[Noisebridge-discuss] seeking circuit idea: frequency to voltage

[jim]

you might get the logarithmic  control by using 
a logarithmic pot rather than a linear pot. 


On Fri, 2009-12-11 at 11:17 -0800, Jonathan Foote wrote:
> In that case, Lamont's circuit should work pretty well. You might want
> a log function in there to match human pitch perception. One of the
> bargraph chips (forget the number) has a log response.
> 
> On Fri, Dec 11, 2009 at 11:10 AM, Michael Shiloh
> <michaelshiloh1010 at gmail.com> wrote:
> > in my case, it is a square-ish wave (555 oscillator) and so is
> > absolutely not a pure (sine) wave.
> >
> > Jonathan Foote wrote:
> >> That's only going to work for a pure tone, and not for speech or music.
> >> Typical audio is harmonically rich, meaning lots of frequencies mixed together.
> >>
> >>
> >> On Fri, Dec 11, 2009 at 12:01 AM, Lamont Lucas <lamont at cluepon.com> wrote:
> >>> Lamont Lucas wrote:
> >>>> Use a 555 as a monostable (one shot) vibrator that kicks out a fixed
> >>>> width pulse every time the input voltage crosses zero (or some higher
> >>>> point at the Vthreshold, you know, one fixed-width pulse per input
> >>>> waveform cycle), then a 741 configured as an integrator after that.
> >>>>
> >>>> As long as your fixed pulse width is less than 1/20000th of a second,
> >>>> and you integrate the output, you should get a output voltage that is
> >>>> linearly related to the input frequency, with higher voltage indicating
> >>>> higher frequency all the way from 0 Hz to 20kHz.
> >>>>
> >>>> that's a 555, 2 caps, and 3 resistors for the monostable vibrator, plus
> >>>> a 741, a resistor and a cap for the integrator.
> >>> Oh, and to go from a linear voltage to a bar graph...hm.  Maybe a
> >>> resistor ladder with LEDs on the legs that will only turn on when the
> >>> voltage at each node in the ladder raises above the forward voltage drop.
> >>>
> >>>
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