Just need some Explanation

Hey all,

The MCP4822 is the DAC for the Anushri and its output is send to the TS912. I just want to know whats the TS912 for and what kind of circuit is it using. Also why is the ref voltage being send to the TS912. Sorry if I am asking too much. I am currently working on my dissertation and I’m using the MCP4822 as well to send pitch cv. So I was just wondering if the TS912 is important to my circuit design.

Cheers, Ashvin.

The MCP4822 can generate an output voltage between 0V and 4.096V.

If we used the 1V/Oct standard, this would give a range of 4 octaves only. It might be ok for a toy like the CVpal, but not OK for Anushri which I thought deserved a larger range.

So instead, Anushri uses a 0.5V/Oct scale internally. Obviously, this needs to be converted to 1V/Oct for the actual CV output.

The bit of circuitry built around the TS912 just multiplies the DAC CV by a factor of 2, and shift it by -4.096V, so that it swings between -4.096V and 4.096V with a 1V/Oct scale. This is standard op-amp stuff, work it out from Kirchhoff.

If you need more than a 4.096V range, you’ll need an op-amp to scale the DAC output.

Thank you my friend thats brilliant!

Alright I have another question. Since I am using the MCP4822 to send CV out. Does it need to be ac or dc coupled ? Also do I need to have a high pass filter or a low pass filter(to remove the noise) ?
Any help would be much appreciated.

Cheers, Ashvin.

The MCP4822 is a D/A Converter so theres not carrier frequency to filter or decouple.

> Does it need to be ac or dc coupled ?

“CV out” doesn’t mean much to me… It can be anything - a LFO, an envelope, a note CV… But I assume this signal can be constant for a long stretch of time. Which means that the signal will have an important DC offset or at least very low frequencies, and you don’t want to remove that. So it’ll have to be DC coupled.

> Also do I need to have a high pass filter or a low pass filter(to remove the noise) ?

Which noise are you referring to?

The noise that MCP4822 produces. They mention this in the MCP4822 data sheet.

Well, you have to be aware that the noise produced by the MCP4822 and your signal will be mixed together. There is no magic way of filtering only the noise without impacting the signal too…

A general rule when designing circuits is that their bandwidth shouldn’t be unnecessarily larger than the bandwidth of the signals they will have to let go through. This keeps noise and EMI low. That’s why you’ll always find a 20pF or so cap in parallel with the feedback resistor in an audio op-amp circuit - because there’s no reason we shouldn’t attenuate frequencies above 50kHz since they are of no use.

You haven’t told me what kind of CV you want to generate, so I have no idea what is your bandwidth requirement. If it’s a triangle or sine LFO, you can low-pass at 1kHz or so. If it’s a note CV or envelope, beware of low-pass filtering since it’ll add an unwanted slowness/portamento.

High pass filtering? I have just explained that DC is absolutely needed for most CV generation applications.

Alright. So I am building a Capacitive Proximity Sensor using electric paint as my capacitor sensor. So the output of the sensor goes to an Arduino. The Arduino then sends the signal to the MCP4822, then the signal goes to an non-inverting amp. The non-inverting amp is to get 0V to 10V range. So you can basically control a filter cut off point using the capacitive proximity sensor. This is my dissertation and it all works nicely rite now. So I am just seeing if I should add any more things to my circuit.

What is the maximum rate of change you expect from your system? From that you can deduce the bandwidth, and from that, the cutoff of the low-pass filter (if necessary).

Alright sorry Pichenettes, but what do you mean by maximum rate of change ?is it how fast the system should react ? If yes than the faster it is the better!(meaning it would be more reactive)

> If yes than the faster it is the better!(meaning it would be more reactive)

Then if you want the fastest possible system, you want the largest bandwidth and you won’t be able to eliminate noise. At all. After all, if the voltage instantly shoots up by 1mV, it could be noise, or it could be a very fast and subtle change performed by the musician, right? So how could we tell them apart?

If you want to eliminate noise, you have to compromise on reaction time. A reasonable assumption would be to say that your output CV won’t change faster than the time taken by your arduino code to process the sensor data and refresh the output. Have you timed that? Or maybe there’s already a low-pass filter built into your sensor, so your data is already band-limited, and you can reproduce this limit on the output. If your sensor cannot detect a fast change, any fast change on the output is surely attributed to noise rather than to the sensor… So you’ll still be able to define a reasonable upper bound on the bandwidth and use this to correctly design your output filter.

I don’t agree with “the faster it is the better”. Good engineering would be: “the faster that still makes sense it is the better”. Beyond that, there is noise and waste.

Pichenettes, Thank you so much ! I’ll do the timing and works things out from there. Thanks your time and mind!
Cheers, Ashvin.

PS- I like the engineering code(“the faster that still makes sense it is the better”.). It makes more sense than mine. haha

Hey Everyone I need a little more explanation, though this has nothing to do with any mutable instruments product. Sorry about this.

For my dissertation I need to explain this circuit. I know the two buffer are there two isolate this circuit. But why is the potentiometer being feedback. Also hows is this circuit different from a normal RC circuit ?
Any help would be greatly appreciated.

Cheers, Ashvin.

Where do you see the feedback?

This is a basic RC filter with a buffer before and after it - so that the resistance of the load or the source doesn’t get added to that of the potentiometer - changing the overall gain or RC constant.

so it is just a simple rc circuit than? and the potentiometer changes the time constant. Why does the output of the potentiometer fed back into the potentiometer? also some people told me that this is an integrator rc circuit ?

> so it is just a simple rc circuit than?

Yes.

> Why does the output of the potentiometer fed back into the potentiometer?

It doesn’t feedback at all. It’s just a standard rheostat circuit.

> also some people told me that this is an integrator rc circuit ?

I’m not sure what is meant by “integrator RC” circuit here.

This is a plain RC circuit . Its transfer function is that of a one-pole low-pass filter (1 / (1 + RCs)) ; not that of an ideal integrator (1 / Cs).

so it this considered as portamento circuit? Thanks Pichenettes! Anyways I would like to buy you a beer sometime!

> so it this considered as portamento circuit?

Yes. Portamento can be done by RC-filtering a control voltage to replace abrupt transitions by slow exponential ramps.

Some people prefer having a linear transition - in which case you need a more complex circuit.