[sdiy] Voltage divider with RC filter - help with the basics, please

[rburnett at richieburnett.co.uk]

Totally agree, Russell.  Most circuit simulation packages will also let 
you enter these laplace transfer functions directly in blocks so you can 
easily examine their responses on bode plots right alongside real 
circuits composed of actual resistors and capacitors.  -Richie,

On 2013-07-23 15:27, Russell McClellan wrote:
> Hey Tom,
> 
> Richie already gave a complete and accurate response, but maybe you'd
> be interested in a standard, general technique for this sort of
> passive network.
> 
> First, I convert everything to "impedances" - this means treated the
> capacitor as an "impeder" with impedance 1/sC.  The units of impedance
> are ohms, and resistors translate directly - their impedance is R.
> Don't worry about what the "s" "means" other than that it depends on
> frequency (it's a fancy math thing called a "laplace variable").  The
> cool thing about impedances is that they behave and combine exactly
> like resistors - you can use the 1/R_parallel = 1/R_1 + 1/R_2 rule for
> parallel impedances and you can use addition to find serial impedance.
>  Using these two tools you can quickly build up your whole network
> into a single equation - this equation is called the "Transfer
> Function" of your system.  Interpreting this transfer function is
> where your lack of formal training may get you into trouble - but, if
> you've correctly made a low-pass filter, your function should look
> like x/(1 + ys) where x and y are combinations of R and C values.  If
> you can't use algebra to turn your equation into that form, it isn't a
> lowpass filter.  As richie mentioned, if it's something like (1 +
> xs)/(1+ys), it could be a shelf filter or an allpass filter (i.e., a
> phasor).  In the lowpass case above, the cutoff is just 1/(y*2Pi).  Of
> course there's deeper reasons why these equations work the way they
> do, but that's probably out of scope for an e-mail exchange.
> 
> The reason I prefer to use this slightly pedantic method to analyze
> even relatively simple networks of capacitors and resistors is I don't
> have very much intuition.  This method wouldn't be used directly by
> experienced folks except in relatively complicated situations.  For
> me, it can be hard to guess whether your network even is a low-pass
> filter just by looking at it.  Calculating the "cut-off" doesn't mean
> much if you don't even know whether it's a low-or-high pass!
> 
> Thanks,
> -Russell
> 
> On Fri, Jul 19, 2013 at 3:26 PM, Tim Parkhurst 
> <tim.parkhurst at gmail.com> wrote:
>> Does the DAC have enough output drive to play nice with the 
>> relatively
>> low input impedance of the 2164's control port? I've always put an
>> op-amp buffer in front of the control port (unless I'm just driving
>> 'manually' from a single pot).
>> 
>> Cheers,
>> 
>> Tim (If I don't ask the stupid questions, who will?) Servo
>> ---
>> "Sire, the church of God is an anvil that has worn out many hammers."
>> - H.L. Hastings
>> 
>> 
>> 
>> On Fri, Jul 19, 2013 at 5:49 AM, Tom Wiltshire 
>> <tom at electricdruid.net> 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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