[sdiy] Odp: Re: Single supply op-amp filter biasing question

[Roman]

2V on R1, leaves (9-2)=7V on RX. So RX is 7/2 times bigger.  or in another words  Ubias = 9V * (R1/(R1+RX))   You will loose a bit of amplitude too, as the divider R1/RX attenuate by 2dB. This can be compensatted by adding some gain to Sallen-Key.   How about blocking the DC at filter output with 22uF, so then you can set the bias at 4.5V with some considerably bigger resistor like 300k, that will not influence neither filter characteristics, nor amplitude in any significant way. It will need quite a few seconds after power up to settle though.   Roman  Dnia 5 maja 2017 17:02 Tom Wiltshire <tom at electricdruid.net> napisał(a):  Thank you very much, Roman.   Where does 3.5 times greater come from? I could see 4.5 times greater, since 9/2=4.5, but how does 3.5 work?   Thanks,  Tom    On 5 May 2017, at 15:25, Roman Sowa <modular at go2.pl> wrote:   yes, you could, but I would just add the "X" resistor straight to 9V and skip 100k trimmer and 22u.  56K in parallel with X is your new value of the first resistor for filter calculations so keep that in mind.   To rise bias by about 2V, you need X to be 3.5 times greater than R1 (let's call that what you have there as 56K). And R1 in parallel with X should be 56k.  Now do your math.   Roman   W dniu 2017-05-05 o 16:03, Tom Wiltshire pisze:   capacitor. But since the signal that I'm trying to retrieve with this  filter is a DC level, using a capacitor (AKA highpass filter) to  connect the input doesn't seem feasible. So could I do something like  this, and how would I calculate the value of X?   electricdruid.net electricdruid.net   Thanks, Tom
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