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wm , 12-09-2024, 07:52 PM

Hello @Robert Feranec , hope all is well. Is Phil on this discord? I have a question about his course - trying to implement the 3rd order Butterworth filter, but not getting the same cut off frequency. Thanks.

Sniper2 , 12-09-2024, 08:00 PM

u sure the op amp is fast enough ?

QDrives , 12-10-2024, 04:41 PM

R4 and C5? You already have about 50dB/dec prior to that.

wm , 12-10-2024, 06:26 PM

The choice is based on Phil's example for his bias generator

wm , 12-10-2024, 06:28 PM

sorry i don't understand. The cut off frequency should be 25kHz

Sniper2 , 12-10-2024, 06:55 PM

Phil is here since there is a phil with the logo

Sniper2 , 12-10-2024, 06:55 PM

also i think you are missing something from aanother sheet

wm , 12-10-2024, 07:46 PM

ok, what is his handle? What am i missing vcom?

QDrives , 12-11-2024, 03:25 AM

Yes, you are missing Vcom.
But if the cut-of frequency should be 25kHz, setting Vcom to Gnd for instance, would make it even lower in frequency.

Sniper2 , 12-11-2024, 05:16 AM

Well isn't that the point?

Sniper2 , 12-11-2024, 05:17 AM

And yes vgnd was missing in my opinion

wm , 12-11-2024, 06:33 PM

sorry i'm confused. i've set vcom to ground and still get the same graph - i'm expecting to see the roll off from 25khz, but seeing it from around 1khz

QDrives , 12-11-2024, 09:01 PM

1) The roll-off is at about 7...8 kHz, it is where the gain line crosses -3dB (or a bit lower since a 1Hz it is already negative).
2) You probably still looking at IN_AA_BUF_OUT and not V(ADC_IN+,ADC_IN-)

Sniper2 , 12-11-2024, 09:07 PM

the AC is gnd centered not Vgnd centered

QDrives , 12-11-2024, 09:12 PM

Do you mean Vcom?

Sniper2 , 12-12-2024, 04:27 AM

Yes

wm , 12-12-2024, 06:52 PM

ok, i'm now looking at both V(ADC_IN+), and V(ADC_IN-) and it still doesn't seem right.

QDrives , 12-12-2024, 07:44 PM

Two things
1) You are not looking at V(ADC_IN+,ADC_IN-), but V(ADC_IN+) and V(ADC_IN-). Big "**difference**" (hint).
2) The cut-off frequency will not be higher than you have at IN_AA_BUF_OUT, it probably will be a little lower.
You may want to check at different places, stating from the input, to see the cut-off frequencies there and adjust the R and C values as needed.

wm , 12-14-2024, 09:19 AM

Thanks for the reply. Sorry i'm new to LTSpice and don't know how to get the difference between the two points. The cut-off frequencies are low even from the input. This is meant to be a completed circuit ready for pcb design. i don't understand what's wrong.

QDrives , 12-14-2024, 08:25 PM

1) To get the difference...
1a) You see the red probe in LTSpice, left click and keep pressed (drag) to the 'reference' signal. You should now see a black probe. Release the mouse button.
1b) I gate you the expression. So right click on the expression on the graph plot. Change the expression to the one I gave.

2) A filter can only reduce the signal strength (except for resonances, as is also visible in your design.)

3) So if the signal is already filtered to below 10kHz in the first stage (R1 + C1), the end cannot have a greater frequency.

Phil , 12-14-2024, 10:18 PM

@wm The issue with your simulation is that you are not AC coupling your input (V2) to the op-amp/filter. Add a series AC capacitor (value can be set to 1 to make it otherwise 'invisible') between V2 and R1 - then you will get the correct frequency response.

Phil , 12-14-2024, 10:24 PM

Phil , 12-14-2024, 10:24 PM

Induranga Randika , 12-14-2024, 10:37 PM

hey, is this from which course of Phils ?

wm , 12-16-2024, 07:12 PM

3rd order Butterworth filter - AC analysis - Mixed-Signal Hardware Design with KiCad

wm , 12-16-2024, 07:30 PM

Thanks Phil! It works, but why does it need the capacitor, i'm using an ac source - is there a dc offset somewhere?

wm , 12-16-2024, 07:31 PM

Thanks for your reply.

wm , 12-16-2024, 07:31 PM

hello, its Mixed-Signal Hardware Design with KiCad

Phil , 12-16-2024, 07:38 PM

I believe the source defaults to 0V DC, even just set to an AC source.

QDrives , 12-16-2024, 07:58 PM

I was wondering that too, but then realized that the power supply is (partly) in the feedback loop too. So yes, it will be 0V for DC and at lower frequencies.

wm , 12-17-2024, 07:18 PM

Sorry @Phil could you please explain this again and how the ac coupling works? Also wondering why the capacitor is not part of the final schematic?