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Captain_Clams
Joined: Jul 02, 2012 Posts: 43 Location: MD, USA
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JingleJoe
Joined: Nov 10, 2011 Posts: 878 Location: Lancashire, England
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Posted: Wed Jul 04, 2012 6:23 am Post subject:
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While that would work the usual method to get say a hi-hat or cymbal sound from a noise circuit is to put it through a voltage controlled filter controlled with an envelope- which is basically a capacitor charged and discharged by a circuit
To calculate the time the noise circuit would operate for, calculate the discharge time of the capacitor; T=RC, R in this case being the resistance of the noise circuit which one cannot simply measure with a multimeter unfortunately.
This method is used in some lunetta circuits to give the circuits in question a voltage starving pulse, resulting in nice weird sounds In my experience something like 1000uF only makes them last for about 5 seconds or less.
You may also want to connect one side of that capacitor to ground, along with the ground connection of your noise circit, then you only need one SPDT switch (single pole double throw) to connect the capacitor either to +V or to the circuit it is intended to discharge through.
The simplicity of this idea for creating some kind of envelope or pulse of sound is appealing, I'm interested to see where this goes _________________ As a mad scientist I am ruled by the dictum of science: "I could be wrong about this but lets find out"
Green Dungeon Alchemist Laboratories |
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Captain_Clams
Joined: Jul 02, 2012 Posts: 43 Location: MD, USA
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Psyingo
Joined: Jun 11, 2009 Posts: 248 Location: Canada
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Posted: Wed Jul 04, 2012 9:10 am Post subject:
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do you want to hear the noise source decay in amplitude over a period of time? |
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JingleJoe
Joined: Nov 10, 2011 Posts: 878 Location: Lancashire, England
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Posted: Wed Jul 04, 2012 12:33 pm Post subject:
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look into the 4066 analogue switch, you may find that as useful for doing what you are doing
However, if you want the light to control the sound, then I believe the conventional method would be to use the light dependant resistor to control a VCF or VCA into which the noise is fed.
What I'd do here is forget about the flip flop and use a comparator to sense the light falling on the LDR, which will switch in the charged capacitor using the 4066, when the light goes above a certain level! Fun times will be had by all
There's a good number of descriptions of comparators around the web which use a comparator as a light sensor in the demonstrations _________________ As a mad scientist I am ruled by the dictum of science: "I could be wrong about this but lets find out"
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Captain_Clams
Joined: Jul 02, 2012 Posts: 43 Location: MD, USA
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Posted: Wed Jul 04, 2012 5:32 pm Post subject:
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Quote: | do you want to hear the noise source decay in amplitude over a period of time? |
Yes I do. Like a drop of water in a frying pan!
Quote: | light dependant resistor to control a VCF... switch in the charged capacitor using the 4066 |
The switch has got to stay closed until the next time you charge the capacitor.
Would this do that? |
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Captain_Clams
Joined: Jul 02, 2012 Posts: 43 Location: MD, USA
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Posted: Wed Jul 04, 2012 7:46 pm Post subject:
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Let me refine what I said in the previous post . I have family at my house and they started to eat watermelon in my ears, so I finished the post in a hurry and got the hell out of there. I don't think I was making any sense.
I just reviewed the 4066 datasheet, and I can't really tell the behavior of the sets of 3 pins that form a switch (one receives the high or low state, and the other two are the signal path to be switched). The way it looks to me is that the circuit is only "closed" during a high state on the switch pin. Immediately after the state goes low, the switch is closed. I may be misunderstanding the datasheet or something you said, but that is not actually the behavior I want. I would like the circuit completion to be maintained even after the falling edge of the trigger pulse.
The thing the match will do is to indefinitely (long after the match burns out) close the switch, until the capacitor dies, which could be 15 seconds or so.
Perhaps this is already what you were advising me for. I am not able to accurately process all of this new information, even when there are no watermelons being eaten rudely in my vicinity. |
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Psyingo
Joined: Jun 11, 2009 Posts: 248 Location: Canada
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Posted: Wed Jul 04, 2012 7:50 pm Post subject:
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I'm not sure how complex you want this, but i have a schematic that combines the elements of a noise source, a vca and a decay envelope that could create this... I could upload it.
It could alternatively be used as a snare/cymbal sound |
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Captain_Clams
Joined: Jul 02, 2012 Posts: 43 Location: MD, USA
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Posted: Wed Jul 04, 2012 8:03 pm Post subject:
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Oh, I see. I found a VCA, which stands for Voltage Controlled Amplifier (?) and it is this item that would cause the signal to sound like it's dying out as opposed to simply stopping abruptly (correct?). I didn't realize that one was necessary - I thought the dying out sound would happen naturally. In this case I am interested in your VCA circuit.
On further inspection: "decay envelope." Is it some combination of this and a VCA that makes a fizzle? I think it would be really good if I could have that schematic to look at as I can't picture how they'd be used in conjunction. Thanks |
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Psyingo
Joined: Jun 11, 2009 Posts: 248 Location: Canada
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Posted: Wed Jul 04, 2012 9:07 pm Post subject:
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heres the schematic. this isnt tested, but i have built all of those circuits seperately. the noise itself was tested at 12v so you may need to tweak r4 and r5 to be smaller. r2 should be a small value, say between 100 ohm and 1k ohm. r1 can be something large like a 100k pot orsomething. |
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JingleJoe
Joined: Nov 10, 2011 Posts: 878 Location: Lancashire, England
Audio files: 14
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Captain_Clams
Joined: Jul 02, 2012 Posts: 43 Location: MD, USA
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Posted: Thu Jul 05, 2012 5:40 am Post subject:
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I went off reading about envelopes, and I don't think that my device will need one. I thought that a capacitor would provide its own sort of exponential envelope. However, I do need a VCA, so thanks Psyingo.
And JingleJoe, the large triangle in your schematic: is it a comparator? If it is a comparator, then would it send a momentary 1 (inverted 0) over to the 4066 while the photoresistor is letting more current through than the other resistor?
The 4066. That's what I still don't get. Does the 4066 stop allowing the current through after the high event ends? I can not decipher its datasheet. |
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Psyingo
Joined: Jun 11, 2009 Posts: 248 Location: Canada
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Posted: Thu Jul 05, 2012 7:24 am Post subject:
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Without an envelope a vca is sort of useless for what you need. If you are just going to power starve the noise with a large power cap, you wouldn't need the Vca, but you may encounter some problems. You may get a loud click everyone the circuit is activated, also the cap is fairly large. On the plus side the circuit would be slightly easier opposed to the one i proposed. With the decay circuit in my schematic you wouldn't need such a large capacitor and the decay time is also adjustable. you're right in that you don't need a proper envelope, but decay is handy for this. |
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JingleJoe
Joined: Nov 10, 2011 Posts: 878 Location: Lancashire, England
Audio files: 14
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Posted: Thu Jul 05, 2012 7:42 am Post subject:
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Captain_Clams wrote: |
The 4066. That's what I still don't get. Does the 4066 stop allowing the current through after the high event ends? I can not decipher its datasheet. |
The 4066 is an analogue switch, basically a regular switch but controlled by voltage, in this case logic or digital voltages of 1 and 0 a switch in the 4066 will be closed when the control input is 1 and open when it is 0.
It has some resistance but it is minor enough not to effect most circuitry.
The triangular component is indeed a comparator, might be an idea to read up on electronic component symbols while you're reading up on synthesizer modules It doesn't produce a pulse but will activate the noise circuit when light falls on the LDR, the large capacitor will discharge through the noise circuit.
Anyway, Just wanted to clarify my idea, psyingos circuit is better in many aspects. _________________ As a mad scientist I am ruled by the dictum of science: "I could be wrong about this but lets find out"
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Captain_Clams
Joined: Jul 02, 2012 Posts: 43 Location: MD, USA
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Posted: Thu Jul 05, 2012 7:46 am Post subject:
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I like the purity of a decay caused by a capacitor losing power. (also the simplicity is good for my inexperience )
For the loud click, I agree. This chart of capacitor discharge time is relevant:
That initial quick dropoff would not be very perceivable as it happens so quickly. POP.
So could a guy stick a voltage regulator after the capacitor so that it rides at a steady voltage until it cannot produce that voltage, and then drops off? That would increase the amount of time it will run for as well. But will it work?
Oh, and Joe:
Quote: | closed when the control input is 1 and open when it is 0. |
I want the circuit to be closed even after the match has burned out. It doesn't sound like the 4066 does that, based on what you said there and what I think ( ) I've inferred from the datasheets. |
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JingleJoe
Joined: Nov 10, 2011 Posts: 878 Location: Lancashire, England
Audio files: 14
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Posted: Fri Jul 06, 2012 2:51 am Post subject:
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Ah okay, I didn't get that when I read your description, a one shot or monostable is what you want then you could still use the same control method, just have the comparator activate the monostable.
The charge and discharge voltage of a capacitor in a monostable circuit can be buffered and used as an envelope. _________________ As a mad scientist I am ruled by the dictum of science: "I could be wrong about this but lets find out"
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Captain_Clams
Joined: Jul 02, 2012 Posts: 43 Location: MD, USA
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Posted: Sat Jul 07, 2012 8:33 pm Post subject:
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Hey, I finally got back after a day of fiddling with a videorecorder (and also wandering around in Baltimore today). I have something to show you! This is all very exciting to me, as it's my first functioning noise circuit. The video consists of simply a short demonstration of a charge and discharge of the 10,000 uF reservoir capacitor into the 555 timer square wave oscillator.
http://www.youtube.com/watch?v=-AiY1ObS4G8
I have a few questions about the "envelope" that the capacitor brings about. First: the capacitor is rated at 25V, and before the circuit is a 5V regulator (to not melt the 555 timer or whatever else might happen, don't want to find out). I would be expecting the "solid line" from this graph I whacked together with a trackpad on a dirtbag Macintosh with spinoff MSPaint "Paintbrush"
Dotted line: unregulated discharge of any capacitor
Solid line: discharge of 25V capacitor w/ 5V regulator
So, is what I'm experiencing with this circuit probably my anticipated solid line?
Second: Why is there only a minimal difference in sound between charging the capacitor for 30 seconds and charging it for 2 seconds? Does it really charge up that fast? It's 10,000 uF, and there's even a small resistor that the capacitor charges and discharges through... I would expect it to take longer than 2 seconds to fully charge.
Anyway, I'm really jazzed up about this project now, and I'm looking forward to making it more exciting over time, with a little help from you fellas, that is |
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inlifeindeath
Joined: Apr 02, 2010 Posts: 316 Location: Albuquerque, NM
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attdestroyers
Joined: Mar 29, 2012 Posts: 47 Location: Malvern, Ohio
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Captain_Clams
Joined: Jul 02, 2012 Posts: 43 Location: MD, USA
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JingleJoe
Joined: Nov 10, 2011 Posts: 878 Location: Lancashire, England
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Posted: Sun Jul 08, 2012 10:23 am Post subject:
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inlifeindeath wrote: | 25V is just the maximum voltage rating of the capacitor. |
Yes, did you acctually charge the capacitor to 25 volts? then discharge it into a voltage regulator? if so you will just get the curve drawn by attdestroyers.
If not, it will just be charged by the 5 volt regulator to 5 volts. _________________ As a mad scientist I am ruled by the dictum of science: "I could be wrong about this but lets find out"
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Captain_Clams
Joined: Jul 02, 2012 Posts: 43 Location: MD, USA
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Posted: Sun Jul 08, 2012 10:29 am Post subject:
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I don't have a multimeter currently (the quality of that Walmart one was inexcusable), but I believe that I had the capacitor charged "all the way up," whatever voltage that may occur at.
The capacitor was not being charged through the voltage regulator, however it was being discharged through one into the main circuit.
I still don't see what could happen to the electrons I colored blue there in the latest graph drawing. Am I looking at it wrong? |
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JingleJoe
Joined: Nov 10, 2011 Posts: 878 Location: Lancashire, England
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Posted: Sun Jul 08, 2012 10:37 am Post subject:
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In short, the electrons in your graph are not there and if they are, they are regulated down to 5 volts.
The capacitor will only charge to whatever voltage you apply to it and even then only to within about 99% of the voltage in a time equal to 5*RC, it never quite reaches 100% of the voltage applied but it gets to within a tiny fraction.
If you discharge say, 25 volts into a 5volt regulator, it will output 5 volts untill the capacitor voltages goes below 5 volts, because that is what they are made to do: take in any voltage up to a certain limit and output thier stated voltage. Below that voltage things get iffy and to avoid damage to the regulator or improper opetation, one should not really power a regulator with less than the regulated voltage +2V. _________________ As a mad scientist I am ruled by the dictum of science: "I could be wrong about this but lets find out"
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Captain_Clams
Joined: Jul 02, 2012 Posts: 43 Location: MD, USA
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Posted: Sun Jul 08, 2012 10:45 am Post subject:
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So when a 5V voltage regulator sees 6V, it passes 5V and kills 1V somehow? As heat perhaps? They do seem to have rather large metal wings on them.
Oh, a more pertinent question to ask would be whether or not there is any way to cause the discharge of the capacitor to happen linearly, or at least a little bit closer to logarithmically.
Either one is better than steady for 10 seconds then fuzzing out all the sudden in a half second. |
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JingleJoe
Joined: Nov 10, 2011 Posts: 878 Location: Lancashire, England
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Posted: Sun Jul 08, 2012 11:23 am Post subject:
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Capacitors charge and discharge with an exponential or logarithmic curve. It's your circuitry which is having unusual behaviour, maybe when it gets below a certain voltage it simply stops working and works fine above that voltage. Most lunettas can work down to 3 volts and when under that voltage they do strange things and some stop working entirely. When you under power something it is not working as it should, it's not good for your chips and unusual currents may be flowing due to the internal transistors inside your ICs functioning improperly, this causing it to drain the capacitor faster as the voltage gets lower and thus turn off the circuit faster. Assuming you are using ICs. _________________ As a mad scientist I am ruled by the dictum of science: "I could be wrong about this but lets find out"
Green Dungeon Alchemist Laboratories |
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