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

[Roman]

Don't be sorry, your questions are one of the reasons I'm still here.  I know it's 2V, because you WANT to move it all up by to 2V, right? We're talking about DC bias so all caps disappear, and 20k becomes a wire, because there's no current flowing to opamp input. So if we want 2V across R1 (56k), we can drive that node from 9V via 3.5xR1 resistance creating nice voltage divider.  So if PWM input is at 0V, this gives exactly 2V at the output as you require. However, when PWM is at 5V,the output is only about 5.9V, not 7V because of attenuation I mentioned. Also mean level is about 4V, not 4.5V as required.  What I should have really said is 2.25 times R1, taking PWM's mean value (2.5V) as voltage divider's base, not 0V. Then output peak-to-peak is even further attenuated, down to 3.46V (-3dB), but now at least it's centered around 4.5V.   Roman  Dnia 5 maja 2017 21:14 Tom Wiltshire <tom at electricdruid.net> napisał(a):  Ok, sorry, but I'm going to have to keep asking more questions!   How do you know there's 2V across R1? I think you're working with some assumption that I don't know about.   I'm guessing it's something like this:   0V at the input means 2V at the output, so there must be 2V at the -ve input of the op-amp. Consequently there's 2V at the +ve input too. Therefore there's 2V at one end of R1 and 0V at the other end. What I don't understand here is where the 20K resistor disappeared to?   Thanks,  Tom   On 5 May 2017, at 18:20, Roman <   modular at go2.pl > wrote:   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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