Cv spec questions

2 quick general spec questions about Mutable Instrument modules : What is the max +/- cv range that can be inputted? How are the circuits protected from over/under voltages?

Thanks!

> What is the max +/- cv range that can be inputted?

What do you mean by maximum?

a. Maximum in the sense of a parameter range (if you send a higher voltage, nothing happens)?

b. Maximum in the sense of not damaging the product?

If you mean a., it depends from module to module and input to input:

  • Level inputs (gain): 0V to +8V.
  • Inputs with built in attenuverters: -8V to +8V.
  • Inputs without built in attenuverters: -5V to +5V.
  • V/O inputs on recent STM32F4 project: -1V to +5V.
  • V/O inputs on earlier project: -3V to +8V.

If you mean b., it’s about -50V to +50V.

> How are the circuits protected from over/under voltages?

The module does not sense the input voltage, but the small current flowing through a large value resistor (100k).

Thank you very much!

I meant answer b.

Are your cv input op amps powered by single side or dual? It isn’t apparent on the schematic.

If you are saying the inverting pin doesn’t see any voltage, I imagine it is staying at a virtual ground, because of the non-inverting pin. Someone said this virtual ground only exists when you stay within the common mode input voltage. They said when you go above that the op amp doesn’t operate as expected and the over volts could damage the op amp. But you are saying it is ok from -50V to +50V, so maybe they are wrong, because you have been using these circuits for awhile.

Thanks again. I’ll keep reading the tutorials.

> Are your cv input op amps powered by single side or dual?

Single.

> It isn’t apparent on the schematic.

It is!

> I imagine it is staying at a virtual ground, because of the non-inverting pin.

It stays at ground potential as long as the output does not saturate - ie, as long as the input voltage is in the expected range. When the output saturates, V-:

  • either tries to go below ground, but then the internal ESD protection diodes on the MCP600x prevent it to go below ground. Excess current will flow into V-, but since this input can accept as much as 2mA, nothing wrong won’t happen unless the input goes above +200V (and before that, the input resistor will have blown because of excess heat dissipation).
  • either raises above ground, but never enough to go above Vdd.

I understand now. Thank you

Sorry. One more question. :slight_smile:

Looking at the Clouds cv input jacks, do the P2s (shorting jack tab) go to ground or they aren’t used?

Thanks!

It’s not connected, but the circuit will behave similarly if you ground it.

Ok, I’m back. Hehe. After studying op amps and learning LTSpice, I came up with this cv input to micro-controller circuit. It uses a rail-to-rail op amp (inverting setup) with a single 3.3V supply and a 1.65V virtual ground.

It has a jack for cv input and two pots. One pot is the offset and the other is for attenuation of the input.

I simulated and studied 4 scenarios and they seem to work :

1) For manual control :
nothing plugged into the input
attenuation pot turned all the way up in the schematic
offset pot=0 to 3.3V
output=0 to 3.3V

2) For -10V to 10V input :
offset pot all the way up in the schematic (3.3V)
attenuation pot turned to .165 ratio. This scales the input down to -1.65V to 1.65V
output=0 to 3.3V

3) For 0 to 10V input :
turn the offset pot halfway (1.65V)
turn the attenuation pot to .33 ratio. This scales the input down to 0 to 3.3V
output=0 to 3.3V

4) For 0 to 5V input :
turn the offset pot halfway (1.65V)
turn the attenuation pot to .66 ratio. This scales the input down to 0 to 3.3V
output=0 to 3.3V

Can you take a look when you get a chance and let me know what you think? Thanks

> Can you take a look when you get a chance and let me know what you think?

  • You’ll need two values of pots (10k and 100k) to make it work.
  • The input does not have a fixed 100k impedance - in your circuit the input impedance depends on the position of the attenuation pot.
  • Unless you don’t have many ADC channels on your MCU, there are many benefits in using one channel for unattenuated CV input, and two ADC channels for the offset and attenuation pots - with the attenuation/offset being done in software. The benefits of this software-based approach include: no loss of resolution for small modulation amounts (the input signal is acquired at full range and only attenuated in software), and a lot of freedom to get custom response curves for the offset and attenuation pot.

I would like to do the attenuation/offset in the software, but the micro-controller can’t input an unattenuated -10V to 10V. Given that I only have 3.3V psu isn’t this unavoidable?

Thanks for the advice!

> The input does not have a fixed 100k impedance – in your circuit the input impedance depends on the position of the attenuation pot.

Could you explain why it is not good practice to have the impedance not fixed at this input? It is still very high, but yes will change with the pot.

Is there a way to have a fixed 100k impedance here, but keep that pot and rest of the circuit?

Thank you, sir.

> I would like to do the attenuation/offset in the software, but the micro-controller can’t input an unattenuated -10V to 10V. Given that I only have 3.3V psu isn’t this unavoidable?

I’m not talking about scaling the input CV to the [0, 3.3V] range. You’ll have to use an op-amp for that, anyway, yes. I’m talking about the two functions performed by your pots. You’ll have more freedom if those two pots are directly read by two MCU pins and if you do the maths in software.

> Could you explain why it is not good practice to have the impedance not fixed at this input?

Let’s say you have a CV generator with the 1k output protection resistor going to module A (your module) and module B (which has a fixed 100k input impedance) through a passive multiple.

Adjusting the attenuator on your module will change the total impedance “seen” by the 1k resistor, so it will slightly affect the amplitude of the signal going into module B.

> Is there a way to have a fixed 100k impedance here, but keep that pot and rest of the circuit?

Yes: the input signal goes through a 100k resistor the wiper of the pot. One terminal of the pot is grounded, the other connected to the V- input of the op-amp.

> Yes: the input signal goes through a 100k resistor the wiper of the pot. One terminal of the pot is grounded, the other connected to the V- input of the op-amp.

Ok. Like this? See picture. I have the jack, then the 100k resistor into the wiper of the pot. Doesn’t the impedance still change though when turning the pot? What size pot would you recommend?

Thanks again so much for the detailed replies! It is really helping a lot and I feel like I’m almost there.

:slight_smile:

With a 10k pot, the worst case scenario is 102.5k input impedance when the pot is in center position (from the wiper, we “see” two 5k resistors to ground in parallel), and the best case is 100k impedance when it is at either extreme setting, so the variation in input impedance is narrow.

Not constant, but still much better than the 100k to 200k you get with your first proposal (and in your first proposal, you can’t change the value of the attenuation pot to a value like 10k because this directly impacts input impedance).

Nice! I like it.

Last concern is the protection we were talking about earlier when the cv input is too much. I’m assuming my limit is near the same as your original circuit, -50V to 50V before damage? Do the ESD diodes or anything else send any of these over or under volts to the power supply that is powering the op amp? The reason I ask is that my 3.3V power is coming from a power output from the micro-controller and don’t think it could handle those.

Thanks again! Enjoy your day.

> Do the ESD diodes or anything else send any of these over or under volts to the power supply that is powering the op amp?

They don’t send “volts”, they steer current to ground or to the power supply. Because your input impedance is about 100k, a voltage of 50V at the input will translate into a current of 0.5mA flowing somewhere in your circuit. 0.5mA is quite small, I really doubt some element in your circuit won’t handle that…

> The reason I ask is that my 3.3V power is coming from a power output from the micro-controller

Really? Which micro-controller is that?

Teensy 3.1

Do you have a link to the schematics of the board? I find it odd that the 3.3V comes from the MCU itself. I would find it more likely that there’s a +3.3V regulator on the board.

yep :

https://www.pjrc.com/teensy/schematic.html

https://www.pjrc.com/teensy/pinout.html

i power this Teensy with 5.34V but using the 3.3V pin to power this op amp project.

I’m having trouble with the new 100k before the 10k pot input revision. Most of the volts from the input jack stay on that 100k resistor and very few are on the 10k pot. Will this be a problem with so few volts actually making it to the op amp through this 10k attenuator? Bad resolution because the entire range going into the op amp is less than a volt.

Looks like the offset pot and other resistors will have to be scaled back too to match this new very small voltage range from the attenuator?