voltage divider inside differential op amp configuration

Question

I'm working on a DIY analog synth based on the CEM3340 chip. When feeding it 12V at VCC it will output waves at different amplitudes that I'll want to scale to 5V to -5V:

Square: 0-10.7V

Sawtooth: 0-8V

Triangle: 0-4V

To scale those waves to -5V - 5V I'm using a differential op-amp config with the values calculated here as seen above:

My question is: using that calculator, I need to keep VRef relatively low, around 2.5V otherwise I start to get negative values for resistors. In my circuit, I have a 10V ref voltage that I would like to use as VRef. Is it possible to change the op-amp config to divide my 10V voltage ref to 2.5V? I've tried to use a voltage divider before R15/R17/R19 but failed. I would also like to keep the config non-inverting. Is something like this possible with only one op-amp unit?

Thank you!

Answer 1

You can replace R15 (R17, R19) and the 2.5V with its Thevenin equivalent: two resistors arranged as a voltage divider to yield 2.5V, with a parallel impedance of R15 (R17, R19). For example, 27K and 81K will drop 10V to 2.5V with an effective resistance of 20K.

However, it doesn’t look like the values shown will generate your desired -5V to +5V swing:

The R16/R12 (etc) divider is converting all inputs to a 0V – 3.3V range. When the input is 0V, so is the negative input of the opamp. So current will flow from the 2.5V source through 20K (R15), and that same current will continue to flow through another 20K (R21) which will therefore generate the same voltage (2.5V) across it, so the output voltage is going to be -2.5V, not -5V.

When the input is at its 3.3V peak, the voltage across R15 is now 3.3-2.5 = 0.8V, and this will also be the voltage across R21. So the output will be 3.3V + 0.8V = 4.1V, not +5V.

What your circuit does is reduce the input to a known swing, 0-3.3V in your case, and amplifies it by 2 (1 + 20K/20K). This will result in a 6.6V voltage swing with some sort of voltage offset.

What you want is a gain of 3 to get your 10V swing, so double R21 or halve R15, something like this:

Answer 2

A simple general formula for the two resistors to create the Thevenin equivalence given the supply voltage Vs and the desired equivalent voltage Vt and resistance Rt (Vs >= Vt) can be found with a bit of algebra:

^{simulate this circuit – Schematic created using CircuitLab}

R1 = \$\frac{V_S}{V_T}\cdot R_t\$

R2 = \$\frac{V_S}{V_S-V_T}\cdot R_t\$

Eg. Rt = 20kΩ, Vs = 10, Vt = 2.5

R1 = 80kΩ

R2 = 26.667k

Answer 3

The source of \$V_{REF}\$ is being loaded by multiple modules, each sinking/sourcing current through its impedance, current which varies with every module's output. This current will cause \$V_{REF}\$ to fluctuate as a function of all three module outputs.

The answer by @td127 is great, but it must be repeated for each individual module, to avoid loading by the others. If you want to share a single 2.5V reference between multiple modules, then you require a zero output impedance source for \$V_{REF}\$. That can obtained with negative feedback, using an op-amp voltage follower:

^{simulate this circuit – Schematic created using CircuitLab}

This reference node REF can be shared with multiple modules, but it can only sink/source 10mA or so. This means that every load upon it must be light. If you have 3 modules sharing this reference, each must sink/source no more than about 3mA. Ideally you should aim for much less than this.

This requires that your choice of resistors such as R15 and R21 must pass no more than 3mA with the module's output at either extreme, +12V or -12V.

For instance, for the square output module, the highest REF current flows when SQUARE_OUT is at -12V, which would place \$+2.5V - (-12V) = 14.5V\$ across R15+R21. This would invoke a current of \$\frac{14.5V}{20k\Omega+20k\Omega}=360\mu A\$, which is fine.