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CV Send Circuit
MUFF WIGGLER Forum Index -> Music Tech DIY  
Author CV Send Circuit
Hey guys,

I'm designing a mixer circuit with a line level send output. The plan is to have CV control over that send output.

The inputs route directly to an output stage (LAB, RAB), with the mixed signal going through a send circuit with CV control. This is the current design I have, hoping some of you can chime in for corrections / improvements etc...

Let me know your thoughts.

what sort of action do you want from the CV input? 0-5V, where 0V is off, and 5V is a gain of 1 (at line level)? the circuit looks good except for the 2164. the input needs a capacitor to ground (check out the datasheet for this), and the 100k at the output will probably be too large for line level. also, you will want to decide on a resistor value for the mode pin.
Thanks guest. The general actions for the cv circuit is:

- If there is no cv input, the potentiometer acts as a normal send attenuator
- With cv input, the potentiometer acts as an attenuator for the incoming cv

The 2164 data sheet suggests 1200pf, although I've seen some schematics with lower and higher values. What is the capacitor value based on? What is it achieving in this circuit?

The data sheet also suggests the leave mode pin open for best noise performance. Also, the output feedback resistor is the same as the input feedback resistance, wouldn't that equal a gain of 0?
for the capacitor, i was referring to R9. you need a capacitor and resistor at the input of the 2164, 500ohm and 560pF. this is for stability, and it doesnt work without it.

im assuming the input is modular level (+/-5V) and you want line level out (+/-1V or so). the gain is R64/R8, which is x5 right now. you want the inverse of that, so id swap those resistors.

the mode pin sets the tradeoff between SNR and distortion. since noise isnt going to be an issue here, id probably use some mode resistance to minimize distortion.

the control pin on the 2164 gives unity gain at 0V, and attenuates for increasing voltages at 33mV/dB. so if you want 60dB attenuation you will need to apply 2V. it never really goes off, it just gets quieter and quieter. so you will need to decide where "off" is in your scheme. is this -60dB attenuation with 0V CV? is this -80dB with 0V? and where is unity gain? is this at 5V CV? and what do you want it to do if it goes above unity gain at the input? does it keep amplifying, or do you want it to cap off at some point?
Ok, that has given me some thought.

If is was -80dB at 0V and unity at 5V, wouldn't any negative voltages be cut off.

Say, a (+5/-5V) LFO is at the cv input, only the positive voltage opens the vca? So technically the 5V coming in is scaled to 0V to 1V? Do I have that correct?
sort of, more negative just means more attenuation, so that end isnt really the issue. the question is what do about voltages greater than +5V? do those increase the signal, or does it just top out at unity gain?

also, with the manual control knob, do you want it to have the same behaviour?
The send maximum amount should be the limit of the mix of the incoming signal A and B, so I wouldn't want the cv input to add gain. Any incoming cv over 5V should top out.

Same with the cv control knob.
ok, so you will need a precision rectifier circuit around that opamp. put a diode in the feedback loop so the anode is at the output. then put a second diode between the output and your feedback resistor, with the cathode at the output (so the diode and resistor are in series). now take the out from the diode and resistor junction. make the feedback resistor around 50k and you will have 80dB attenuation at 5V CV.

if you put your pot between 0V and -12V, and increase its resistor to the opamp to 220k you should get a decent offset.
Ok, hopefully I understood you correctly. What type of diode is necessary?

What confuses me is why would I want to limit the range from 0V to 5V? Wouldn't I want the incoming cv signal to simply be scaled to a range of +1/-1V for line level? The potentiometer is just the attenuator for that cv:
- fully CW, the cv is +1/-1v
- half open, the cv is +0.5/-0.5v
- fully CCW, the cv is 0/0v.

From my understanding, with the rectifier circuit, the potentiometer would offset the incoming cv positively, not attenuator it.

Here is the updated schematic.

the diodes can be any small signal diode, 1N4148, etc.

D4 is in the wrong place. remove it, and connect D3 to R61. then take D4 from output to the inverting pin of the opamp, so it is facking the other direction from D3. so D3 conducts for positive output voltages, and D4 conducts for negative output voltages.

the pot needs to connect to -12V, so when its turned all the way to the -12V side, the output goes to +2.5V, and shuts off the VCA. when its turned all the way to 0V, the output is 0V and the VCA is full open.

R63 should be 100k.

Like this?
sorry, i know this is difficult using words, i should probably just draw a picture. D4 should go from pin2 to pin1 on the opamp (and be reversed from how its currently facing). so for current flowing into the junction at pin2, all of that current flows through D4 to the opamp output, and pin2 is held at 0V. for current flowing out of the juncture at pin2, that current will need to get sourced through D3, and will therefore go through R61, making the usual voltage gain of an inverting amplifier.
Ok, I looked up a precision rectifier and followed the 'improved' one shown on wikipedia. Hopefully that is what you meant.

Is the pot -12v / GND orientation correct?

With the pot closed, the incoming cv will reach 2v or so with half wave rectification, and with the pot open, the full cv signal of 5v will come through with full offset. Is that correct?

the diodes look correct now, but the pot is backwards (at least if i read the arrow correctly). if you want full CCW to be VCA off, and full CW to be VCA on, then the wiper needs to be at the 0V side for full on.
Right, I understand. As you mentioned earlier, and that data sheet states. Increasing voltages at 33mV/ dB attenuate the signal.

Simuation A:

When the pot is fully CW (open), there is still a incoming signal of 2V (plus), shouldn't it be 0V? So that the out going signal is unattenuated as the pot is completely open?

Simulation A

Simulation B:

When I change it to +12v / -12V on the pot, and completely open the pot (CW), its 0V, allowing the signal through at unity.

Simuation B
it all depends upon the functionality you want. when the pot is at -12V for situation A, the full signal comes through identical as situation B. but, that back half of the pot, between 0V and +12V, doesnt do a whole lot in simulation B.

you are also using a bipolar CV in the simulation. with simulation A, a bipolar CV will go from full open to well past full closed when the pot is at -12V. when it is at 0V, it saturates at full open for any positive CV, and then closes off for any negative CV.
Correct me if I'm wrong. Based on an LFO bipolar cv input, what simulation A is doing:

- Pot fully open (0V): unity gain, with attenuation every positive signal of the LFO.
- Pot fully closed (-12V): allows the full range of the LFO to open and close the vca.

Reading through my previous posts, and now that I understand that the higher the incoming voltage, the more attenuation. 0V is unity (vca is open) and 3.3v is -100dB (Max for the 2164 vca).

What I'm after, is:

- Pot fully open CW: - the positive voltage of the incoming cv opens and closes the vca between 0 - 3.3v.
- Pot fully closed CCW; - complete attenuation of the LFO cv. The vca is closed at 3.3v.

Then if there is no cv input, the pot at fully CW, allows the send signal through, and full CCW, no send signal


When I patch an LFO into a vca, it only opens up when the voltage is positive. And doesn't do anything when the negative voltage

So, I sketched up my understanding of what should be happening. As the incoming cv signal goes positive to 3.3v, the vca should open all the way to unity. Then when the incoming cv goes negative, the vca should remain closed and stay closed until the incoming cv has gone positive again.

Then as the pot is turned CCW, the vca troughs reduce until completely closed.

ok, if you want it to saturate at the top end (which it essentially does inside of the 2164), you can either add a 3V zener in the feedback loop, or add some resistance to the output, and just have the opamp saturate. i would probably do the latter, but it doesnt matter too much.

to get the offset you want, add another 220k to -12V

one more thing to consider, is what you want it to do when no external signal is applied. as it stands, this is equivalent to a 0V CV, and therefore the VCA is closed regardless of the pot position. you can either add more range to the pot, or normal the CV jack to +12V through another 100k resistor.
The simulation you sent me was the the previous simulation A I sent you. Did you send me an updated example?

Adding 220k to the -12v doesn't do anything? No offset.

This simulation, does what I want (-3.2v - opens the vca when there is positive voltage from the incoming cv), however I need to offset it as shown in my diagrams above.
sorry, that didnt work, here is a new one:
Ok, thank you. That works, except for when there is no cv.

As suggested, I added a 100k resistor and +12v, and when the pot is fully CCW, the vca is still open. It only goes to 1.9v.

I somehow feel this is NOT the 'usual' way to implement vca's as i seen on other schematics a difference. They don't use rectifiers (diodes), for example Mutable Instruments Veils schematic is completely different.
the veils VCA is linear (uses another VCA cell for linearisation), and it also has a different knob setup than yours. it has a gain knob that switches duty between CV attenuation and VCA gain, depending upon whether a CV is plugged in. it also has a power-law setting.

since you want hard limits for VCA on/off, some form of clamping is required.
Ok. Its good to learn more about the differences in application.

Knowing that I basically want a VCA for my send output, is there another circuit you would recommend?

Also, how would I get the pot to close the vca competlwy to 3.3v when there is no cv input? With the last simulation, it only goes to 1.9v
i dont have a good reference for stuff to read, wish i did. im betting other folks might have suggestions here. i mostly just look at what other people do and read datasheets.

as for VCA toplogies, there is exponential and linear, and every so often you see a power-law circuit. there can be other types, all it means is the mapping between CV and amplitude, so in theory it could be anything.

linear: out = in*CV
expo: out = in*e^CV
power: out = in*CV^n (n usually set with a pot)

mixing boards and such typically use expo VCAs, as your ear hears volume exponentially (which is why we talk about loudness in decibels). a lot of synth circuits use linear VCAs, as the CVs are easily made exponential (RC decay). and for things like tremolo or feedback, i find linear to give a better "feel", as i typically want less subtlety than exponential gives (either i want signal or not). to determine which works for you, you can set up a circuit with log pots and lin pots, and see which attenuates the volume how you like.

for a send, an expo VCA would probably be the more common thing to use, but as i use sends for effects, i might go with linear, but thats really a preference thing. if you want expo, what you have is fine. i noticed on your last simulation you dropped the feedback resistor to 39k. if you want more range, you need to increase this. my original calculation of the 50k was based on a 2.5V "off" voltage. if you want to go with 3.3V, youll need to increase that to 66k.
Ok, will definitely read up on all of this.

Thanks again for your help guest.
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