[sdiy] Protection Device?

Tim Ressel timr at circuitabbey.com
Tue Jun 19 23:36:49 CEST 2018

the issue is this: the uP input cannot tolerate voltages much beyond its 
rails, so the max range is -0.3V to +3.6V (in my case). The inputs must 
be clamped that range.


On 6/19/2018 2:22 PM, O Gillet wrote:
> The CV input circuitry of pretty much all my modules look like this:
> https://imgur.com/a/Xz5YN0X
> The op-amp has R2R outputs and is single-supply, powered by 3.3V.
> Benefits:
> - The V- input of the op-amp is a summing point - you can attach there
> as many CV inputs or pots as you want. It's very helpful in some
> situations to have both a CV input and a pot to control a parameter!
> - If you connect an external signal straight to an MCU ADC input, the
> range of the CV will be necessarily equal to the range of the
> microcontroller ADC input. It's probably OK if you use a +5V
> microcontroller and want 0V to correspond to the minimum value of the
> parameter, +5V to the maximum value of the parameter... But what if
> you want the range of the parameter to be -5V to +5V ; or 0V to +8V?
> Or what if your microcontroller is powered by +3.3V? The schematic I
> have posted covers all these cases - you just change the resistors to
> get different scale/offset values.
> - The capacitor acts as a 1-pole low-pass filter which removes some of
> the high frequencies in the CV signal - providing cleaner readouts.
> - ADCs do not like being driven from sources with high ouptut
> impedances. For example, the ADCs on the AVR want a source impedance
> of 10k or lower - to rapidly charge the S&H capacitor which is part of
> the ADC circuit. The output of the op-amp works as a very low
> impedance source for the ADC - and will simultaneously leave you in
> control of the input impedance of your module (100k is an implicit
> standard for Euro modules). Standardizing all input impedances to 100k
> is great - it allows consistant behaviour when using passive modules.
> - Of course the op-amp input will never "see" extreme external
> voltages, because you're not directly exposing any gate to the
> external world - every input voltage goes through the 100k resistor.
> If you connect +50V to the CV input, well, that's only 0.5mA flowing
> through the input resistor (and 25mW dissipated by the resistor).
> So yes, an op-amp might look a bit too much, but it kills a whole
> flock of birds with one stone.
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--Tim Ressel
Circuit Abbey
timr at circuitabbey.com

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