One of the BEST, CLEAREST EE lectures I've seen and heard in a long time. Thank you very much. You did an excellent job. I'll look at your books. Again, thank you. (KZ1G)
Why did you choose to scale the cap values instead of resistor values for setting f0 here? In earlier videos you made what sounded like a good argument for adjusting the resistors (greater variety in standard values,) but here it seemed like it was a case of wanting to show an alternate way to do the math, and a little bit of making things easier to write out.
A designer can scale as needed. Indeed, nothing says you can't adjust both resistors and caps. Whatever works best. That said, I do like to show calcs using the least amount of "friction". There are a lot of resistors here, and most of them will be the same value, so practically speaking, it's less cluttered and moves faster. Nothing magical here. ;-)
You could replace the input resistors on the two integrators with rheostats. This is discussed in my op amps text book. (It's free- see links in video description).
Thank you very much for your text book and this playlist , your coverage of the topics was really nice you in text book give example for fixed G and Q and here cover the variable circuit so thank you for your effort. I have tried this circuit with variable Q and G I have made a calculation to make my fo = 40.972Khz (f1 = 39.5 Khz , f2 = 42.5 Kz) with B.W = 3Khz but I have made a f-Scaling to the resistors instead of capacitor the final results given that : 1 ohm R = 3978 ohm 1F C = 1nF and I want unity gain so Rq = Rk = 54300 ohm (with Q = 13.657) the problem that the circuit didn't work until i adjust Rk to equal = 96.5K and Rq = 30K (This value gives a unity gain and when i decrease Rk it gives more Gain) so what's the problem with my calculations ?
To start with, there appears to be a problem with the initial calc for f0. I don't get the same value using your R and C. Beyond that, are you using Fig 11.44 from the text as your guide? Remember, for bandpass, gain = kQ. Also, there is a TINA-TI simulation of this circuit in the playlist. That might help: czcams.com/video/cJORc8inK9Q/video.html
@@ElectronicswithProfessorFiore yes iam using fig 11.44 in the text book My R = 3.9kOhm for any 1 ohm resistor C = 1nF with Rq = 30K and Rk = 96.5K i got the narrow BPF with fo = 40.9KHz (Q = 13.5) with Unity Gain , i made this circuit on breadboard and worked fine the question about that Rq and Rk calculation it's the not same as in text book so that made my confused I have see the simulation in the next video on playlist and in will try to make simulation to show what's problem with Calculation
Thanks for this video - I've been trying to understand this circuit for a while and this is the closest I have come to getting it, BUT, I still don't get it! I'm obviously missing something, please try and explain to me :-) If you have a high pass response, i.e. low frequencies have been removed, and that is then fed into another stage, how can the low frequencies be restored? As a previous person commented - there are assumptions made, and given what you say it seems logical, but I still don't get it, help!
Your wording might shed light on the issue. When you say "low frequencies have been removed", be careful. In reality, low frequencies are being attenuated at a specific rate (6 dB per octave). They are not being removed completely. If you now boost them at the mirror rate, you'll get back to where you started (ignoring any additional noise or distortion).
In that case, I suggest that you subscribe to this channel if you haven't already. I have a bunch of videos covering EQ and similar circuits in the pipeline, including multi-band and parametric EQ. BTW, I wouldn't use a 4558 as there are much nicer op amps out there!
Using a quad is a popular technique. For example, if you're using TL081 BiFET op amps, you could use a single TL084 quad instead. Much easier to layout. The only downside is that typically quads do not have DC offset null connections, so if you need that capability, you're going to have to go with singles. Offset nulling is described in another video that covers DC output offset.
But why does the output of the summing filter have a high-pass response? It seems you assumed this in the beginning and never came back to it..! Helpful otherwise, thank you!
Yes and no. It's a circular system. We use that as a starting point, we assume that's what we get. As we proceed through the analysis, we are curious to see if that's what comes back at that point. If it does, then we know the original assumption was correct (and it does, so it is).
One of the BEST, CLEAREST EE lectures I've seen and heard in a long time. Thank you very much. You did an excellent job. I'll look at your books. Again, thank you. (KZ1G)
Really appreciate the way you approached the integrator curve and combined it with filter responses. Thanks a lot!
Great talk coach. You helped me with your walkthrough, as well as filter symbols.
I like the active approach
Phew, that was a fair amount. I think I need to see the current flows to get a better sense of how some of that is working. Thank you.
Thanks, nice job. I have not used this filter topology before.
Very clear. even for a non technician. thanx a lot
Why did you choose to scale the cap values instead of resistor values for setting f0 here? In earlier videos you made what sounded like a good argument for adjusting the resistors (greater variety in standard values,) but here it seemed like it was a case of wanting to show an alternate way to do the math, and a little bit of making things easier to write out.
A designer can scale as needed. Indeed, nothing says you can't adjust both resistors and caps. Whatever works best. That said, I do like to show calcs using the least amount of "friction". There are a lot of resistors here, and most of them will be the same value, so practically speaking, it's less cluttered and moves faster. Nothing magical here. ;-)
nice video man, it was very helpful. How could I implement a variable resistor to dynamically change the cutoff frequency?
There's an example of this in my accompanying Op Amp's text. You can download it for free (see the description for details). No strings attached.
awesome... what need to change to move the peak over the audio frecuency 20-20khz??? like cutof knob
You could replace the input resistors on the two integrators with rheostats. This is discussed in my op amps text book. (It's free- see links in video description).
Thanks man, this was super helpful.
Great video! The only caveat is that the sound of the marker on paper gave goosebumps everytime like an anti ASMR 😅
Thank you very much for your text book and this playlist , your coverage of the topics was really nice you in text book give example for fixed G and Q and here cover the variable circuit so thank you for your effort.
I have tried this circuit with variable Q and G
I have made a calculation to make my fo = 40.972Khz (f1 = 39.5 Khz , f2 = 42.5 Kz) with B.W = 3Khz
but I have made a f-Scaling to the resistors instead of capacitor
the final results given that :
1 ohm R = 3978 ohm
1F C = 1nF
and I want unity gain so Rq = Rk = 54300 ohm (with Q = 13.657)
the problem that the circuit didn't work until i adjust Rk to equal = 96.5K and Rq = 30K (This value gives a unity gain and when i decrease Rk it gives more Gain) so what's the problem with my calculations ?
To start with, there appears to be a problem with the initial calc for f0. I don't get the same value using your R and C. Beyond that, are you using Fig 11.44 from the text as your guide? Remember, for bandpass, gain = kQ.
Also, there is a TINA-TI simulation of this circuit in the playlist. That might help:
czcams.com/video/cJORc8inK9Q/video.html
@@ElectronicswithProfessorFiore
yes iam using fig 11.44 in the text book
My R = 3.9kOhm for any 1 ohm resistor
C = 1nF
with Rq = 30K and Rk = 96.5K i got the narrow BPF with fo = 40.9KHz (Q = 13.5) with Unity Gain , i made this circuit on breadboard and worked fine
the question about that Rq and Rk calculation it's the not same as in text book so that made my confused
I have see the simulation in the next video on playlist and in will try to make simulation to show what's problem with Calculation
Thanks for this video - I've been trying to understand this circuit for a while and this is the closest I have come to getting it, BUT, I still don't get it! I'm obviously missing something, please try and explain to me :-) If you have a high pass response, i.e. low frequencies have been removed, and that is then fed into another stage, how can the low frequencies be restored? As a previous person commented - there are assumptions made, and given what you say it seems logical, but I still don't get it, help!
Your wording might shed light on the issue. When you say "low frequencies have been removed", be careful. In reality, low frequencies are being attenuated at a specific rate (6 dB per octave). They are not being removed completely. If you now boost them at the mirror rate, you'll get back to where you started (ignoring any additional noise or distortion).
pls show us to build 5 band equalizer 4558 op
In that case, I suggest that you subscribe to this channel if you haven't already. I have a bunch of videos covering EQ and similar circuits in the pipeline, including multi-band and parametric EQ. BTW, I wouldn't use a 4558 as there are much nicer op amps out there!
Can I use a quad opamp for all of this or do I need four separate opamps?
Using a quad is a popular technique. For example, if you're using TL081 BiFET op amps, you could use a single TL084 quad instead. Much easier to layout. The only downside is that typically quads do not have DC offset null connections, so if you need that capability, you're going to have to go with singles. Offset nulling is described in another video that covers DC output offset.
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But why does the output of the summing filter have a high-pass response? It seems you assumed this in the beginning and never came back to it..! Helpful otherwise, thank you!
Yes and no. It's a circular system. We use that as a starting point, we assume that's what we get. As we proceed through the analysis, we are curious to see if that's what comes back at that point. If it does, then we know the original assumption was correct (and it does, so it is).
Could you write any f'ng smaller?
if only there were an object with plastic lenses that you could put in front of your eyes