1V audio burst every OTHER pushbutton

[dougy83]

thanks!! so is that extra bit of sketching the 48V option? I literally know nothing about what you just drew out, but it seems that the circuit is built around a 9V input either way, so I can use a 9V battery or have the 48V option drop it down to 9V for the remainder of the circuit anyway? And what would the max RMS voltage of the output be with the pot all the way up? (I am no doubt showing my absolute ignorance here...)

Yes, 9V either way. Because I didn't write any values for the resistors at the output, there's no way of knowing the output level. Also, because you don't know the mic amp output impedance, you won't know the level after summing either. The maximum output level is 18Vpp, which 27 dBu - this is without the attenuator. With the attenuator, you'll have some level below this. As an example, the maximum output level will be 6Vpp (17 dBu) with a 10k pot at max., and two 10k resistors forming the attenuator, with no load. Adding a load will affect this (likely dramatically)

 

[devin sheets]

yeah, i really don't understand any of that, i'll just have to rely on a friend of mine to put it together and decide what's best I guess... I know that the mic is very similar to a Shure SM58 if that helps at all, and I think that line input voltage is typically 1.23V rms or no more than 2V peak to peak. My plan is just to plug in the mic, get a "medium" level at the mixer gain stage (about -30dB max), and then bring up the pot until the burst is almost peaking the input (wherever that may be). I guess it doesn't matter about the max output voltage too much except possibly for protection against accidentally turning it up way too much and damaging the mixer input?

 

[audioguru]

1) The Shure SM58 is a dynamic type (coil and magnet like a small speaker) and has a rated impedance of only 150 ohms that will overload the weak high impedance output from the Cmos squarewave 10kHz generator.
2) You do not want to blast the 10khz tone into the dynamic mic because the mic diaphragm will vibrate like crazy at the 10kHz (causing severe distortion to lower frequencies) and the other mics might hear its 10kHz.
3) As mentioned earlier, you must filter out the harmonics from the squarewave 10kHz so they do not beat with the digital audio mixer sampling frequency (41kHz?).

 

[dougy83]

1) The Shure SM58 is a dynamic type (coil and magnet like a small speaker) and has a rated impedance of only 150 ohms that will overload the weak high impedance output from the Cmos squarewave 10kHz generator.
2) You do not want to blast the 10khz tone into the dynamic mic because the mic diaphragm will vibrate like crazy at the 10kHz (causing severe distortion to lower frequencies) and the other mics might hear its 10kHz.
3) As mentioned earlier, you must filter out the harmonics from the squarewave 10kHz so they do not beat with the digital audio mixer sampling frequency (41kHz?).

Some simple patches to address these issues:
1) put 1k resistors in line with the microphone. This is assuming that the preamp has a high input impedance. If the attenuator is has two 4k7 and a 10k pot, the voltage at the preamp is 1.43Vp (2.85Vpp).
2) with the 1k resistors in place, the microphone will reproduce less of the 10kHz waveform.
3) add an RC filter. There should be a good LPF on the CODEC/ADC module to remove anything above half-Nyquist, so this shouldn't be an issue.

 

[AnalogKid]

Many console mic inputs have an input impedance selectable down around 200-600 ohms for dynamic mics. This impedance provides mechanical damping of the diaphram through back-emf loading, and sets the tone or color of the mic's audio performance. My guess is that you can't count on a high console input impedance, and that a true mixer will be necessary. This is why I keep asking about bringing the tone burst in on a separate channel.

ak

 

[Tony Stewart]

with AC coupling of the 10kHz signal and impedance ratio to select the injected signal, it should not load down the mic in the low band and easily be passively mixed at 10kHz . The circuit that dougy suggested is easy to follow and consumed very little current. It can be driven from 48Vdc with a 9V zener that draws maybe 2x more current than the CMOS quad NAND ( few mA) with appropriate series R and C or if you prefer , a zener to drop the voltage for an LDO input to output V of your choice.

 

[audioguru]

We have all heard dynamic earphones playing high music frequencies coming along but the person wearing and playing them is still too far away to see them. They are receiving only a few volts of signal about like what this Shure dynamic mic will be receiving. Sure it will be heard. The diaphragm of the mic will probably be forced beyond its maximum normal excursion which will severely distort lower frequencies that are picked up.

But without using a power amplifier for the high impedance 10kHz circuit to drive the low impedance mic then nothing will happen.
Maybe the mic should have inductors in series so that the 10kHz signal is loaded with their higher impedance? The inductors will also reduce the 10kHz level fed to the mic.

I think this 10kHz during sustain-off idea will cause a very loudly played piano chord to cause the muting to occur by mistake. How loud can a piano be played? Is there a limit? I don't think so.

 

[devin sheets]

audioguru,

you bring up really valid concerns (but of course, i feel like i'm reading Chinese when you guys talk). I understand enough to see your point about the 10kHz tone moving the mic capsule... so how about this: what if it isn't a mic? There are piano "pickups" which operate much like electric guitar pickups, they are a long bridge-looking thing that is suspended about 1/4" above the low piano strings, and they pick up the vibration of the metal string within the pickup's magnetic field. So, given that these pickups are not "acoustic" in any way (no moving parts, no sound is created aside from metal objects vibrating close to them) would a 10kHz tone affect it? An example pickup would be the Helpinstill, have you heard of these?

I'm a piano player myself and when it comes to the acoustic loudness of a piano, there is somewhat of a maximum loudness... even slamming a key down with an object won't produce a louder sound at some point - the piano mechanics won't allow it, and most classical musicians realize that the sound is compromised anyway at these louder volumes and don't play up in that range.

I have a simple electrical meter (can't claim to really know how to use it though), is there a way for me to find out the impedance of the mic, or this other pickup I could use?

 

[audioguru]

A simple meter measures DC resistance. You need to know AC impedance instead for most audio things.
I know that a guitar magnetic pickup has a very strong resonance at about 5kHz that is damped and reduced in level with a load impedance or resistance of about 47k ohms. Most guitar amplifiers have an input impedance of 1.5M ohms so that the peak is +14dB for the "twang" sound in a guitar speaker that has poor high frequency response.
I never knew that piano magnetic pickups are available. It would solve the problem of the 10kHz being overloaded by the low impedance mic and would not be vibrated by the 10kHz tone. I suspect the piano magnetic pickup has a line level output so the 10kHz level might not be high enough.

 

[devin sheets]

actually I think the pickup outputs mic level, it does not require phantom power, but also ignores it just like an SM58 mic, and when i put it on the piano and play, it outputs similar level as the mic.

Question about triggering: do I only need one magnet to trigger the sensor? Does the sensore automatically open after the magnet moves away, or do they require an opposite polarity magnet to re-open? Also, are there sensors which only trigger when the magnet is moving at certain speeds? I don't know how much is possible here, but it would be neat to be able to have the sensor only kick in when the release movement on the pedal is above a certain speed near its finish (the dampers only really cause the thump when they are moving down fast. Sometimes piano players slowly mute the strings as a fade effect, and this compression trick might sound awkward in the middle of that...). If this isn't possible to achieve in a simple way, then maybe I shouldn't bother with it?

 

[devin sheets]

What XLR pins should I be using to lash all of this together? I don't know which pins supply the 48V, or how exactly to do the balanced output...

AK, it would be so much more convenient to combine it on the same channel, because otherwise i'll have to use up a total of 40 channels for piano instead of 30, and I still have to mic up string quartets, vocalists at most of the pianos, several other odds and ends inputs and have space for reverb returns and stuff like that... the Yamaha CL5 is a pretty crowded desk already with 30 piano channels, unless I want to use my CL1 to submix the piano channels and bring them into the CL5 stereo plus a subwoofer and reverb send... rather not do that.

Does everyone agree that, no matter what the max voltage of the pot is, I can just start at zero and slowly bring it up until the 10kHz tone is 30dB above wherever my piano pickup level is? I just thought that the max output should be no more than 1.4V rms for safety to the preamp, but then again someone pointed out to me that the pre is seeing 48V phantom anyway... I don't know enough about this.

 

[Les Jones]

I had been assuming that the phantom power was an extra wire with 48 volts with a reasonably low source resistance. There is not an extra wire. The 48 volts is injected into each leg of the balanced pair via a 6.8 k resistor. so the most current you could take from one leg if you decide on 9 volts to power the CMOS IC would be about 5.7 mA ( (48 - 9)/6.8 mA). I found this information here. https://en.wikipedia.org/wiki/Phantom_power
This would mean that all that was needed to obtain the 9 volts would be a 9 volt zener diode plus a resistor so that the zener did not clamp the audio signal. This would mean if a 1 K resistor was used you would only be able to draw less than 5 mA from each leg. A 1K resistor would be OK to use with a dynamic mic but a higher value would probably be needed with a pickup coil. Using a choke instead a resistor would help with this problem but would add the possible problem of hum pickup unless the choke had a closed magnetic circuit. (A 9.1 volt is probably what you would use.) Dougy83's CMOS IC solution is the only one that could be powered this way. (NE555s would take too much current.)

Les.

 

[dougy83]

Question about triggering: do I only need one magnet to trigger the sensor? Does the sensore automatically open after the magnet moves away, or do they require an opposite polarity magnet to re-open? Also, are there sensors which only trigger when the magnet is moving at certain speeds? I don't know how much is possible here, but it would be neat to be able to have the sensor only kick in when the release movement on the pedal is above a certain speed near its finish (the dampers only really cause the thump when they are moving down fast. Sometimes piano players slowly mute the strings as a fade effect, and this compression trick might sound awkward in the middle of that...). If this isn't possible to achieve in a simple way, then maybe I shouldn't bother with it?

No opposite polarity magnet required for a sensor to turn off. The rest depends on the sensor you get. You can get an analogue sensor (output varies with strength of magnetic field near it), from which you can trigger on the rate of change. Alternately you could use two ON/OFF sensors at different locations to accomplish the same thing.

I had been assuming that the phantom power was an extra wire with 48 volts with a reasonably low source resistance. There is not an extra wire. The 48 volts is injected into each leg of the balanced pair via a 6.8 k resistor. so the most current you could take from one leg if you decide on 9 volts to power the CMOS IC would be about 5.7 mA ( (48 - 9)/6.8 mA). I found this information here. https://en.wikipedia.org/wiki/Phantom_power
...
(NE555s would take too much current.)

I didn't actually pay any attention to the word "phantom" before (I just assumed 48V supply). From what you've said, many hall effect sensors will draw too much current to run from the phantom power supply also.

 

[devin sheets]

What about using phototransistors and and LED? I'd have a top and bottom phototransistor with a lit LED resting in front of the bottom one in nominal position (or just past it?), and then the LED moves up past both, then begins to descend passing first in front of the top phototransistor then the bottom one a short time later. There would need to be some kind of pot or slider which can set a time from 0-500ms only over which the circuit is triggered...

 

[Les Jones]

You only need one photo transistor. The LED and photo transistor would be aligned when the pedal is in the up position. A better way would be to have the LED and photo transistor fixed and a piece of thin metal fixed to the pedal that interrupted the light between them. This system is available ready made as one assembly for example this item.
**broken link removed**
Making the position of the sensor adjustable would allow you to set it to the best position so the tone burst occurred just before the pedal reached the up position. The fact that you want to use the phantom supply would rule out this method and the hall switch due to it's very limited current capacity. You would probably have to use a reed switch as this would not consume any current.

Les.

 

[devin sheets]

that's a cool piece.

the only issue with putting the sensor at the top position is that many times, players don't depress the pedal all the way when they play... i'd need it to be placed where it would be triggered almost near the end of the cycle, and then of course the problem is how to distinguish between the LED's up and down passes.

 

[Les Jones]

The beam of light (Actually infra red.) is probably only about 2mm diameter. So if the edge of the piece of metal that interrupted the beam only passed beyond the beam by a few mm then providing the pedal moved down from the rest position so the beam was not interrupted then when it came back up to interrupt the beam the tone burst would be generated.

Les.

 

[Tony Stewart]

if it makes contact ( any thump), an electret mic is simplest reliable solution with a CMOS variable linear gain of 200 , diode peak detector , variable threshold and a Schmitt trigger. It would take someone with experience and the tools , 10 minutes to design it with a CD4049 with 6 resistors, a pot, 3 caps, 1 diode and 1 chip with 5~15V supply.
I'm on the road, so no tools. https://www.google.ca/search?q=cd40...h=539#tbm=isch&q=cmos+inverter+audio+amp+4049

 

[tomizett]

Just to throw another option into the mix (no pun intended), the CL5 has GPIO (general-purpose input-output) lines - 5 inputs and 5 outputs. These are a simple electrical interface to which you could connect a switch or relay, and use it to trigger macros etc.
I'm no expert on the Yamaha automation systems, but it *may* be possible to use that to fire a burst of the desk's oscillator into your compressor key (thereby avoiding the audio path altogether).
The only catch with that, of course, is that the GPIO connectors are local to the surface - if you're using a stagebox then you may have to run additional lines from stage. On the other hand, I'm sure some other manufacturers have GPIO on thair stage boxes, maybe Yamaha do to?

On the other topic, the input impedance of the preamps is not stated in the manual, but is usually "a few kilo-Ohms", maybe between 3 and 10k. Standard pinning for mic XLRs (to save you looking it up) is 1=ground, 2=signal hot, 3=signal cold.

(edited for typos)

 

[devin sheets]

The GPIO thing is really interesting! Unfortunately, I don't think the Yamaha stageboxes for the CL consoles have GPIO, it seems to be only a local feature, so I'd have to make home runs about 300ft each. Also, there is only one oscillator (maybe two now) but I'd need 10 of them, or a way to change the patching accordingly every time someone plays a pedal... I doubt that kind of programming is possible from what little I know about it. I'll look into it more though, maybe something will come up that I'm not thinking about yet!

Tony, if you get a chance, can you spell out your ideal a bit more? I'm a total beginner and need a lot of spoonfeeding...

So at this point I'm wondering if there's a generous person out there who'd be willing to do a quick sketch of this circuit thus far, with a top and bottom photo-interrupter where both are blocked at the nominal pedal position, both are exposed at full depression, and then the top one is blocked first upon release, followed shortly by the second one near the end of the release, with a selectable timeframe of 0-.5s within which the release must happen for there to be a trigger (also rejecting other combinations such as a half-release where only the bottom one was exposed during the cycle), a selectable delay of the 10kHz burst from 0-.5s once it is triggered, the selectable volume control from 0-1.4V RMS on a balanced output, the necessary circuitry for sine-ifying the burst and limiting it to 5ms in length, the ability to use the 48V Phantom Power from the audio mixer to power the circuit, and then some kind of summing circuitry to be able to combine the 10kHz burst with the signal from the piano pickup or mic without either one affecting/damaging the other (the summing feature can be left out if it is too much trouble, and I can just use an additional channel on the mixer if necessary).

Any takers?? I really appreciate everyone's help so far!