[sdiy] Voltage divider with RC filter - help with the basics, please
rburnett at richieburnett.co.uk
rburnett at richieburnett.co.uk
Fri Jul 19 15:43:14 CEST 2013
Hi Tom,
You just need to work out the resistance "seen by the capacitor". So
if you put yourself in the place of the capacitor C in your example,
then it will see Ra in parallel with Rb in parallel with Imp. The
equation you need to find R is: 1/R = (1/Ra) + (1/Rb) + (1/Imp)
(You can also substitute Ra with [Ra + Rsource] if the source impedance
of whatever is driving it cannot be assumed to be zero.)
You then use this "total parallel R value" in combination with your
chosen capacitor C to give you the -3dB break frequency for your lowpass
filter.
As you said, you can already work out the DC gain of the "frequency
independent" potential divider part by assuming the capacitor is
effectively open-circuit at 0Hz.
The total response is the combination of the "frequency independent"
gain and the 1st-order LPF response. i.e. It will have a flat passband
gain determined only by the resistors up to the frequency where the pole
resulting from the parallel resistance and capacitance starts to kick
in. At this point the response will be a further 3dB down, and will
ultimately continue to fall by 6dB for every octave. (PS. You can put a
resistor in series with the capacitor to introduce a zero if you ever
want to kill off the pole at some higher frequency to make a shelving
filter!)
You can also make cheap half-decent 2nd and 3rd order LPFs without
buffers simply by cascading 1st-order RC filters, provided you make the
impedance of each stage about 10 times that of the preceeding one.
Hope this helps,
-Richie,
On 2013-07-19 13:49, Tom Wiltshire wrote:
> Hi All,
>
> I've got the following situation:
>
> http://www.electricdruid.net/FilterDivider.png
>
> It's a voltage divider, with a cap across the lower resistor. It will
> feed another input, which will also have an input impedance, which is
> going to affect things.
>
> Now, what I already know is that for a voltage divider, the voltage
> out is Rb/(Ra+Rb) * Vin, and also that for a simple RC filter, the
> cutoff would be 1/2PiRC - in this case Ra and C. But I also know that
> putting them together isn't that simple.
>
> How do I combine these two to make a voltage divider filter so that I
> can work out how to get 80% volume with cutoff at 100Hz, for example?
> Furthermore, if I feed a low impedance, that'd affect the situation
> too, so how do I allow for that?
>
> My specific case is a MCP4822 DAC (which outputs 0V to 4.095V)
> feeding a SSM2164 CV input, which only accepts 0V to 3.3V, and has an
> impedance of 5K. So I need a voltage divider that gives me 3.3/4.095 =
> 0.806 of Vin. Fc of the filter isn't decided yet, but think CV
> filtering, so probably <1000Hz.
>
> Some pointers would be appreciated. I mostly do pretty well, but
> there's times where some formal education in this stuff would be
> helpful, and that's when I start asking you all!
>
> Thanks,
> Tom
>
>
>
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