VCA theory question

[juniorhifikit]

So as not to distract from the other vca and That threads, I thought I would ask my mundane question in it's own thread.

After reading the groovy thread here about vca's and current summing, and reading this datasheet for the 2180 vca from That:

http://www.scottgreiner.com/2180data.pdf

my question was: why scale the control voltage with a buffer amp? I understand that the vca's control ports want to see a very low source impedance (under 50 Ohms) and that the output of some of the DAC's I've seen are in the 10's of KOhms; so an opamp output like a 5534 would work there (0.3 Ohms). But in their application notes, they suggest taking the DAC's 4.98V output (with a 5V supply) and scaling it down to .776 volts. So then this 0V to .776V control voltage is divided into .5dB steps (8bit DAC) to divide up the 127dB gain range of the vca. Why not use the 4.98V directly - why scale it down?

Here's the application note - page 3 is where they discuss the scaling of the control voltage:

http://www.scottgreiner.com/vcaGAINan102.pdf

For my own application, I'm planning on using a 10bit DAC to get 0.1dB steps until down around -100dB

[JR.]

I'm not sure I understand the question but perhaps I'm too close to this.

The simple answer is full scale VCA CV range is only hundreds of mV, while the dynamic range of most DACs are several Volts, and simple panel controls a couple tens of Volts.

Scaling up the CV to better fit the dynamic range of the medium you are working in, minimizes all errors related to those mediums, whether it's digital LSB, or ground errors and DC offsets in opamps. Back in the '70-80s when I was designing VCA based products I commonly used opamps with 15 mV offest voltages for CV crunching, so scaling down after that was required for any reliable repeatability and consistency.

Modern opamps are much better, thankfully.

JR

[mediatechnology]

Why not use the 4.98V directly - why scale it down?

Like John I'm not sure I understand the question either. But if you want to apply 4.98V directly to Ec+ or Ec- it will cause problems. Ec is limited to less than one Vbe or the junctions in the gain cell will become reversed biased. Take a look at the Ec values here: -90 dB is 540 mV: http://www.thatcorp.com/datashts/2180data.pdf#page=4

A 4.98V 10 bit D/A converter is going to have steps around 5 mV. With the VCA having a scale factor of ~ 6.1 mV/dB each step will be about 0.8 dB. By scaling the 4.98V output down to the range Ec requires you can gain a lot of resolution with a small step size.

[juniorhifikit]

Thanks guys. I missed that spec of the CV range being in the millivolt range.

I'm still a little unclear about the CV buffer circuit in regards to what I've been able to understand from the application document. With both CV ports tied to ground I'm getting unity gain through the VCA, as stated in the document. If I leave the positive CV port tied to ground, as per their example circuit, would a negative voltage at the negative CV port result in gain reduction? Or a positive voltage...? As stated in the datasheet:

For pin 2 positive or pin 3 negative, the output current will be scaled larger than the input current. For pin 2 negative or pin 3 positive, the output current is scaled smaller than the input. The scale factor between the output and input currents is the gain of the VCA. Either pin 2 (Ec+) or pin 3 (Ec-), or both, may be used to control gain. Gain is exponentially proportional to the voltage at pin 2, and exponentially proportional to the negative of the voltage at pin 3. Therefore, pin 2 (Ec+) is the positive control port, while pin 3 (Ec-) is the negative control port.Figure 8. Gain vs. Control Voltage (Ec-) with Temp (°C)

I'm a little unclear.

BTW, thanks for your patience with my ignorant questions!

[JR.]

The two gain control ports can be thought of as + and - inputs that sum together to control the VCA gain.

For simple applications where you aren't calling for much gain range it is cheaper/easier to just leave one control port grounded and drive just the other one. For console fader automation where you are commanding larger amounts of gain (actually attenuation) it is better for the VCA performance to split the control voltage in half and drive both control ports in opposite directions.

The explanation for why this is better requires looking inside the VCA and how it works. TMI for here now.

I think we pretty much beat to death how to optimally buffer the VCA control voltage in one of my VCA sum discussions. There are some subtle noise/distortion mechanisms associated with this CV buffer noise and source impedance. Some of these errors show up as a modulation term, so XX dB below any signal, and hard to even see on a good test bench, but if it's worth doing its worth doing better..

JR

[juniorhifikit]

Thanks!

I also just spoke with someone at THAT and they walked me through the math. It all makes sense now.

[juniorhifikit]

Well, almost makes sense...

I'm still trying to keep up with everything you all have left in the dust. Always a problem for noobies.

It seems from That's datasheets and design notes, that the maximum CV for a 2180 VCA is less than a volt, positive voltage on the negative port = gain reduction, and vice-versa for the positive CV port. Their graph shows a voltage range of -540 to +180mV to get -90 to +30 dB (on the positive CV port), but their math and schematic examples are a little vague for my noob-ness. They show a control voltage from a DAC of just under 5V max, and then doing the math get a buffered .776 volts max. That all works out fine, except with the inverting input of the buffer amp as shown in the schematic, I get a negative .776 volts. Biasing the gain range of that with the negative supply, and adding it to the negative CV port as shown ends up with massive gain (too much). It seems quite a bit of "interpretation" of their example data needs to be done.

If my understanding is correct, I need a negative CV, biased by V+ to get a mostly positive and slightly negative range, and applied to the negative CV port;

or I need a positive CV, biased by V- to get a mostly negative and slightly positive range, and applied to the positive CV port.

Have I got this right?

[JR.]

The effective CV is the total of voltages applied to + and - CV pins simply added together, with the sign of the CV pin factored in. So counting the sign, if we send +18mV to both control pins it becomes 18-18 or 0V for unity gain. OTOH +18mV on the + pin and -18mV on the - pin, is 18 - (-18) or 18 + 18 = 36 mV.

0V DC is 0 dB gain. For modest amounts of boost or cut, which is only tens of mV total, you can apply the control voltage to only one of the CV pins with no problems.

A little inside baseball.. these CV pins are actually connected to transistors inside the VCA so if the control voltage gets more than a diode drop away from 0V in either direction it not very good for the internal audio path symmetry. Therefore for large gain commands (such as a fader or gate commanding 60-70 dB of cut, it is much preferred to split the control voltage in half and apply one half to one control pin, and the other half in the other direction to the other CV pin. You still have the full CV generated between the two pins, relative to each other, but the internal VCA path is happier.

For modest gain control commands like for a compressor, you can generally get away with only driving one pin and leaving the other grounded.

I hope this helps.

JR

[juniorhifikit]

The effective CV is the total of voltages applied to + and - CV pins simply added together, with the sign of the CV pin factored in. So counting the sign, if we send +18mV to both control pins it becomes 18-18 or 0V for unity gain. OTOH +18mV on the + pin and -18mV on the - pin, is 18 - (-18) or 18 + 18 = 36 mV.

0V DC is 0 dB gain. For modest amounts of boost or cut, which is only tens of mV total, you can apply the control voltage to only one of the CV pins with no problems.

So it seems the examples on the That datasheets and design notes are general at best. I'm going to have to put on my thinking cap...

A little inside baseball.. these CV pins are actually connected to transistors inside the VCA so if the control voltage gets more than a diode drop away from 0V in either direction it not very good for the internal audio path symmetry. Therefore for large gain commands (such as a fader or gate commanding 60-70 dB of cut, it is much preferred to split the control voltage in half and apply one half to one control pin, and the other half in the other direction to the other CV pin. You still have the full CV generated between the two pins, relative to each other, but the internal VCA path is happier.
For modest gain control commands like for a compressor, you can generally get away with only driving one pin and leaving the other grounded.

I hope this helps.

JR

That's very good information and helps a lot. I seem to remember you mentioning this in the VCA current summing thread. I have noticed that many currently available automation offerings do the single CV port, but I guess it can be done better.

[juniorhifikit]

A little inside baseball.. these CV pins are actually connected to transistors inside the VCA so if the control voltage gets more than a diode drop away from 0V in either direction it not very good for the internal audio path symmetry. Therefore for large gain commands (such as a fader or gate commanding 60-70 dB of cut, it is much preferred to split the control voltage in half and apply one half to one control pin, and the other half in the other direction to the other CV pin. You still have the full CV generated between the two pins, relative to each other, but the internal VCA path is happier.

JR

Don't we still have to pass through 0V regardless of which ports we use in order to use the full range of the VCA? Unless the two buffers are offset so that only one side at a time passes through 0V...