Can my mate solder 8 piezo contacts to one jack cable?

[DanOwens]

[i]Hi guys
I'm new, but have always been a bit of a lurker. Finally have to ask a questions as I'm totally stumped.

I need a jack cable with a LOT of female inputs into one cable (we're talking at least 8). I want to be able to shove a load of contact mics into one jack cable. Going through a mixer isn't an option, and volume control over each of the contact mics isn't important.

How on earth can I do this? If I need to hack something together with my soldering iron thats cool but if there was something out there already that does what I need it'd be way better (my soldering skills aren't exactly top notch).

Thanks
Liam.[/i]

Thanks chaps,

Dan

[Silent Fly]

[i]If[/i] I understand correctly, Liam wants a device/cable that takes 8 inputs (high impedance, 1/4” unbalanced jacks), mix the signals and sends out in1+in2+...+in8.

He doesn’t need to change the signal levels so the output is simply the mix of the inputs.

If he needs to use all the inputs at the same time, the only way to do it efficiently and without signal and/or tone loss is use an active mixer.

Edited April 20, 2011 by Silent Fly

[icastle]

There's no device (or lead) on the market that I've seen that'll do this.

If volume control is (as Liam has already stated) unimportant then I guess the easiest option would be to use a small box with 9 sockets wired in parallel (8 inputs and one output).

It's rather crude though and there are a number of practicalities around impedance and size to consider.
If we knew what the ultimate goal is we might be able to suggest a better solution.

Edited April 20, 2011 by icastle

[DanOwens]

Silent Fly > He has given me very little info, but yes I guess he wants to sum a load of piezos to one jack cable and he's hoping it'd be as simple as soldering them in parallel.

icastle > I've sent him this link and I might send him my login so he may just come back and offer some more info.

Just out of curiosity, could you explain how the impedance will negatively effect this setup (I know it will, and that was my first thought, but I don't know why...).

Dan

[flyfisher]

Hmm. Interesting question . . . it would be helpful to know the application and where the 8 piezo mics are going to be placed.

I've never heard of connecting multiple mics together, except via a mixer of course, and my immediate concern would be how they might interact with each other.

If these are the devices I think they are, then they are basically the same as those piezo "sounders" often used in cheap products that need to make some sort of noise. Basically, if mechanical energy is applied to the device (e.g. sound vibrations) then you'll get a small voltage waveform out (i.e. behaves like a microphone), but if electrical energy is applied to the device (e.g. an audio signal) then the device will vibrate and emit a sound.

So, if you connect a number of devices together (in parallel) then the mechanical vibrations applied to one device will "energise" the other devices and cause them to vibrate. With eight devices connected in parallel, there will be all sorts of complex interactions between them as they are simultanously energised by the vibrations from the sound source and the electrical outputs from the other devices.

I'm not really sure what will happen (without a bit of experimentation) but I don't think I'd be holding out much hope for a quality audio signal being picked up.

I'm not sure of the impedance of these devices but I assume it's pretty high. If all the devices are the same impedance then eight in parallel will have an impedance of 12.5% of the unit impedance.

But, as mentioned above, I don't think this will be the real problem. Still, it would be a simple thing to try.

[Silent Fly]

[quote name='DanOwens' post='1205928' date='Apr 20 2011, 03:04 PM'](...)

Just out of curiosity, could you explain how the impedance will negatively effect this setup (I know it will, and that was my first thought, but I don't know why...).

Dan[/quote]

The simpler way of describe the problem is this. Let's assume that only on pickup generates signal (pu#1) the other pickups (pu#2 to up#8) are totaly silent. If a generator (pu#2-8 in this case) produces zero voltage it is equivalent to its internal resistance (i.e. output impedance).

In other words the circuit is:

pickup #1: produces signal and it has an output impedance of Z
pickup #2 to #8 are like 7 loads in parallel from the pu#1. The total load that pu#1 sees is Z/7.

If you apply the [url="http://en.wikipedia.org/wiki/Voltage_divider"]voltage divider[/url] formula the attenuation is:

Vout/Vin = 1/8

which is roughy -18dB. This before applying any real load to the pickups like for example an amp input.

Things are slightly more complicated because of the capacitive and inductive effects but you get the idea.

[MarkBassChat]

Dan,

I'm not sure what you mean by piezo-mics. Is it a set of piezo transducers e.g. for a drum kit, or is it for a guitar? Connecting piezo transducers in parallel is widely used by piezo bridge manufacturers. If you look at this: [url="http://www.shopping.com/Graph-Tech-Graph-Tech-Ghost-LB-63-Floyd-Bridge-with-Piezo-Pickups-Gold/info"]http://www.shopping.com/Graph-Tech-Graph-T...ckups-Gold/info[/url] you will see that 6 piezo transducers are connected in parallel and the output goes to a preamp. There are two conditions when you do it: the wires have to be very short (15 cm max), and the input impedance of the preamp has to be very high (e.g. 10 MOhms). Otherwise it will sound bad. There are no such problems like described above (one transducer vibrating cause another transducer vibrate), or voltage drops down. Voltage divider is when you have two components in series and here you have 6 components in parallel. Make a search on piezo buffers or Ghost piezo bridges. Some companies make buffers where every piezo element has a separated input (e.g. Ghost - for a MIDI preamp) but ,as you can see with other Ghost bridges, it is not required. Just two days ago I installed a piezo bridge in my colleague's guitar and it sounded great - almost like an acoustic guitar. But remember what is needed: short wires and high impedance buffer.

Mark

Edited April 20, 2011 by MarkBassChat

[Silent Fly]

[quote name='MarkBassChat' post='1206347' date='Apr 20 2011, 08:09 PM']Dan,

I'm not sure what you mean by piezo-mics. Is it a set of piezo transducers e.g. for a drum kit, or is it for a guitar? Connecting piezo transducers in parallel is widely used by piezo bridge manufacturers. If you look at this: [url="http://www.shopping.com/Graph-Tech-Graph-Tech-Ghost-LB-63-Floyd-Bridge-with-Piezo-Pickups-Gold/info"]http://www.shopping.com/Graph-Tech-Graph-T...ckups-Gold/info[/url] you will see that 6 piezo transducers are connected in parallel and the output goes to a preamp. There are two conditions when you do it: the wires have to be very short (15 cm max), and the input impedance of the preamp has to be very high (e.g. 10 MOhms). Otherwise it will sound bad. There are no such problems like described above (one transducer vibrating cause another transducer vibrate), or voltage drops down. Voltage divider is when you have two components in series and here you have 6 components in parallel. Make a search on piezo buffers or Ghost piezo bridges. Some companies make buffers where every piezo element has a separated input (e.g. Ghost - for a MIDI preamp) but ,as you can see with other Ghost bridges, it is not required. Just two days ago I installed a piezo bridge in my colleague's guitar and it sounded great - almost like an acoustic guitar. But remember what is needed: short wires and high impedance buffer.

Mark[/quote]
I am sure that your experience in connecting in parallel multiple piezo pickups has been positive – only you can tell it.

However, there is one thing on which I respectfully disagree: there is a voltage divider. It is true, as you said, that a voltage divider is when you have al least two passive components connected in series. [i]This is exactly what happens.[/i]

The 7 pickups that do not generate signal are connected in parallel but for the [url="http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem"]Thévenin's theorem[/url] they are equivalent to 7 impedances in parallel. The equivalent impedance is Z/7. If you look at [url="http://en.wikipedia.org/wiki/Voltage_divider"]Wikipedia[/url] figure 1, Z2=Z/7 and Z1=Z (where Z is the impedance of each transducer), Vin is the voltage generated by pickup #1. For the Thévenin's theorem and the voltage divider equation, Vo/Vi = 1/8 regardless the value of Z.

Having said that, if there is a voltage drop, it doesn't necessarily mean it is a bad thing. If the piezo transducers can generate enough voltage, the voltage drop can be within reasonable limits. It can also be compensated by increasing the gain on the preamp. Moreover, considering they are piezo transducers, the inductive effects are probably zero. The high output impedance also offers some form of isolation from capacitive effects.

I am convinced that an active mixer would sound better than a simple parallel connection but I must admit that I never tried an A/B comparison.

Edited April 20, 2011 by Silent Fly

[MarkBassChat]

I cannot argue with the Thévenin's theorem . Today I installed a piezo bridge in a bass guitar (4 transducers connected parallel). And the signal was so strong that I had to use a voltage divider on the output . Otherwise the bridge was overdriving the preamp. Of course an active buffer will sound better because it causes that very high output impedance of the transducer "is seen" by the following preamp stage as a low impedance (which e.g. allows to use longer wires).

Mark

[DanOwens]

So Liam is in awe of your genii. To the point where he's asked me to ask you (Max mainly but everyone please pitch):

How much for a simple mixer with 50 inputs? No level control, just summing 50 piezo sources.

Dan

[Silent Fly]

[quote name='DanOwens' post='1206459' date='Apr 20 2011, 09:39 PM']So Liam is in awe of your genii. To the point where he's asked me to ask you (Max mainly but everyone please pitch):

How much for a simple mixer with 50 inputs? No level control, just summing 50 piezo sources.

Dan[/quote]

"Simple" and "50 high impedance sources" are somehow not entirely compatible

I would recommend Liam to contact me directly to discuss the details. My email address is in my signature.

Just as a personal curiosity, what is the source(s) of the 50 piezo?

Edited April 20, 2011 by Silent Fly

[Owen]

[quote name='Silent Fly' post='1206533' date='Apr 20 2011, 10:36 PM']"Simple" and "50 high impedance sources" are somehow not entirely compatible

I would recommend Liam to contact me directly to discuss the details. My email address is in my signature.

Just as a personal curiosity, what is the source(s) of the 50 piezo?[/quote]

It is not personal curiosity. We all want to know!

[Silent Fly]

[quote name='owen' post='1206537' date='Apr 20 2011, 10:40 PM']It is not personal curiosity. We all want to know![/quote]

[DanOwens]

[quote name='owen' post='1206537' date='Apr 20 2011, 10:40 PM']It is not personal curiosity. We all want to know![/quote]

Me too! Liam builds lots of intallations. His last one was a bunch of LDRs wired into MaxMSP to generate music whilst the sun set. This, I imagine, might be similar. Either that or he's trying to construct a pickup system for our vibraphone player.

Dan

[icastle]

[quote name='flyfisher' post='1206205' date='Apr 20 2011, 06:20 PM']If these are the devices I think they are, then they are basically the same as those piezo "sounders" often used in cheap products that need to make some sort of noise. Basically, if mechanical energy is applied to the device (e.g. sound vibrations) then you'll get a small voltage waveform out (i.e. behaves like a microphone), but if electrical energy is applied to the device (e.g. an audio signal) then the device will vibrate and emit a sound.

So, if you connect a number of devices together (in parallel) then the mechanical vibrations applied to one device will "energise" the other devices and cause them to vibrate. With eight devices connected in parallel, there will be all sorts of complex interactions between them as they are simultanously energised by the vibrations from the sound source and the electrical outputs from the other devices.[/quote]

Simplest solution there would be to stick a diode on each piezo to stop the voltage from one piezo from energising it's partners...

[flyfisher]

Would that work with an AC waveform though? Could easily add a bias circuit I guess but that moves things away from a passive solution . . .

[icastle]

[quote name='flyfisher' post='1206935' date='Apr 21 2011, 11:49 AM']Would that work with an AC waveform though? Could easily add a bias circuit I guess but that moves things away from a passive solution . . .[/quote]

A GP signal diode (1N4148 for example) should cope with an AC waveform without clipping it (1N400n series are the rectifiers).

Thinking about it a little further, a piezo can't create enough voltage to drive another piezo.
I don't know what the threshold is, but it could be that seperating the piezos into two 'clumps' of 4 and running it as a stereo pair might circumvent the need for extra components (and complexity).

[DanOwens]

I've sent this to Liam. I know he was hoping for a 'gaffa will do it' response.

Dan

[icastle]

[quote name='DanOwens' post='1207550' date='Apr 21 2011, 09:28 PM']I've sent this to Liam. I know he was hoping for a 'gaffa will do it' response. [/quote]

Gaffa will do it - as long as he adds a few other bits and pieces as well...

[DanOwens]

[tauzero]

[quote name='Silent Fly' post='1206423' date='Apr 20 2011, 09:10 PM']However, there is one thing on which I respectfully disagree: there is a voltage divider. It is true, as you said, that a voltage divider is when you have al least two passive components connected in series. [i]This is exactly what happens.[/i]

The 7 pickups that do not generate signal are connected in parallel but for the [url="http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem"]Thévenin's theorem[/url] they are equivalent to 7 impedances in parallel. The equivalent impedance is Z/7. If you look at [url="http://en.wikipedia.org/wiki/Voltage_divider"]Wikipedia[/url] figure 1, Z2=Z/7 and Z1=Z (where Z is the impedance of each transducer), Vin is the voltage generated by pickup #1. For the Thévenin's theorem and the voltage divider equation, Vo/Vi = 1/8 regardless the value of Z.[/quote]
I don't see that as being a potential divider - what you have are 8 impedances in parallel (actually nine, as there will also be the amplifier input impedance), with a voltage being generated across them. Therefore the voltage will be the generated voltage. Otherwise you will be trying to make one point in a circuit be simultaneously two different potentials.

[MarkBassChat]

I changed my mind and I agree with SilentFly. At least I understand the problem with 2 transducers. Maybe he couldn't explain it well (for me) but if you imagine a piezo transducers as a voltage source with some impedance, one piezo is feeding the other through its impedance. And these two transducers have their source impedance connected in series (this info was missing). So on the output you get V/2 voltage. And there are no two potentials (at the same time). One voltage is with one transducer, and the other is when you connect two transducers (or more) in parallel.
However, in real world it's not a problem. Piezo transducers have such a strong signal that it is a common practice to connect them in parallel.
The problem with the voltage divider is solved by FET buffers (separate buffer for each transducer). So if you want to do it correctly, you have to make a buffer for each transducer. But this is a huge cost. So you may solder transducers in groups (each 6) and then mix the groups. I'm sure SilentFly knows how to do it.

Mark