I am in search for a way to only limit the negative slew rate while leaving rising parts of the signal untouched. The signal will be anything from Subaudio till 20kHz, with potentially arbitrary waveforms and the amount of slew should be changable (via potentiometer, control voltage or current).

I found multiple ways of slewing both rising and falling edges, but nothing to slew only falling edges. I thought about some way to do it with positive and negative peak detectors and so on, but this gets quickly too complicated and to big for my intended applications (mostly a ±12V Audio environment).

The resulting slew can be linear or logarithmic, it doesn't really matter, for me it is more about getting unstuck with this topic.

The added slew is displayed in red (with different fall rates), the original waveform is the square wave.

Variable Slew on the falling part of a wave only

  • \$\begingroup\$ You can use diodes to create independent slew rates for rising and falling edges. Each diode would be in series with a different resistor. \$\endgroup\$ – mkeith Apr 29 '17 at 16:02

Something like this, perhaps? The diodes effectively give you a different R value for different input edges in this RC lowpass filter.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Oh yeah – it is so easy. I made a short LTSpice sim which yields this: i.imgur.com/AgrF7hO.png I think with small adjustments this should do what I want. Thanks a lot! \$\endgroup\$ – ato Apr 29 '17 at 16:11
  • \$\begingroup\$ Adding a bit of active circuitry can let you linearize it as well, by using the resistor as a component of a current source instead of a conventional filter. \$\endgroup\$ – Hearth Apr 29 '17 at 16:12
  • \$\begingroup\$ Also, if this answered your question, click the "accept answer" button--that will take it off the list of unanswered questions, as well as help people with the same problem find it. \$\endgroup\$ – Hearth Apr 29 '17 at 16:16
  • \$\begingroup\$ For my applications a logarithmic response is enough, but somebody else (or my future me) might be happy about this suggestion, thanks! \$\endgroup\$ – ato Apr 29 '17 at 16:19

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