[sdiy] Voltage divider with RC filter - help with the basics, please
Steve Lenham
steve at bendentech.co.uk
Fri Jul 19 15:34:34 CEST 2013
Hi Tom,
I've had to do this recently and it isn't instinctively obvious. I would
suggest the following:
First, ignore the cap and establish your division ratio. Unless the
input impedance of the next stage is very much bigger than the
impedances in the divider, you will need to calculate
Rb' = the parallel combination of Rb and the load impedance
to give you the true value of the bottom leg of the divider. If your
load impedance is variable or not well-known then so will your division
ratio be - in that case you either need to buffer the signal or ensure
that the divider impedance is jolly low.
Then use your formula:
Vout/Vin = Rb'/(Ra + Rb')
to calculate appropriate values for Ra and Rb. To avoid having to keep
recalculating Rb', I suggest fixing Rb at something sensible and varying
Ra until you get the desired division ratio.
You now need to know the output impedance of your potential divider.
This is equal to the parallel combination of the top and bottom legs, i.e.
Rout = Ra in parallel with Rb'
Note that we use Rb' and not Rb, to make sure that the load impedance is
taken into account.
It is this output impedance that combines with the capacitor to form an
RC filter:
cutoff freq = 1/2pi*C*Rout
so you can then choose a value for C which gives the cutoff frequency
you want.
Hope this helps!
Cheers,
Steve L.
On 19/07/2013 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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