5V, according to the voltage/current curve in the data sheet.

I don't understand this? Why would I still see the 0-10V wave? The zener is 30V which is borderline, but they are cheap and I have a bunch of them to play with. The impedance is mostly from the scope now that I have pulled the resistor. I could add it back but it didn't seem to matter and I was trying to simplify as much as possible. Maybe I went to far. The voltage is in the correct range, the wave form just seems wrong. I expected cut off peaks, but everything looks the same, just with a lower range.

[AG6QR]

If you have a low enough impedance 0-10V sine wave, you can connect anything across it, including a 5V zener, and you'll still see a 0-10V sine wave.  The current won't be sinusoidal, though, and you might burn up your zener, depending on its ratings.

I don't understand this? Why would I still see the 0-10V wave?

That's the definition of a low impedance voltage source.  Whatever you connect to it, it maintains constant voltage, and sources or sinks current as necessary to maintain that voltage. 

Of course, in the real world, all voltage sources have nonzero impedance, and finite ability to source or sink current.  And real-world wiring or parts will fail, sometimes spectacularly, if current limits are exceeded.

[TerminalJack505]

I'm guessing that since you are using just one diode it looks like you're getting only half of the peak-to-peak output that you're expecting.  Put two zeners back-to-back and see if you get something more like what you're expecting.

Having just one zener means that it will clamp at about 700mV in one direction and the zener's breakdown in the other.

You also need to factor-in the piezo element's capacitive reactance.  You might plug the equivalent circuit of a piezo into a simulator and put a zener parallel to it and see what happens.  You will notice that the output is different than if you just put an ideal AC voltage source across a zener diode.  (Which would probably fry the zener, by the way.  So don't do that unless you limit the current.)

[tnadys]

If you have a low enough impedance 0-10V sine wave, you can connect anything across it, including a 5V zener, and you'll still see a 0-10V sine wave.  The current won't be sinusoidal, though, and you might burn up your zener, depending on its ratings.

I don't understand this? Why would I still see the 0-10V wave?

That's the definition of a low impedance voltage source.  Whatever you connect to it, it maintains constant voltage, and sources or sinks current as necessary to maintain that voltage. 

Of course, in the real world, all voltage sources have nonzero impedance, and finite ability to source or sink current.  And real-world wiring or parts will fail, sometimes spectacularly, if current limits are exceeded.

Sorry, reading comprehension error, I though you were talking load. So Basically you overload the zener which reduces its effect. I'm not sure that is the situation in my case because it seems to be working as advertised. I just have an issue with the shape of the curve and I don't want to move on until I understand why. My results are completely consistent so I guess I'll just keep plugging away until I can find the pattern. Maybe I will use a better signal. In the end a scope is not that much different than a debugger, which is cool. To bad you can't step into parts

[tnadys]

I'm guessing that since you are using just one diode it looks like you're getting only half of the peak-to-peak output that you're expecting.  Put two zeners back-to-back and see if you get something more like what you're expecting.

Having just one zener means that it will clamp at about 700mV in one direction and the zener's breakdown in the other.

You also need to factor-in the piezo element's capacitive reactance.  You might plug the equivalent circuit of a piezo into a simulator and put a zener parallel to it and see what happens.  You will notice that the output is different than if you just put an ideal AC voltage source across a zener diode.  (Which would probably fry the zener, by the way.  So don't do that unless you limit the current.)

No, the problem is that the shape of the + side of the signal isn't what I expected. I first noticed it when I added the zener to a half wave rectifier. I keep saying clip, but clamp is maybe a better term for this crowd. I expected the zener to clamp/clip/remove the top of the wave (anything over the 5V breakdown) which would leave obvious marks on the curve. What I am getting is a curve that looks similar but is within the 5V range (down from 30V). One thing I am thinking is that because of the signal is swinging in and out of the breakdown voltage the curve is getting distorted. I thought I read somewhere there were issues with this but I'm not really sure.

I'm pretty new to this so I apologize if this is stupid but isn't the piezo element's capacitive reactance the point of them? I am using it precisely because they are good for studying rectifiers, zeners and filters. It was an easy source of a dirty signal. maybe to easy in hindsight.

[David_AVD]

You really need to show a schematic of what you're talking about here.  Everyone is guessing based on your words.

[Tac Eht Xilef]

And, if possible, a picture of the waveform that is worrying/confusing you so much...

[tnadys]

You really need to show a schematic of what you're talking about here.  Everyone is guessing based on your words.

[TerminalJack505]

I'm guessing that since you are using just one diode it looks like you're getting only half of the peak-to-peak output that you're expecting.  Put two zeners back-to-back and see if you get something more like what you're expecting.

Having just one zener means that it will clamp at about 700mV in one direction and the zener's breakdown in the other.

You also need to factor-in the piezo element's capacitive reactance.  You might plug the equivalent circuit of a piezo into a simulator and put a zener parallel to it and see what happens.  You will notice that the output is different than if you just put an ideal AC voltage source across a zener diode.  (Which would probably fry the zener, by the way.  So don't do that unless you limit the current.)

No, the problem is that the shape of the + side of the signal isn't what I expected. I first noticed it when I added the zener to a half wave rectifier. I keep saying clip, but clamp is maybe a better term for this crowd. I expected the zener to clamp/clip/remove the top of the wave (anything over the 5V breakdown) which would leave obvious marks on the curve. What I am getting is a curve that looks similar but is within the 5V range (down from 30V). One thing I am thinking is that because of the signal is swinging in and out of the breakdown voltage the curve is getting distorted. I thought I read somewhere there were issues with this but I'm not really sure.

I'm pretty new to this so I apologize if this is stupid but isn't the piezo element's capacitive reactance the point of them? I am using it precisely because they are good for studying rectifiers, zeners and filters. It was an easy source of a dirty signal. maybe to easy in hindsight.

The capacitive reactance will contribute to the signal's attenuation when you put a non-trivial load across it. 

I don't know if that alone can account for what you're seeing, however.  If you can find a datasheet for your piezo element then you can get a better idea if that's the case or not.

[David_AVD]

A bigger picture would be better next time.  lol

So that diagram shows a piezo, zener and resistor all in parallel.  Now you need to explain what you're feeding it with and what you're observing.

EDIT: For symmetrical clamping, that you really want is two zeners in series, with their anodes together.

[tnadys]

I'm guessing that since you are using just one diode it looks like you're getting only half of the peak-to-peak output that you're expecting.  Put two zeners back-to-back and see if you get something more like what you're expecting.

Having just one zener means that it will clamp at about 700mV in one direction and the zener's breakdown in the other.

You also need to factor-in the piezo element's capacitive reactance.  You might plug the equivalent circuit of a piezo into a simulator and put a zener parallel to it and see what happens.  You will notice that the output is different than if you just put an ideal AC voltage source across a zener diode.  (Which would probably fry the zener, by the way.  So don't do that unless you limit the current.)

No, the problem is that the shape of the + side of the signal isn't what I expected. I first noticed it when I added the zener to a half wave rectifier. I keep saying clip, but clamp is maybe a better term for this crowd. I expected the zener to clamp/clip/remove the top of the wave (anything over the 5V breakdown) which would leave obvious marks on the curve. What I am getting is a curve that looks similar but is within the 5V range (down from 30V). One thing I am thinking is that because of the signal is swinging in and out of the breakdown voltage the curve is getting distorted. I thought I read somewhere there were issues with this but I'm not really sure.

I'm pretty new to this so I apologize if this is stupid but isn't the piezo element's capacitive reactance the point of them? I am using it precisely because they are good for studying rectifiers, zeners and filters. It was an easy source of a dirty signal. maybe to easy in hindsight.

The capacitive reactance will contribute to the signal's attenuation when you put a non-trivial load across it. 

I don't know if that alone can account for what you're seeing, however.  If you can find a datasheet for your piezo element then you can get a better idea if that's the case or not.

I have read a lot of papers and data sheets on these things and the only thing I learned was that a complete understanding was beyond my current abilities. The general consensus from the non-PHD level papers is that they are a source of relatively high ac voltage that bears some non-obvious relationship to how hard they are hit. They are like crystal oscillators that you can hit  .  They are usefull for studying other things though, because they are so sloppy.

[tnadys]

A bigger picture would be better next time.  lol

So that diagram shows a piezo, zener and resistor all in parallel.  Now you need to explain what you're feeding it with and what you're observing.

EDIT: For symmetrical clamping, that you really want is two zeners in series, with their anodes together.

it's a piezo? it's the source that is being observed, or rather the zener effect on the piezo is being observed. Taking out the resistor has very little effect on the signal (when connected to a scope) so basically it is a piezo that is being regulated down to 5V by a zener.

[tnadys]

A bigger picture would be better next time.  lol

Sorry, I rushed, because I am frustrated that two components that can only be connected in one way even needs a diagram. Either way, I am about ready to give up. I asked a simple question about the zener effect on an AC signal and no-one seems to know the answer. I'll probably learn more figuring it out myself anyways.

[sleemanj]

Sorry, I rushed, because I am frustrated that two components that can only be connected in one way even needs a diagram.

If fully specifying your problems makes you angry and frustrated, electronics engineering isn't for you.

[David_AVD]

The real issue at play here is that you haven't been able to clearly communicate to others what you're trying to achieve, what you've tried and how you measured/observed the issue(s).

I mean the above in the nicest possible way.   

[tnadys]

Sorry, I rushed, because I am frustrated that two components that can only be connected in one way even needs a diagram.

If fully specifying your problems makes you angry and frustrated, electronics engineering isn't for you.

Been a programmer for a long time so detailed specs aren't new to me, but two pages of restating how two components that can only be connected one way and ending with a diagram is a little much. In fact reversing the diode doesn't even change the answer just the sign. I can understand that maybe the answer is complicated but the question wasn't. The frustration comes from trying to come up with new ways of saying the same thing. I honestly started to wonder if you guys are just messing with me because a forum for EE's shouldn't need a circuit diagram for this question. Then again maybe i'm just really horrible at asking questions and assume to much, either way it has been a big waste of everyone's time

[tnadys]

The real issue at play here is that you haven't been able to clearly communicate to others what you're trying to achieve, what you've tried and how you measured/observed the issue(s).

I mean the above in the nicest possible way.   

Quite possible. I've been a programmer so long that I think of English as my second language.

[AG6QR]

Been a programmer for a long time so detailed specs aren't new to me, but two pages of restating how two components that can only be connected one way and ending with a diagram is a little much.

But look through the thread.  Where did you say it was two components?  Certainly not in your first post, when you could have specified the problem.  Nor in the first response to the posts that first asked for clarification.  And when you finally gave us a diagram, it shows three components, not two.  So are you now telling us your schematic is wrong?

[Tac Eht Xilef]

... but two pages of restating how two components that can only be connected one way and ending with a diagram is a little much.

You were saying?

That is, in fact, the usual way of connecting a zener to a voltage source. Your way isn't uncommon for certain purposes but, absent of any other explanation, is a little unusual...

If you're getting so very frustrated by questions asked in good faith so that you can help us to help you, I'd suggest walking away and coming back to this thread tomorrow.