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omissis
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omissis
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 Posted: Wed Apr 04, 2012 9:58 am Post subject:
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Hi again
any hint about it?
Thx again
M
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JingleJoe
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Posted: Wed Apr 04, 2012 12:58 pm Post subject:
Re: Resistors and CV mixing
Subject description: a beginner's question |
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| omissis wrote: | What am I doing wrong in this formula? Could anybody give me advice?
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No. You haven't supplied enough information.
That resistor voltage summer should work fine, but no one has any idea what you are doing or whats going wrong.
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omissis
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| Posted: Wed Apr 04, 2012 2:20 pm Post subject:
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Hi
It's just a doubt and nothing is broken (at the moment) 
I'm not so skilled when trying to explain a diagram; this cv mixer should be able to mix voltages and currents.
My aim is to calculate the correct value of N, so I'm using a nodal analysis.
As a hardware component, this CV mixer has resistors even on the I signals; the formula I'm using takes for useless the resistors on the I path.
Thing is ,I just can't understand what's the role of the resistors on the I signals...can you help?
Thanks
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Dave Kendall
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| Posted: Thu Apr 05, 2012 9:55 am Post subject:
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I'll take a shot here, but without knowing what's either side of the circuit in your diagram, it's difficult to help much.......
*If* point N is followed by an opamp wired as an inverting summer, then the resistors on the left of point N determine the voltage of each of the sources (V1, V2 etc) at the output of the opamp.
The opamp will present the sum of those voltages at its output (VOUT), but inverted. For argument's sake, make the feedback resistor 100K between the opamp's OUT and its NEG in.
If V1 = 1 Volt, R1 = 100K, and no other voltages are being passed through R2, R3 and R4, then the output (VOUT) is -1V.
If V2 = 2V, and R2 = 50K, and no voltages are being passed through R1, R3 and R4, then VOUT = - 4V .
If V1 and V2 are *both* connected with 1V and 2V through R1 and R2 respectively, then VOUT = -5V
Hope this helps some...... Someone smarter than me at all this can probably explain it better, but that's how input resistors work with a simple inverting summer. If you already knew this, or if there's a completely different circuit after N then ignore what I wrote 
cheers,
Dave
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omissis
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| Posted: Fri Apr 06, 2012 12:11 pm Post subject:
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| Dave Kendall wrote: | | I'll take a shot here, but without knowing what's either side of the circuit in your diagram, it's difficult to help much....... |
Thanks for the friendly reply
Voltage makes its way straightly into the drawn chip iG00153 VCO to control the pitch.
The puzzle for me is in the lower two wires, named "I"
They are actually currents, the VCAs generating them are OTA based and do work as attenuators for the key velocity and aftertouch voltages (the attenuation is controlled by a slider sending 0 to 10V).
Question is I probably can get rid of the value in the 56K resistors on the "I" lines in my formula, but I could be caught wrong by those resistors: it's unusual for me to have them on a current path and my guess is the resistors on the I lines are there to stabilize the signal.
Please help, I just want to understand.
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Uncle Krunkus
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| Posted: Sat Apr 07, 2012 5:24 am Post subject:
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In a voltage summer like this, and again, this is assuming it is a voltage summer, (what else could it be).
It only sums voltages, so I think the "I" may be misleading. What if it just stands for Intensity, of the velocity and aftertouch, which are outputting voltages, just like the other two inputs?
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omissis
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| Posted: Sat Apr 07, 2012 10:32 am Post subject:
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| Uncle Krunkus wrote: | In a voltage summer like this, and again, this is assuming it is a voltage summer, (what else could it be).
It only sums voltages, so I think the "I" may be misleading. What if it just stands for Intensity, of the velocity and aftertouch, which are outputting voltages, just like the other two inputs? |
I wish it could be as you say but's fairly crazier than that
Can't be voltages, as the VCAs are OTAs, they get voltage (from the keyboard) and translate it into current, unlike opamps do .
My only question is : being them currents, the resistors R3 and R4 are useless to get the sum (which I can get anyway) because current is the same before and after resistors, but hell why place them if they're useless?
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EdisonRex
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| Posted: Sat Apr 07, 2012 11:52 am Post subject:
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but hell why place them if they're useless?
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because you can't have current without some voltage too?
V=IR or V/R = I or I=V/R
If you're modulating current you probably have a constant voltage.
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Dave Kendall
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| Posted: Sat Apr 07, 2012 12:31 pm Post subject:
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Hi all.
As the output of the resistor summing node is going into an IG00153 VCO chip maybe this diagram will help......
http://www.jhaible.de/cs80_vco.gif
cheers,
Dave
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omissis
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| Posted: Sat Apr 07, 2012 4:27 pm Post subject:
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| Dave Kendall wrote: | Hi all.
As the output of the resistor summing node is going into an IG00153 VCO chip maybe this diagram will help......
http://www.jhaible.de/cs80_vco.gif
cheers,
Dave |
Thx but more likely it's this one you were looking for, as the signal goes into a dedicated vibrato input to the VCO's pitch, so to give an expo flavor to the modulation, rather than linear ( remember that the ig00153 VCOs are V/Hz, current driven)
http://www.jhaible.de/jh_cs80_vibrato_guess.pdf
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omissis
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| Posted: Sat Apr 07, 2012 4:37 pm Post subject:
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| EdisonRex wrote: | | Quote: |
but hell why place them if they're useless?
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because you can't have current without some voltage too?
V=IR or V/R = I or I=V/R
If you're modulating current you probably have a constant voltage. |
No doubt
In facts I get
V = (V1 / R1 + V2 / R2 + I3 + I4) / (1/R1 + 1/R2)
So could the resistors be for another purpose , like co-operate with the capacitor to create a differentiator for DC bias removing?
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EdisonRex
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| Posted: Sun Apr 08, 2012 4:00 am Post subject:
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| Quote: |
In facts I get
V = (V1 / R1 + V2 / R2 + I3 + I4) / (1/R1 + 1/R2)
So could the resistors be for another purpose , like co-operate with the capacitor to create a differentiator for DC bias removing?
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If you consider that both I3 and I4 have a V constant associated with them, you would then be using those resistors to convert modulating current back to a varying voltage, would that not be the case? Your equation does not take that into account, as the resistors R3 and R4 have no voltage constant applied to them. If you assume a constant voltage at the current source, then passing varying current through those resistors you will have a varying drop in voltage across those resistors, Ohm's law being relative.
So it'd be more like
V = (V1 / R1 + V2 / R2 + (Vc / I3) + (Vc / I4) / (1/R1 + 1/R2 + 1/R3 + 1/R4)
Or so I'd have thought. Vc cannot be non existent, and it can't be zero.
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