# [sdiy] Single supply op-amp filter biasing question

Tom Wiltshire tom at electricdruid.net
Sat May 6 00:17:36 CEST 2017

```Ah, yes, thanks Walker, that makes sense.

On 5 May 2017, at 21:41, Walker Shurlds <walkershurlds at gmail.com> wrote:

> No current goes into the op-amp. No DC current goes through the capacitor. So no DC current goes through the 20k resistor either, and therefore it has the same DC voltage at either end.
>
> Hth,
> Walker
>
>
> On May 5, 2017, 14:17 -0500, Tom Wiltshire <tom at electricdruid.net>, wrote:
>> 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.
>>>
>>> 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?
>>>
>>>
>>> Thanks, Tom
>>>
>>>
>>>
>>
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