Xanthene
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« Reply #45 on: August 09, 2007, 12:59:01 PM » |
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Colin, can you be more specific as to what the + is? I'm really not following this. Let's start from the first noninverting buffer. Where does 3OUT go from there?
Sorry man, I just want to make sure I get this right. Thanks man!
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expanoncolin
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« Reply #46 on: August 09, 2007, 08:47:25 PM » |
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Colin, can you be more specific as to what the + is? I'm really not following this. Let's start from the first noninverting buffer. Where does 3OUT go from there?
Sorry man, I just want to make sure I get this right. Thanks man! Let's go backwards. H11F3 positive input still be connected to a 470ohm resistor, which should be connected to the output of a noninverting op amp (you have the connetions for those right). The input of that noninverting op amp is, at the moment, the center lug of that pot you wired up, with one end to Vref. Disconnect that center lug from the noninverting op amp input. In it's place, put a voltage divider pot - a pot with one side lug to 9V, one side lug to ground, and the center lug now to the input. Tell me what range of resistance you get now. -Colin
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expanoncolin
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« Reply #47 on: August 10, 2007, 10:05:40 AM » |
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I'm removing the pot right? So I have the 3OUT detached, Vb is out of the equation and I have 4IN+ detached. I remove this pot from the equation. Now you want me to put a pot in place that has one side to 9V and the other side to Ground and the center to 4IN+. But where does 3OUT come in play? Would it go to the center lug? I'm sorry if I'm not getting this but my theory is lacking when it comes to this stuff. Please come to me for graphic design needs so I can feel that I'm redeeming myself  you may grow to regret that. -Colin
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Xanthene
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« Reply #48 on: September 10, 2007, 08:24:37 AM » |
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I haven't tried this yet, but I will be this week. (Vacation and all)
Hopefully you're still with me. hehe
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Xanthene
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« Reply #49 on: September 10, 2007, 08:26:18 AM » |
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Now if I get you right, this would give me 4 pots in total no?
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expanoncolin
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« Reply #50 on: September 10, 2007, 09:40:24 AM » |
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Now if I get you right, this would give me 4 pots in total no? I'm not sure, I can't decide what we were talking about right now... -Colin
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Xanthene
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« Reply #51 on: September 10, 2007, 07:23:45 PM » |
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Haha. Well you said to take the center lug from the amplitude pot and connect it to a voltage divider pot. No?
On top of that I have a waveform and rate pot. This is where I'm confused. Sorry man.
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expanoncolin
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« Reply #52 on: September 10, 2007, 09:20:46 PM » |
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Haha. Well you said to take the center lug from the amplitude pot and connect it to a voltage divider pot. No?
On top of that I have a waveform and rate pot. This is where I'm confused. Sorry man. The voltage divider was just to test your H11F3 setup. -Colin
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Xanthene
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« Reply #53 on: September 11, 2007, 08:54:09 AM » |
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Ah ok, I'll try that. I'll let you knwo.
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stopmakingsteve
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« Reply #54 on: November 04, 2007, 08:39:18 AM » |
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I've been thinking about doing a sequencer to vary resistance for a two connection exp pedal... while I've never actually played around with a 4051 or a 4040, I've been reading this page and I think I grasp the logic. http://milkcrate.com.au/_other/sea-moss/#seqcould you modify this sequencer circuit so that one side of the exp pedal jack connects to the common point on the 4051, then have X0 - pot - other exp pedal side for the gates X0 - X7 ? For a two connection exp pedal jack, you're just putting resistance between the tip and the sleeve, right? This seems like it would do the same thing, and then you'd be able to cycle through 8 pots at a variable rate. Am I doing this right? It almost seems too simple. I need to buy some parts and start playing around with them.
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expanoncolin
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« Reply #55 on: November 04, 2007, 11:25:16 AM » |
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The circuit you posted looks more like an oscillator and sequencer built together, not a circuit for a varying resistance. You want something more like this: http://milkcrate.com.au/_other/sea-moss/08_custom.gifBut with the output connected to an H11F3 or vactrol as discussed elsewhere in this thread. That circuit outputs a series of settable voltages, which would vary the LED brightness and thus the resistance. -Colin
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stopmakingsteve
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« Reply #56 on: November 04, 2007, 02:05:03 PM » |
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The circuit you posted looks more like an oscillator and sequencer built together, not a circuit for a varying resistance. You want something more like this: http://milkcrate.com.au/_other/sea-moss/08_custom.gifBut with the output connected to an H11F3 or vactrol as discussed elsewhere in this thread. That circuit outputs a series of settable voltages, which would vary the LED brightness and thus the resistance. -Colin yeah, that schematic is basically what I was trying to describe  Couldn't you just hook the exp pedal connections up like I labeled it here? So instead of varying resistance to vary voltage to vary resistance, you'd just be varying resistance directly.
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expanoncolin
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« Reply #57 on: November 04, 2007, 03:28:54 PM » |
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Those pots are voltage dividers, the output of the 4051 just cycles through the voltages set by the pots, not cycling through resistances. The pots are set up to simply output a voltage.
-Colin
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danoisefactory
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« Reply #58 on: November 11, 2007, 03:55:28 AM » |
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Hi again, It's getting winter with shorter days and longer nights, so i'm back at my workbench again and with http://experimentalistsanonymous.com/di ... %20LFO.gifI succeeded in making an automated feedbackloop. Hoorah! You should have seen me when i breadboarded the schematic and the led went on-off haha! I'm really happy since i never made a lfo, its easy, but its someting new so always exciting. But it work and control a feedbackloop in my distortion++, it's pretty harsh at this moment but i don't mind. One question, which bits in the schematic make / control the speed of the LFO? There's the 47k resistor and the 100k pot which i can control, but i want it to be slower but can't figure out if and what to change. Anyway, thanks colin for you help a long time ago! Now i'm gonna sit and be proud of myself. 
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expanoncolin
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« Reply #59 on: November 11, 2007, 10:28:20 AM » |
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Hi again, It's getting winter with shorter days and longer nights, so i'm back at my workbench again and with http://experimentalistsanonymous.com/di ... %20LFO.gifI succeeded in making an automated feedbackloop. Hoorah! You should have seen me when i breadboarded the schematic and the led went on-off haha! I'm really happy since i never made a lfo, its easy, but its someting new so always exciting. But it work and control a feedbackloop in my distortion++, it's pretty harsh at this moment but i don't mind. One question, which bits in the schematic make / control the speed of the LFO? There's the 47k resistor and the 100k pot which i can control, but i want it to be slower but can't figure out if and what to change. Anyway, thanks colin for you help a long time ago! Now i'm gonna sit and be proud of myself.  Congrats - blinking lights are always fun to watch. The 1 uF cap can be increased for a lower overall range. You can replace the 100k pot with a larger value to get a larger range, but you might want to use an audio taper pot. -Colin
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