[sdiy] how to calculate cutoff in an OTA filter?

[Chris McDowell]

Thanks, Guy! Yes this was my error.

> Depending on exactly what you mean by 1/gain

I say 1 / gain because the voltage gain of the divider, in my case of 100k and 220, is 0.002195 etc etc. To adjust the gm given in the LM13700 datasheet, we have to "undo" that gain, so instead of just gm = 19.2*Iabc, we have total_gm = 19.2*Iabc*(1/divider gain), otherwise gm would get smaller but the correction needs to be the other way. Maybe this is what you said anyway and I'm talking to the wind lol. 

thanks again for working through that with me. any expansion on this topic is welcome if anyone wants to do so. 

Cheers, 
Chris 



> On Dec 11, 2020, at 1:25 PM, Guy McCusker <guy.mccusker at gmail.com> wrote:
> 
> Depending on exactly what you mean by 1/gain (I would just call this
> value the gain of the input attenuator -- some fraction quite a lot
> lower than 1) this seems right. You just want to know how many amps
> out per volt in to the OTA block taken as a whole. I think what was
> confusing you in the previous calculation was that you weren't
> dividing by the input voltage to get the transconductance in
> amps-per-volt (aka mho or Siemens).
> 
> 
> On Fri, Dec 11, 2020 at 7:05 PM Chris McDowell <declareupdate at gmail.com> wrote:
>> 
>> you need to take into account the voltage divider at the input of the OTA.
>> 
>> 
>> ah, okay, I thought I understood that, but was not doing anything about it! it appears to simply be, for the 13700 at least:
>> 
>> gm = 19.2 * Iabc * (1 / gain of input attenuator)
>> 
>> Sound reasonable? Spreadsheet seems to like it ¯\_(ツ)_/¯
>> 
>> 
>> On Dec 11, 2020, at 12:38 PM, Tom Wiltshire <tom at electricdruid.net> wrote:
>> 
>> I think Guy means that if you’re looking for the unity-gain point, you need to take into account the voltage divider at the input of the OTA. There’s a big attenuation going in, so the gain required to achieve unity gain overall will depend on the component values you choose to bring the input down to the right level. However, it *won’t* depend on the input level itself.
>> 
>> Is that right, Guy?
>> 
>> Tom
>> 
>> 
>> 
>> On 11 Dec 2020, at 18:18, Chris McDowell <declareupdate at gmail.com> wrote:
>> 
>> How would I go about this without making the cutoff frequency seem voltage dependent? If I derive gm from Vin/Iout, then the resulting frequency is dependent on the input voltage, but we know it is not. Is it just some algebra I'm overlooking to separate it out?
>> 
>> 
>> 
> 
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