Modifying input impedance of Boss OC2 to use as piezo buffer

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PRR

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> a piëzo as roughly a current source in series with an internal capacitance

More useful to imagine a voltage source in series with a cap.

Or a current source with a cap across it.

A current source is an infinite impedance, so it does not matter what you put in series.

The underlying principle of a piezo seems to be deflection causes voltage, so a voltage source appears to be the more natural model for the mechano-electric action.

Of course VS+C and IS||C are duals, equivalent, so on the electric side can be modeled either way.

Charge-amps have been used in condenser microphones. They are also a favorite tool for photo-diode work.
 
This is interesting, I wonder what the downside is as I have never seen this in a PU amp stage yet?
Since I first found out about this approach, I too have wondered why it wasn't widely used for audio. (Though it seems to be used in various other fields.)

I don't have an answer, though. I never really found a downside, at least in this era of cheap JFET input opamps.

I have always regarded the behavior of a piëzo as roughly a current source in series with an internal capacitance (would you agree?), so the arrangement would make sense.
I found a couple of ways to make sense of it.

One way: most of the old electronics textbooks show a piezo equivalent circuit consisting of a voltage source in series with a small capacitance. Usually source resistance is assumed to be zero or near zero.

If you short this out, the current through the capacitance will be the usual i = C dv/dt.

The derivative tells you the current will be proportional to frequency, i.e., rising at +6 dB/octave. Feeding this into an integrator gets you a flat frequency response.

I was not satisfied as to where that equivalent circuit came from, so I looked at physics textbooks instead of electronics textbooks. The physical reality is that when you bend / deform a piezo element, it generates an unbalanced electric charge across its two electrodes. The amount of charge depends on the amount of deformation. There is also a small capacitance between the piezo electrodes.

The physics textbooks stop there. But consider the capacitor equation Q = CV, from which V = Q/C.

So if you deform the piezo and place a charge Q on its own internal parallel capacitance, you get a voltage which is proportional to the amount of charge, i.e., to the amount of deformation of the piezo.

So the traditional recipe of feeding a piezo into a very large load resistance amounts to making sure that the charge generated by the piezo is not drained off by whatever the piezo is connected to. Basically making sure the RC product is greater than the duration of one cycle of audio at the lowest frequency of interest.

On the other hand, if you short the piezo, you get an amount of charge dQ flowing through it, and this happens during a time dt which is the duration of one audio cycle (period T). So the current dQ/dt is equal to dQ/T.

Since T = 1/f (where f is the frequency), that means the current through the piezo is f dQ. dQ itself is proportional to how much deformation the piezo experienced, so basically, we have a shorted current which is proportional to the frequency, and to the amount of deformation of the piezo.

Once again, if you run that current into the input of an integrator, with its 1/f frequency response, you end up with a flat frequency response...

You can scale the integration capacitance to set the amount of output voltage, and put a resistor in parallel with the integration capacitor to set the lower - 3dB frequency.

I discovered this approach about twenty years ago, because the piezo elements I was trying to use for motional feedback in a subwoofer did not have much capacitance, and they needed to respond accurately down to 10 Hz to keep the servo feedback stable.

This meant that the "feed it into a large enough resistance" approach resulted in very large input impedance, and chronic hum and noise problems. Feeding the piezo into a dead short (integrator input) worked much better for me.

-Gnobuddy
 

PRR

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Note that a piezo is also a bunch of resonances and a "speaker".

So open/short termination should give different internal responses (damped, undamped).

I'm inclined to think undamped is more "exciting" and damped is more "accurate".

There may be another side to this. As you say it is a Charge effect and that is not yet intuitive to me.

And as you say, 19 cent opamps can be wired both ways for cheap trials.
 
For my purposes (wide band acceleration sensor), I needed to ensure no piezo mechanical resonances within the bandwidth of interest (10 Hz to 5 kHz or so).

I used a generic disc (bimorph) piezo, which was fairly floppy on its own, with a first (bending) resonance around 2 kHz, if I'm remembering correctly after all these years. To stiffen it up, I machined a mount for it that supported it rigidly all the way around the edge. Mounted like this, the first mechanical resonance was up around 5 kHz.

Piezo material being dense and stiff, it would take sizeable electromagnetic or electrostatic forces to affect its motion. Very little electric power is generated by bending. So I suspect (but am not sure) that there is little or no mechanical damping from the induced electricity.

-Gnobuddy
 
Gnobuddy & PRR,
agreed with all you said.

Anyway... I still have difficulties to put this all into at least some kind of equivalent circuit.

My reason for calling it a current source is that a piezo is actually non-conductive, and I made this non-scientifical thought experiment:
All that a piezo can do is shift charges from the one side to the other. If you could bend it infinitely, you would have a DC source. I you could bend it infinitely fast, you had infinite current. Vice versa, Voltage applied to the piezo results in deformation.
So, in my mind, the element should tend to hold against mechanical force as long as the charge can not be balanced, meaning no current flows. Applying more force will make the unbalance (or voltage) higher as long as current is allowed to flow. Isn't that the reason why you get a spark in a piezo lighter?

This ability to build up nearly deliberate voltage combined with a high resistance reminds me of the definition of a current source.

Or am I completely mistaken?
 
My way of looking at it: a current source can supply a DC current indefinitely into a load. Since charge is current multiplied by duration, it can also supply an indefinite amount of charge (given enough time). When the load drains off some charge, the current source does not run out - it keeps supplying more charge.

A deformed piezo is different in a couple of ways: one, the amount of charge caused by the deformation is finite - if you drain it off, no more charge appears. Two, you cannot get DC current to flow out of it at all (as you pointed out, it is non-conductive). All you can get is a brief pulse of current. Three, if you vibrate it (AC charge) and short it, the amount of current rises proportional to frequency.

So the piezo doesn't behave much like a current source - but its DC behaviour is exactly like a capacitor that's been charged. And it's AC behaviour is also exactly like a capacitor with a fluctuating AC charge on it.

I think that mental picture is closest to the physical reality. A piezo acts like a capacitor, which can be charged by physically deforming it.

I think today's piezo ceramic discs are kissing cousins to ceramic disc capacitors even when it comes to the basic construction - a ceramic dielectric, metallized on both sides. The only major difference is that the dielectric is deliberately chosen to be piezo-electric in one case. (And sometimes ceramic caps exhibit unwanted, accidental piezo-like behaviour, too.)

Does that make sense?

-Gnobuddy
 
so...
I've put some capacitance across the pup and sure enough when I plug into my oc2 not only does it not distort but it also gives me all of my low frequencies into the 1Mohm Z

Lowest frequency on the violin is 198hz so I've used a total capacitance of around 1500pf (including the pickup and cable capacitance). Basically makes my fiddle a similarish signal level and output impedance as an electric guitar, ideal!

Talk about thread drift, ended up having very little to do with modifying oc2s in the end. Cheers guys
 
Basically makes my fiddle a similarish signal level and output impedance as an electric guitar, ideal!
Excellent! The perfect happy outcome!

Talk about thread drift, ended up having very little to do with modifying oc2s in the end.
True, but it did give you the answer to the real question you were asking, the meta-question one layer above "How do I modify my Boss OC2?"

That meta-question was "How do I get full bandwidth and no noticeable distortion out of my violin, while running it through my Boss OC2?" :)

-Gnobuddy
 

PRR

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Joined 2003
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Lowest frequency on the violin is 198hz so I've used a total capacitance of around 1500pf (including the pickup and cable capacitance). Basically makes my fiddle a similarish signal level and output impedance as an electric guitar, ideal!

So you have 1/4th the too-hot level and -3dB at 111Hz on a 198Hz instrument. In the lab, it is down about 1dB on the lowest note. The real violin (and speaker and room and player) are +/- many dB, and the player strives to get the desired note-balance. So as long as he/she is not fighting overload or steep bass cut, it should work out.
 
So you have 1/4th the too-hot level and -3dB at 111Hz on a 198Hz instrument. In the lab, it is down about 1dB on the lowest note. The real violin (and speaker and room and player) are +/- many dB, and the player strives to get the desired note-balance. So as long as he/she is not fighting overload or steep bass cut, it should work out.

I thought the rolloff would be -6dB per octave but as I was nearly an octave above the -3db I'd be pretty much flat. How do you work out, mathematically, where "pretty much flat" is?

The helmholtz chamber resonance of a violin is around 300hz and soundboard resonance tuned around A440hz so yes, there is a roll off of lower frequencies on an acoustic instrument. My thought was maybe our ears are accustomed to that frequency response and I might want to achieve a similar response by fine tuning the capacitor values. We'll see where intuition leads...

What I definitely want to lose is a bit of top end as there are some unnaturally loud sounding higher harmonics. If I use a small coil in series will I achieve this or is there another passive solution?
 
What I definitely want to lose is a bit of top end as there are some unnaturally loud sounding higher harmonics. If I use a small coil in series will I achieve this or is there another passive solution?
A resistor between the direct output of the piezo and the additional voltage reduction capacitor you've added should do the trick. Together, they will form a low-pass filter.

I think the resistance value you want will be determined by the corner frequency of your choice, and the combined series capacitance of your (420 pF) piezo and 1500 pF external cap. Call it 330 pF.

So a 47k resistor will start to roll off everything above 10 kHz. A 100k resistor will start to roll off around 5 kHz. A 220k resistor, around 2.2 kHz.

I suspect you'll find a resistor value somewhere in that range (47k - 220k) will do what you want. Maybe try it with a 250k or 500k pot, adjust to suit your ears, then measure and replace with a fixed resistor.

Thanks for the violin-related numbers! Fascinating stuff. (I know something about guitars, but next to nothing about violins.)

-Gnobuddy
 
Thanks for that Gno, I'll give it a go. Question - why 330pF? have you made a mistake there or am I missing something?
Series capacitors combine like parallel resistors - the combined capacitance is less than either individual one. They obey the same formula as for parallel resistors.

So in this case, Ctot = (420 x 1500)/(420+1500) = 328.125 pF

Since we don't care about 1% accuracy, Ctot = 330 pF, close enough for our purposes.

My 1500pF is the total capacitance of my piezo and added cap, is this the capacitance I should be using for my calculation?
Ah. So the added capacitance is only 1080 pF, not 1500 pF? Okay, in that case there will be a minor correction:

Ctot = (420 x 1080)/(420+1080) = 302.4 pF

Call it 300 pF, rather than 330. Only a 10% change, not enough to matter in this case (10% frequency change is less than 1 decibel change from a first-order low pass filter, and we can't really hear less than 1 dB).

LTSpice seems to agree - see attached image. The piezo is modelled as an AC voltage source in series with a 420 pF capacitor. The load is 1080 pF. Inserting a 100k series resistor will cause treble to be 3 dB down a little above 5 kHz.

I do not know the violin audio spectrum well enough to know if that will be too little, just enough, or too much filtering. But you can find out easily enough what sounds right to you, by changing the value of that 100k resistor.

-Gnobuddy
 

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Do I need to take the capacitance of my jack cable into consideration as well then? It's about 250pf <snip>
That one is in parallel with the 1080 pF "padding" cap you added, effectively increasing it to 1330 pF.

As before, the change from 1080 to 1330 pF will make a rather small small audible effect in what comes out of the low-pass filter - less than 2 dB.

I'll do it with a pot then I can tweak!
Yeah, I think that is the only way to find out what actually sounds best to you - it's going to depend on too many factors to calculate, it has to be subjective in the end.

-Gnobuddy
 
I think these capacitors are in parallel, the capacitance of the piezo is across the voltage source?
This is a case of Thevenin vs Norton equivalent circuits for the piezo.

Both are mathematically correct, but as discussed earlier, the Thevenin circuit is a flat-frequency voltage source in series with a capacitor, while the Norton circuit is a rising-with-frequency current source in parallel with a capacitor.

I find it easier to think in Thevenin terms, so that's what I put into LTSpice. (Come to think of it, I don't even know how to generate a rising-with-frequency current source in LTSpice.)

But if I got it right, it doesn't matter which mathematical model you prefer to think in terms of - they will both behave exactly the same way as far as oscilloscopes and signal generators and spectrum analyzers are concerned! :)

The downside to the "series resistor to roll off treble" approach is the one you have hinted at - source resistance is no longer negligible at high frequencies, so cable capacitance and input capacitance of your Boss pedal starts to matter. Also there may be slightly higher noise and hum pick up, though if you are using a fairly short length of good quality shielded cable between violin and Boss, I don't think you'll have a problem.

The best way to find out is by trying it. If you don't like the result, it's easy enough to remove that series resistor.

(If you end up not liking the simple solution, you could instead build an active lowpass filter into a little diecast aluminium FX pedal box, and connect that between the output of your Boss pedal and whatever comes next in the audio chain. But it's a much more complex solution than just one single resistor.)

-Gnobuddy
 
Swapping the sommer cable with a 6m fairly standard guitar cable of unknown capacitance I can hear the HF is attenuated. The 3.5m sommer cable sounds totally fine to me with 5Mohm Zin, infact if anything I'd like to remove a bit more HF.

so...
I've put some capacitance across the pup and sure enough when I plug into my oc2 not only does it not distort but it also gives me all of my low frequencies into the 1Mohm Z

Lowest frequency on the violin is 198hz so I've used a total capacitance of around 1500pf (including the pickup and cable capacitance). Basically makes my fiddle a similarish signal level and output impedance as an electric guitar, ideal!

I wonder how these observations of your's do match, as in fact it doesn't matter where to put the additional capacitance - maybe next to the pickup, maybe in parallel with the 1st active stage (the OC2 in your case), maybe in the cable - or in every possible combination of these.

OTOH and generally spoken, I am totally biased against any long cable between a high impedance piezoelectric audio source and an amplifier's input. Fortunately there's a way to place an active buffer next to the AF source without the urge of also placing the battery next to it. Let's see how they do it in two-pin terminated electrete microphone capsules: They arrange a high input impedance open drain FET stage next to the transducer (which also are capacitive sources, same as your piezo violin PU) and it's load resistor into the amplifier/transmitter (or similar). Thus the cable's hot lead carries both the AF signal and the buffer's supply current - and is of comparable low Z.

What about this idea?

Best regards!

Btw, I'v also been educated in playing the violin in my youth and was a great admirer of Jerry Goodman...
 
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