0
\$\begingroup\$

I'm into DIY module-making for modular synthesizers. I came across this stripboard layout from the musician Look Mum No Computer for a module that will take an input signal (usually ±5V range of any shape via the "CV IN" jack), pass it through an ADC, and return high/low outputs corresponding to bit value, which can be used for various applications in modular synthesis.

stripboard layout

I've made the circuit and it operates great. Just one problem, when the input signal drops into negative voltage, the ADC does not generate any outputs, which is an inconvenience as the module only works half the time, in essence.

My idea for fixing this issue was to add a bridge rectifier to the input signal coming in from the "CV IN" jack, thus ensuring that the input signal always stays positive. Although I am unsure exactly how I would go about adding this. The circuits for bridge rectifiers that I have looked at in the past require a positive/negative input and generate a positive/negative output, and "grounding" these negatives to each other negates the effect of the bridge rectifier. The jacks are also mounted to a metal plate, which is grounded to the GND line. For clarification, the bridge rectifier would be added between the "CV IN" jack and the green wire.

So my question is, can you add a bridge rectifier to this circuit and if so, how exactly? Or is there another way to essentially take the absolute value of the input signal coming in via the "CV IN" jack?

If clarification is needed, please ask, will be happy to provide more info.

\$\endgroup\$
3
  • 4
    \$\begingroup\$ Alternatively, AC couple and reference/offset your input signal so the the input's mid-scale becomes your ADC's Vref/2 (likely 2.5V). Scale the input to 2.5V +/- 2.5V. \$\endgroup\$
    – Chris Knudsen
    Dec 7, 2020 at 20:47
  • 1
    \$\begingroup\$ This sounds like a job for offset, either via AC coupling or if you need coverage down to DC, via an op-amp. Really though the goal is a bit indistinct - this kind of thing can be fun to play with but numeric simulation is going to be a faster route to any useful behavior, either to use as is, or to recreate in hardware if you really want to. \$\endgroup\$
    – Chris Stratton
    Dec 7, 2020 at 20:52
  • \$\begingroup\$ @Jtaft121 Please use the circuit diagram tool when posting, thanks \$\endgroup\$
    – Voltage Spike ♦
    Dec 7, 2020 at 21:03

1 Answer 1

Reset to default
4
\$\begingroup\$

Don't bother with the rectifier, just move the whole signal up into the ADC's range.

Like this:

schematic

simulate this circuit – Schematic created using CircuitLab

  • C1 lets your audio circuit "float up" so that is is no longer +-5V but rather is entirely above 0V.
  • R1 and R2 move your audio to be "riding" on a 2.5V offset.
  • R3 plays together with R1 and R2 to reduce the audio from 10V peak to peak to just 5V peak to peak.

If I haven't bungled it somewhere, that'll reduce the level of your audio to fit your ADC and move it to where the ADC can properly sample it.

The bridge rectifier would "chop off" a part of your audio signal. Nothing between -0.7V and 0.7V would get through. Quiet bits of your music would be "lost" to the ADC.

\$\endgroup\$
3
  • \$\begingroup\$ Ah I hadn't thought of using an offset, this makes much more sense for this application. Thank you for providing the above circuit. It turns out that a capacitor value of 100uF works out perfectly for a input signal of +-5V range. The reason for the capacitor value change is because the input signal via CV IN isn't quite an audio signal, its much lower in frequency, typically under 10Hz. \$\endgroup\$
    – JTaft121
    Dec 7, 2020 at 22:30
  • \$\begingroup\$ Along this line of thinking, I suppose a larger capacitance would allow for preservation of a 0-5V output for really low frequencies. \$\endgroup\$
    – JTaft121
    Dec 7, 2020 at 22:38
  • \$\begingroup\$ The circuit was assuming an audio signal and thus was ac coupled. You really want dc coupling for CV signals. I’d use an op-amp for that. \$\endgroup\$
    – Kartman
    Dec 8, 2020 at 1:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.