AE Modular conversion of Grids


#1

Hello all,

just got an ae modular ( https://www.tangiblewaves.com/ ) and now i am trying to convert grids from eurorack to ae modular. In ae modular, all voltages are between 0 to 5V, there are no negative voltages.

So this is one CV input from Grids. As far as i can tell it is some kind of inverting summing amplifier and it needs a negative 5V at one of it’s inputs to work properly.

The problem is: AE Modular doesn’t know negative voltages. So instead of an inverting summing amplifier i will use a non inverting one and then correct the inverted readings in the code.

So i came up with this (i found it on the internet to be honest) which is a non inverting summing amplifier:

So basically it works as expected. The Problem is: when the CV input is not connected then the summing doesn’t work anymore. So one could just use a switching jack which ties the cv input to ground in case noting is connected, but we can’t use these for ae modular because it uses even smaller connectors (the ones found on an arduino board for example)

So i came up with this solution:
image

In case there is nothing connected to the input then the whole input is tied to ground via R7. Comparing to R2, R7 is so small that it doesn’t affect the summing that much which should be ok.
But the problem here is: R7 needs to be small to not affect the summing too much, but when it is small it will suck a lot of current from the cv input when there is something connected.

So to solve this issue i came up with this:

Now the pull down resistor R6 is so big that the current it sucks from the input should be small enough and it also doesn’t affect the summing any more because of the buffer.

So this circuit finally works, but at the cost of having another opamp.

The question i now have is:

Is this complete bullshit or does this sound like a reasonable approach?
Is there maybe a better solution which doesn’t need the second opamp, or maybe no opamps at all?


#2

Note that the LT6002 and MCP6002 are different beasts… but it works for this sim because they both have rail-to-rail outputs and work with low supply voltages!

The LT6002 is sluggish (32kHz gain-bandwidth product? what???)

Here’s a simple inverting summing circuit that does not require a negative supply.

(see 2 posts below for the correct, working schematics).


#3

Thanks for your reply.

But for example when POT = 2V, CV = 1V, then the original Grids Circuit will put out:

-100 * ( 2/100 + 1/100 - 5/100 ) = 2V which is then read by the atmega.

So basically 5V - ( Vpot + Vcv )

which the circuit you posted doesn’t, it will give me 4.5V (still have to read up why and check my simulation).

The last circuit posted by me above will give me Vpot + Vcv = 3V.

So if i then just do 5V - 3V in the atmega (which is the inversion my circuit can’t do) i will also end up with a 2V reading in the atmega, so that should be fine?

Or am i completely wrong here?


#4

Decent intro to what’s going on here: https://ocw.mit.edu/courses/media-arts-and-sciences/mas-836-sensor-technologies-for-interactive-environments-spring-2011/readings/MITMAS_836S11_read02_bias.pdf#page=5

The output of @pichenettes’s circuit (with bias voltage Vb) is Vb + R6/R3(Vb-Vpot) + R6/R12(Vb-Vcv)… since R3=R6=R12, output is 3 * Vb - (Vpot + Vcv). A bias voltage of 2.5V here (5V * 100k/(100k+100k) is why you get 7.5-(2+1)=4.5V.

if you wanted the output to be 5V - (Vpot + Vcv) you could set the bias voltage to (5/3)V which you should be able to do by changing R1 to 200k!


#5

Thanks for correcting me on this!

My simulation appeared to work because I had a node labelled VPOT and the other POT :confused:

For sake of clarity, here is the correct schematics:

50


#6

Thank you both, with the correct bias voltage it finally works.

But there is still a problem left.

If there is nothing patched into the cv input then R12 and Vcv are no longer part of the equation, so i would get 1,66 + 1* (1,66 - 2 ) = 1,32.

I cannot use a switching jack so that cv is connected to ground when nothing is patched in because i have to use this type of connectors:
image

So for that reason i added the second opamp in my schematics which sets cv to 0V when nothing is connected:

If i just do it like this:
image
Then the CV resistor would be 200k in the equation when CV is not connected, so that will give me the wrong results. I could lower R11 to 1k so that the error it adds to the equation is reasonable small, but then i think it will suck too much current from the cv input if something is connected, right?


#7

Ahhh finally got it! I needed to see those connectors :slight_smile:

I can’t think of a single op-amp solution (for now).


#8

So thanks again @all for the help. I think i will then use that design.