Hey guys,
I'm required to design and build a graphic equalizer (15 band) which provides a +/-10dB boost/cut for each band across the audio spectrum
After doing alot of research I came across this Constant Q topology on sound whistles based off of the research paper(linked below). It posed some really nice advantages and simplicity compared to other topologies using the MFB band-pass filter.

I proceeded to derive the TF for the MFB band-pass filter (Didn't upload it as it's in a notebook...difficult to scan. It's correct though as I got the standard result)

I proceeded to design for the 15 bands for a Q of 4 and recorded my results in the excel sheet.

Following this I designed all the filters and simulated them and the response was exactly what I wanted, you can view them in the image shown below or in the Multisim file in the rar file(Filter Simulations). As you can see, there were no overlaps in the pass-bands of the filters.

After this I decided to integrate the filters with the rest of the circuitry as was done on sound whistles....and it worked...sort of. I simulated it for full boost and cut for alternating bands. Here is where I get confused(Oh btw sound whistles and the research paper didn't provide any circuit analysis so I went ahead and did it in the attached pdf. There you can see my analysis for boost and cut as well as the output of the Vu2a amplifier, based on the diagrams I provided. My lecturer said he thinks the 10 and 100 ohm values Rod had were probably an error and should be 10k and 100k, so I used 10k in my derivation.

Took the diagrams from here:
http://sound.whsites.net/project84.htm

Different project but same circuitry just the components were positioned a little differently.

Firstly this:

"There is one thing of special note in this circuit. R6 (39k as shown) determines the maximum amount of boost and cut, and if you wanted to, you can make it variable. With the filter circuits shown below, 39k allows a boost and cut of 12dB - which is about right in most installations. A value of 10k will allow a maximum of a little over 5dB. Any value between these limits will provide the optimum for a given environment, and this can be preset. This is a very useful feature, and one that I believe is unique to this circuit."

How does a value of 10k provide a gain of "A little over 5dB", shouldn't it be 0dB? as 20lg(10k/10k) = 0, I used the 25Hz test circuit located in the rar file as well to play around with the value of R6. 39k gives me about 13dB, 100k gives me 20dB(Makes sense), 31.6227766k doesn't give me 10dB(Gives me 12.3dB) however which doesn't make sense to me since 20lg(3.16227766) should give me 10dB. Using 10k does give me a little over 5dB as Rod Elliot said on sound whistles, which leads me to wonder, unless I've made a mistake in the circuitry in multisim, is it just based on tuning? Like do I just have to play around with this resistor value to get the gain I want?
Mathematically it doesn't make sense to me.


Okay that's that. So I decided to play around with the values of R6 in the Input Output Stage + Filters (2) file on multisim, a value of 39k gives me the 10dB value for full boost and cut for all the bands except the first and last band (25Hz and 16kHz) and I can't wrap my mind around why thats happening. Not sure where I went wrong.

Any Help would be much appreciated guys, Thanks in advance!


Research Paper:
https://pdfs.semanticscholar.org/3292/c31cdc4df5fdb09f55800c767be249930e89.pdf

Reference:
http://sound.whsites.net/project75.htm

 

Hello,

What circuit do you want an analysis for?
Post an image of the circuit and make sure it is complete.
Dont post circuits that have lines that dont go anywhere

 

Hello,

What circuit do you want an analysis for?
Post an image of the circuit and make sure it is complete.
Dont post circuits that have lines that dont go anywhere

Oh, these are the diagrams I did the analysis for:
Source: http://sound.whsites.net/project75.htm

Where the Sig node from the first image connects to the Sig node in this image above.

I did wrote it out in the attached pdf. I used H(s) to represent the transfer function of the MFB bandpass filter. You'll see my final result there.

My analysis starts from Page 3 in the pdf. I modelled the pots as two resistors R1 and R2 whose values change based on the position of the pot.

I dont understand exactly from that equation I got in the end how that op-amp U2-B determines the overall gain of the circuit as stated in the original post.

Oh and the overall circuit connecting everything is in the Multisim file named Input Output Stage + Filters (2)

I just posted these images as it's a bit clearer to see than in the Multisim file

 

Hi,

That's much more clear now, much easier to follow, very good. The other diagrams were spewed about

Did you analyze the circuit with one of those bandpass sections?
The kind of function you are looking for is a function that includes R5 and R6 so you can observe the change in output with change in ratio R6/R5, or reduce that stage to just a gain G=R6/R5 and then you just need to look at what happens with change of gain G. So the R5,R6 op amp stage is just a voltage gain stage and ideally you'd want to see Vout=Vin*G and that would tell you right away what G does (and hence the effect ofR6/R5).

Maybe i dont understand the question because if you do a circuit analysis you end up with a function that almost always includes every resistor in the circuit, and that would include R6/R5. That's unless of course they cancel out completely and that might mean that stage doe not do anything. But if the designer was competent he would have seen that and eliminated it, so it's not likely that stage cancels out.

Doing this with one BP stage probably isnt too hard, but of course we need to use care in getting the equations written properly or we get unreal results.
Later, adding another stage will show us how two stages interact with different settings, then later still adding a third stage would show how three work together with various settings. That will be more complicated but numerically it should not be too unreasonable using regular floating point.

My advice would be to try again and see if you can get a transfer function that includes R5 and R6 or just replace that stage with a gain G and then get a function that includes G. You also need to include a way to change the pot setting in the equation so that you can observe what happens with say middle, high, and low settings at least.
I'll also try this and see if i can get something reasonable, either later tonight or tomorrow sometime.

[LATER]
With the center frequency for that one stage set to close to 900Hz, i get this:
(0.5*(3.552749205937289*10^15*s^3*R22^2*R6+1.421199158357922*10^19*s^2*
R22^2*R6+1.9895395553248817*10^23*s*R22^2*R6-3.5527492059372886*10^20*s^3*R22*R6+
2.8420288231245649*10^28*s^2*R22*R6-1.9895395553248818*10^28*s*R22*R6-
3.5570093075101283*10^24*s^3*R6-2.8424694247024493*10^33*s^2*R6-1.9933466941826624*
10^32*s*R6-1.9899354736963915*10^33*R6+3.5527847334293481*10^24*s^3*R22^2+
1.4212133703495056*10^28*s^2*R22^2+1.989559450720435*10^32*s*R22^2-
3.5527847334293484*10^29*s^3*R22-1.4212133703495055*10^33*s^2*R22-1.9895594507204353*
10^37*s*R22-3.5570448776032038*10^33*s^3-1.426471039039483*10^37*s^2-
1.9933666276496041*10^41*s-1.9899553730511286*10^42))/(1.776445658663313*10^15*s^3*
R22^2*R6+7.1062800344636795*10^18*s^2*R22^2*R6+9.9480956885145538*10^22*s*R22^2*R6
-1.7764456586633131*10^20*s^3*R22*R6-1.4211849560299754*10^28*s^2*R22*R6-
9.9480956885145524*10^27*s*R22*R6-1.7786006679650272*10^24*s^3*R6-
1.5066514353232027*10^29*s^2*R6-9.9672714632128968*10^31*s*R6-9.9501151511434939*
10^32*R6+1.7764634231198998*10^24*s^3*R22^2+7.1063510972640242*10^27*s^2*R22^2+
9.9481951694714383*10^31*s*R22^2-1.7764634231198998*10^29*s^3*R22-7.1063510972640241*
10^32*s^2*R22-9.9481951694714392*10^36*s*R22-1.7786184539717071*10^33*s^3-
7.1327400645243848*10^36*s^2-9.9673711359275272*10^40*s-9.9502146522950061*10^41)

where R6 is the R6 in the schematic, and R22 is the pot upper section and the lower section is 100000-R22 because the pot is 100k.
This allows me to change R22 to change the pot setting, and change R6 to change the 'gain' setting.
What this shows is that with R22=50k (pot in exact center) the gain is close to 1, and it DOES NOT matter what R6 is set to. With R6=40k or R6=80k i still get a gain of 1.
However, setting R22=0 and R6=40k i see a center frequency gain of about 9, and with R6=80k i see a center frequency gain of about 18 (approximate).
So it appears that R6 changes the center frequency gain but only when the pot is set to boost and probably when it is set to cut too, except the gain is less than 1 for that i would suppose.

Of course this can also be seen in a simulator i would think.

[LATER LATER]
I was redrawing the schematic to include both sections in one, and i noticed that if we look at this in block diagram form then we can probably eliminate the R5, R6 op amp section. That section uses R5 and R6 to set the center gain, but if we replace that with a wire then we can use R4 and R7 to set the gain instead.
So if R5=10k and R6=40k originally, then replacing that section with a wire and making R4=40k and R7=40k we should see the same results.
The limits analysis should be the same too except now the R4 op amp will see a higher output rather than the R5,R6 op amp, but it will be the same so if it worked in the R5, R6 section it should work in the R4 section just as well.
So to recap, R7=R4, and R4 takes on the original value of R6, and replace the R5, R6 section with a wire.

 

Hi,

That's much more clear now, much easier to follow, very good. The other diagrams were spewed about

Did you analyze the circuit with one of those bandpass sections?
The kind of function you are looking for is a function that includes R5 and R6 so you can observe the change in output with change in ratio R6/R5, or reduce that stage to just a gain G=R6/R5 and then you just need to look at what happens with change of gain G. So the R5,R6 op amp stage is just a voltage gain stage and ideally you'd want to see Vout=Vin*G and that would tell you right away what G does (and hence the effect ofR6/R5).

Maybe i dont understand the question because if you do a circuit analysis you end up with a function that almost always includes every resistor in the circuit, and that would include R6/R5. That's unless of course they cancel out completely and that might mean that stage doe not do anything. But if the designer was competent he would have seen that and eliminated it, so it's not likely that stage cancels out.

Doing this with one BP stage probably isnt too hard, but of course we need to use care in getting the equations written properly or we get unreal results.
Later, adding another stage will show us how two stages interact with different settings, then later still adding a third stage would show how three work together with various settings. That will be more complicated but numerically it should not be too unreasonable using regular floating point.

My advice would be to try again and see if you can get a transfer function that includes R5 and R6 or just replace that stage with a gain G and then get a function that includes G. You also need to include a way to change the pot setting in the equation so that you can observe what happens with say middle, high, and low settings at least.
I'll also try this and see if i can get something reasonable, either later tonight or tomorrow sometime.

[LATER]
With the center frequency for that one stage set to close to 900Hz, i get this:
(0.5*(3.552749205937289*10^15*s^3*R22^2*R6+1.421199158357922*10^19*s^2*
R22^2*R6+1.9895395553248817*10^23*s*R22^2*R6-3.5527492059372886*10^20*s^3*R22*R6+
2.8420288231245649*10^28*s^2*R22*R6-1.9895395553248818*10^28*s*R22*R6-
3.5570093075101283*10^24*s^3*R6-2.8424694247024493*10^33*s^2*R6-1.9933466941826624*
10^32*s*R6-1.9899354736963915*10^33*R6+3.5527847334293481*10^24*s^3*R22^2+
1.4212133703495056*10^28*s^2*R22^2+1.989559450720435*10^32*s*R22^2-
3.5527847334293484*10^29*s^3*R22-1.4212133703495055*10^33*s^2*R22-1.9895594507204353*
10^37*s*R22-3.5570448776032038*10^33*s^3-1.426471039039483*10^37*s^2-
1.9933666276496041*10^41*s-1.9899553730511286*10^42))/(1.776445658663313*10^15*s^3*
R22^2*R6+7.1062800344636795*10^18*s^2*R22^2*R6+9.9480956885145538*10^22*s*R22^2*R6
-1.7764456586633131*10^20*s^3*R22*R6-1.4211849560299754*10^28*s^2*R22*R6-
9.9480956885145524*10^27*s*R22*R6-1.7786006679650272*10^24*s^3*R6-
1.5066514353232027*10^29*s^2*R6-9.9672714632128968*10^31*s*R6-9.9501151511434939*
10^32*R6+1.7764634231198998*10^24*s^3*R22^2+7.1063510972640242*10^27*s^2*R22^2+
9.9481951694714383*10^31*s*R22^2-1.7764634231198998*10^29*s^3*R22-7.1063510972640241*
10^32*s^2*R22-9.9481951694714392*10^36*s*R22-1.7786184539717071*10^33*s^3-
7.1327400645243848*10^36*s^2-9.9673711359275272*10^40*s-9.9502146522950061*10^41)

where R6 is the R6 in the schematic, and R22 is the pot upper section and the lower section is 100000-R22 because the pot is 100k.
This allows me to change R22 to change the pot setting, and change R6 to change the 'gain' setting.
What this shows is that with R22=50k (pot in exact center) the gain is close to 1, and it DOES NOT matter what R6 is set to. With R6=40k or R6=80k i still get a gain of 1.
However, setting R22=0 and R6=40k i see a center frequency gain of about 9, and with R6=80k i see a center frequency gain of about 18 (approximate).
So it appears that R6 changes the center frequency gain but only when the pot is set to boost and probably when it is set to cut too, except the gain is less than 1 for that i would suppose.

Of course this can also be seen in a simulator i would think.

[LATER LATER]
I was redrawing the schematic to include both sections in one, and i noticed that if we look at this in block diagram form then we can probably eliminate the R5, R6 op amp section. That section uses R5 and R6 to set the center gain, but if we replace that with a wire then we can use R4 and R7 to set the gain instead.
So if R5=10k and R6=40k originally, then replacing that section with a wire and making R4=40k and R7=40k we should see the same results.
The limits analysis should be the same too except now the R4 op amp will see a higher output rather than the R5,R6 op amp, but it will be the same so if it worked in the R5, R6 section it should work in the R4 section just as well.
So to recap, R7=R4, and R4 takes on the original value of R6, and replace the R5, R6 section with a wire.

Thanks a bunch man, I think I see where you're going, let me see if I can explain what I understand from what you did (or atleast what I think I understand)

Oh I saw you mentioned that I should include R5 and R6 in my analysis, A is actually R6/R5 in my working, I forgot to mention that earlier haha sorry about that. (Page 4 of the pdf)

Okay So doing so, if I replace H(s) with the transfer function of the MFB Bandpass filter, i'd see that the gain A does in fact affect the center frequency gain, It then makes it G = 1/2 *(R3/R1) * (R6/R5).

Now you said you used 900Hz as your center frequency, by eyeballing it I believe your R3/R1 in the filter stage would have worked out to be 4.5, therefore when you said you used R6 as 40k, then this will in fact make the overall center frequency gain 9 and hence if R6 is 80k then the center frequency gain will be 18. And like you said this does NOT change as long as the pot is not centered (There must be some boost or some cut), centering the pot just zeros everything out.

Now the [LATER LATER] part is what I don't quite understand, you said that we can eliminate the R5,R6 op-amp section and instead set the gain with the R4,R7 section.
"So if R5=10k and R6=40k originally, then replacing that section with a wire and making R4=40k and R7=40k we should see the same results."
How is this so?

Also I still don't get how setting R6 to be 10k here http://sound.whsites.net/project75.htm should give a gain of 5dB....I still think it should be 0dB based on his designs.

Anyways I'll run the simulations a bit later removing R6/R5 and replacing it with a wire (Its 5:41 AM where I am lol) and i'll post the results.

 

Thanks a bunch man, I think I see where you're going, let me see if I can explain what I understand from what you did (or atleast what I think I understand)

Oh I saw you mentioned that I should include R5 and R6 in my analysis, A is actually R6/R5 in my working, I forgot to mention that earlier haha sorry about that. (Page 4 of the pdf)

Okay So doing so, if I replace H(s) with the transfer function of the MFB Bandpass filter, i'd see that the gain A does in fact affect the center frequency gain, It then makes it G = 1/2 *(R3/R1) * (R6/R5).

Now you said you used 900Hz as your center frequency, by eyeballing it I believe your R3/R1 in the filter stage would have worked out to be 4.5, therefore when you said you used R6 as 40k, then this will in fact make the overall center frequency gain 9 and hence if R6 is 80k then the center frequency gain will be 18. And like you said this does NOT change as long as the pot is not centered (There must be some boost or some cut), centering the pot just zeros everything out.

Now the [LATER LATER] part is what I don't quite understand, you said that we can eliminate the R5,R6 op-amp section and instead set the gain with the R4,R7 section.
"So if R5=10k and R6=40k originally, then replacing that section with a wire and making R4=40k and R7=40k we should see the same results."
How is this so?

Also I still don't get how setting R6 to be 10k here http://sound.whsites.net/project75.htm should give a gain of 5dB....I still think it should be 0dB based on his designs.

Anyways I'll run the simulations a bit later removing R6/R5 and replacing it with a wire (Its 5:41 AM where I am lol) and i'll post the results.

Hi,

Ok sounds good. Just remember that i assume that R5=10k always. That makes R6 alone control the gain and that helps reduce the circuit as mentioned.

If you look at these stages carefully you will see that they are somewhat isolated from each other current wise. What i mean by that is that the current from some of the nodes does not matter, as long as it never exceeds the output ability of the op amp. What this means is that we can redraw the circuit in block diagram form, and so there could be reductions. As it turns out, it looks like that one reduction works because if we double the R4 op amp gain and halve the R7 op amp gain then the output of the R7 op amp is the same, but the output of the R4 op amp is doubled, so to compensate with the R6 op amp we'd have to halve that too. These changes would give us the same result as not changing anything, presumably.
Now since we wanted a gain in the R6 op amp anyway, why not increase the R4 op amp gain and decrease the R7 op amp gain to force the R4 op amp output to be what we want for the output of the R6 op amp. That's the key to reduction. It's simply combining the gains to get the same outputs at each op amp output as we had originally. This is a technique called "block diagram reduction" and theoretically the entire circuit is reduced to one single block when we find the transfer function, however it would be too hard to implement in that form so we need to keep some blocks anyway.

For a simple example, if we have a 2x gain stage immediately followed by a 4x gain stage, the total gain out of the final output is 8x so we can replace that 'block' with a single gain of 8x, and that could be one op amp vs two op amps. There are many more tricks like this too although not all can be used for all circuits as it depends on the original connections and the possible ways to implement the individual blocks.

I had 'eyeballed' that reduction so it's good to test it to be sure.
In this case we have a gain of 1, another gain of 1, and a pickoff from the first gain with a gain of 4. Thus if we make the first gain 4, we eliminate the gain of 4 section, but then the second gain of 1 will be too large so it has to be reduced by a factor of 1/4 which is what R7=40k does. The feedback resistor does not change, so the other gain (from the pot) stays the same. This makes the gain 4*1/4 which is again 1, and the gain of the R6 output (now replaced by a direct connection to the R4 output) again with gain of 4 so everything is as original.

You might also note that some of the research papers are not aimed at the best possible implementation, their main goal is to convey the theory behind the thesis. That means that parts count is not always a big deal as long as ONE implementation proves the idea works. If the title was "Minimum Parts Count Equalizer" that would be different, but it appears it's just another way to do it.
This is especially true when the author wants to make a point about something somewhere in the circuit, such as how the gain there changes things. It would be harder to have to explain that you have to change R4 and R7 to set the gain then to just say that you have to change R6 to set that...much easier to point out.

[LATER]
I took another look at the proposed reduction, and see that the original gain was not really 4 it was -4. It still might work but now definitely needs to be tested.
Ideally, the other two op amp stage gains would have to be flipped too instead of inverting, they would have to be non inverting. That may or may not be possible.
The change in phase with the current reduction possibility may or may not mess up the response so it would have to be checked carefully.

 

Hi,

Ok sounds good. Just remember that i assume that R5=10k always. That makes R6 alone control the gain and that helps reduce the circuit as mentioned.

If you look at these stages carefully you will see that they are somewhat isolated from each other current wise. What i mean by that is that the current from some of the nodes does not matter, as long as it never exceeds the output ability of the op amp. What this means is that we can redraw the circuit in block diagram form, and so there could be reductions. As it turns out, it looks like that one reduction works because if we double the R4 op amp gain and halve the R7 op amp gain then the output of the R7 op amp is the same, but the output of the R4 op amp is doubled, so to compensate with the R6 op amp we'd have to halve that too. These changes would give us the same result as not changing anything, presumably.
Now since we wanted a gain in the R6 op amp anyway, why not increase the R4 op amp gain and decrease the R7 op amp gain to force the R4 op amp output to be what we want for the output of the R6 op amp. That's the key to reduction. It's simply combining the gains to get the same outputs at each op amp output as we had originally. This is a technique called "block diagram reduction" and theoretically the entire circuit is reduced to one single block when we find the transfer function, however it would be too hard to implement in that form so we need to keep some blocks anyway.

For a simple example, if we have a 2x gain stage immediately followed by a 4x gain stage, the total gain out of the final output is 8x so we can replace that 'block' with a single gain of 8x, and that could be one op amp vs two op amps. There are many more tricks like this too although not all can be used for all circuits as it depends on the original connections and the possible ways to implement the individual blocks.

I had 'eyeballed' that reduction so it's good to test it to be sure.
In this case we have a gain of 1, another gain of 1, and a pickoff from the first gain with a gain of 4. Thus if we make the first gain 4, we eliminate the gain of 4 section, but then the second gain of 1 will be too large so it has to be reduced by a factor of 1/4 which is what R7=40k does. The feedback resistor does not change, so the other gain (from the pot) stays the same. This makes the gain 4*1/4 which is again 1, and the gain of the R6 output (now replaced by a direct connection to the R4 output) again with gain of 4 so everything is as original.

You might also note that some of the research papers are not aimed at the best possible implementation, their main goal is to convey the theory behind the thesis. That means that parts count is not always a big deal as long as ONE implementation proves the idea works. If the title was "Minimum Parts Count Equalizer" that would be different, but it appears it's just another way to do it.
This is especially true when the author wants to make a point about something somewhere in the circuit, such as how the gain there changes things. It would be harder to have to explain that you have to change R4 and R7 to set the gain then to just say that you have to change R6 to set that...much easier to point out.

[LATER]
I took another look at the proposed reduction, and see that the original gain was not really 4 it was -4. It still might work but now definitely needs to be tested.
Ideally, the other two op amp stage gains would have to be flipped too instead of inverting, they would have to be non inverting. That may or may not be possible.
The change in phase with the current reduction possibility may or may not mess up the response so it would have to be checked carefully.

Ohh I think i understand what you're saying.
I Removed the opamp and it seems to work the same way for full boost and cut, however for intermediate values of boost/cut the response goes crazy (with or without the op-amp), I don't know if I made a mistake in the simulation or what but maybe you can have a go at the simulation and see what responses you get.
For example, id set a pot to 75% ( Half way max-boost) and the maximum value isn,t halved, sometimes other bands are affected.


Another thing, by doing what we previously did there, how do we design for a specific maximum boost or cut?
I saw he used the same theory for setting max gain in other project (http://sound.whsites.net/project84.htm#reference)
Here he said:
"U1B is a summing amp, and it takes its input from the combination of the input, and the output signal from the CUT bus - this comes from the pots used as the level control for each frequency band. The combined signal is summed again by U2A, this time with the signal from the BOOST bus added. The signal drive to all filters is performed by U2B, the gain of which determines the maximum boost and cut allowed. As shown, The circuit will provide about +/-14dB, and the response is completely flat with all pots centred. Reduce the range by reducing the value of R8 (39k) - a value of 10k gives 6dB of boost and cut."

That part still doesn't make sense to me. It doesn't correlate with the analysis we did here...but yet it works for him and this is what's confusing me

 

This is what I get for:
25Hz pot @75%
40Hz pot @25%
63%pot @50%

Notice the almost flat response for 63Hz but it's not at 0dB

 

There is another topology for a graphic equalizer that I think is easier to implement, and has less interaction between bands.

http://www.geofex.com/Article_Folders/EQs/paramet.htm

 

Ohh I think i understand what you're saying.
I Removed the opamp and it seems to work the same way for full boost and cut, however for intermediate values of boost/cut the response goes crazy (with or without the op-amp), I don't know if I made a mistake in the simulation or what but maybe you can have a go at the simulation and see what responses you get.
For example, id set a pot to 75% ( Half way max-boost) and the maximum value isn,t halved, sometimes other bands are affected.


Another thing, by doing what we previously did there, how do we design for a specific maximum boost or cut?
I saw he used the same theory for setting max gain in other project (http://sound.whsites.net/project84.htm#reference)
Here he said:
"U1B is a summing amp, and it takes its input from the combination of the input, and the output signal from the CUT bus - this comes from the pots used as the level control for each frequency band. The combined signal is summed again by U2A, this time with the signal from the BOOST bus added. The signal drive to all filters is performed by U2B, the gain of which determines the maximum boost and cut allowed. As shown, The circuit will provide about +/-14dB, and the response is completely flat with all pots centred. Reduce the range by reducing the value of R8 (39k) - a value of 10k gives 6dB of boost and cut."

That part still doesn't make sense to me. It doesn't correlate with the analysis we did here...but yet it works for him and this is what's confusing me

Hello,

To follow your work i would need to see every value you are using in the circuit.
I assume you are using the values on the schematic we saw earlier, but for the band pass sections i would need to see values for
R1, R2, R3, C1, C2
for each stage you are using. Also, state the center frequency for each stage along with the component values so i could verify.
Hopefully you can do one stage at a time, or maybe three stages if needed, but whatever stages you are using i would like to see the values you are using for those five components for all the BP stages.

 

Hello,

To follow your work i would need to see every value you are using in the circuit.
I assume you are using the values on the schematic we saw earlier, but for the band pass sections i would need to see values for
R1, R2, R3, C1, C2
for each stage you are using. Also, state the center frequency for each stage along with the component values so i could verify.
Hopefully you can do one stage at a time, or maybe three stages if needed, but whatever stages you are using i would like to see the values you are using for those five components for all the BP stages.

These are the values im using for my bandpass sections.
And yup all other values are the same as the schematic.

 

These are the values im using for my bandpass sections.
And yup all other values are the same as the schematic.

Hello again,

Ok thanks. I tried the 100Hz band filter and it worked pretty well. Did you have more than one BP filter in your schematic? Did you try just one filter at a time?
Also, do you use LT Spice?
I am suspecting that maybe your simulator is not handling this right, what one are you using now?

I was able to prove that the original proposed solution for eliminating the R6 op amp section DID NOT work. It did not work and the effect was that it would only allow a 'cut' mode operation and no 'boost' mode, so that' solution is completely out of the picture now.
HOWEVER, there is another way to get rid of that R6 op amp section and that is to again replace it with a wire, then convert the BP filter which is currently an inverting BP filter into a non inverting BP filter and probably set the pass gain to 4 (currently your choice of values has it at 1). I did not experiment with the gain yet except for a gain of 2 which allowed a similar operation to the original circuit possibly with a different pass gain. I'll get back to this soon though.

For now i guess it would be better to concentrate on why you are seeing some unstable operation. I am guessing the simulator is having a hard time handling the circuit, but it would help to know the simulator, and maybe we should both use LT Spice as that's free and works pretty well. Right now i am using MicroCap but i could convert to LT Spice.

I should also point out that i was using ideal op amps, not real world models. I do that to test the basic theory of the circuit itself first so we know if it even works right in theory because if it does not work in theory then there's very little chance that it will work with real op amps that have phase shifts and gain variations and offsets and what not. Once it passes the theory test, then i move to real world models that simulate op amps we can actually buy and note any differences.

 

Hello again,

Ok thanks. I tried the 100Hz band filter and it worked pretty well. Did you have more than one BP filter in your schematic? Did you try just one filter at a time?
Also, do you use LT Spice?
I am suspecting that maybe your simulator is not handling this right, what one are you using now?

I was able to prove that the original proposed solution for eliminating the R6 op amp section DID NOT work. It did not work and the effect was that it would only allow a 'cut' mode operation and no 'boost' mode, so that' solution is completely out of the picture now.
HOWEVER, there is another way to get rid of that R6 op amp section and that is to again replace it with a wire, then convert the BP filter which is currently an inverting BP filter into a non inverting BP filter and probably set the pass gain to 4 (currently your choice of values has it at 1). I did not experiment with the gain yet except for a gain of 2 which allowed a similar operation to the original circuit possibly with a different pass gain. I'll get back to this soon though.

For now i guess it would be better to concentrate on why you are seeing some unstable operation. I am guessing the simulator is having a hard time handling the circuit, but it would help to know the simulator, and maybe we should both use LT Spice as that's free and works pretty well. Right now i am using MicroCap but i could convert to LT Spice.

I should also point out that i was using ideal op amps, not real world models. I do that to test the basic theory of the circuit itself first so we know if it even works right in theory because if it does not work in theory then there's very little chance that it will work with real op amps that have phase shifts and gain variations and offsets and what not. Once it passes the theory test, then i move to real world models that simulate op amps we can actually buy and note any differences.

Woa thats great news.
1 filter does not even work for me much less when I use all 15 filters LOL
So you're getting whatever gain you want then?
I also didnt use ideal opamps in my simulation...which could have been the problem as you said.

I used multisim 12.
I never used LT spice before but I'll get it so that we're both on the same platform.
Btw i re-analyzed the circuit and got a transfer function for it which I'll post. Was wondering if it was the same thing you got when you derived yours earlier.

 

It was a bit longer but it reduced to this once i let
r8=r7
R4=R3
R4=R8
R10=r11
A=R6/R7

Oh and just to note RB=R-RA...rb is the cut resistor and ra is the boost resistor when modelling the pot of resistance R

 

This is a 10 channel circuit, but you can add 1 more IC to make it 15. I do not know the dB of it.

https://www.eleccircuit.com/10-channels-graphic-equaliser-by-la3600/

 

This is a 10 channel circuit, but you can add 1 more IC to make it 15. I do not know the dB of it.

https://www.eleccircuit.com/10-channels-graphic-equaliser-by-la3600/

I saw these earlier in my research stages, but for the requirements of the project I have to build it using discrete analog components...so everything from scratch

Woa thats great news.
1 filter does not even work for me much less when I use all 15 filters LOL
So you're getting whatever gain you want then?
I also didnt use ideal opamps in my simulation...which could have been the problem as you said.

I used multisim 12.
I never used LT spice before but I'll get it so that we're both on the same platform.
Btw i re-analyzed the circuit and got a transfer function for it which I'll post. Was wondering if it was the same thing you got when you derived yours earlier.

Oh when I said the filters didn't work I mean entirely as a system, by themselves they work, but once I integrate it with the boost/cut circuitry the gain doesn't work as calculated. Same goes for when I use all 15 filters.

 

The datasheet of that IC should have the inner circuit with a few components missing. You can check it for reference. To me this is a circuit that only changes the voltage and current on every channel.

 

It was a bit longer but it reduced to this once i let
r8=r7
R4=R3
R4=R8
R10=r11
A=R6/R7

Oh and just to note RB=R-RA...rb is the cut resistor and ra is the boost resistor when modelling the pot of resistance R

Hello again,

My transfer function was in post #4 but i did not make an attempt to get it into a more manageable form because all i wanted to do was check the validity of the theory that R6/R5 had a direct effect on the midband gain of a BP stage. After numerous tests now it looks like it does.
Yes i get whatever gain i set for.

For the pot setting at mid range (equal upper and lower resistances of 50k each) i see a tiny variation in gain over frequency, but it's so little it's not worth mentioning being less than 0.1db. I've never seen it vary more than that and probably not even that much, but again i am using ideal op amps which i will change soon in the simulation.

I'll see if i can get a transfer function in a more manageable form. The first time i did it i threw a very general method at it and so it came up a little more complicated. Viewing the circuit and doing each piece separately it should come out simpler. I'll try to get to this later today.

Are you going to use LT Spice next? I hope so, for one i think it is better and secondly if you create the schematic you can post it here and i can download and begin simulating right away without having to draw the circuit all over again. Would be nice

Oh one more thing, in your last post your transfer function contains H(s). Is that the BP transfer function or what?

 

Hello again,

Ok i got some usable results finally. Note also there are component labels that repeat
so i had to change the first R3 to R33. If you use the very first op amp stage you will
also have to adjust other labels too because the BP section uses R1 R2 R3.

Using one BP section, the midband gain is:
G= ((AAAA*R23*R3+2*R1*R23+2*R1*R22)*R4+2*R1*R22*R23)/((AAAA*R22*R3+2*R1*R23+2*R1*R22)*R4+2*R1*R22*R23)

where R23=100k-R22 as before as those two make up the pot, and R4=10k (as are all the other 10k resistors), and
where "AAAA" same as A is the ratio R6/R5 (not -R6/R5) and i used four A's to make it easy to see where that gain is located in the equation.
What this says is that the gain is not exactly R6/R5 but it is close.
For your 100Hz example, using R1 and R3, we get in the pot center:
G=1
and for the pot at one extreme (boost) we get:
G=A+1
and for the other pot extreme (cut) we get:
G=1/(A+1)

so it's not related to A it's really related to A+1.

There was a short cut used to get to this highly theoretical result, and that was to make the two 10 ohm resistors equal to zero.
With the two being 10 ohms we'd get a much messier result, which is hard to comprehend. Also, the value of the last coupling capacitor was allowed to go to infinity in order to simplify the solution. This is valid because the value in real life is chosen to pass all frequencies of interest with little or no attenuation.
This solution makes it extremely clear how the midband gain is related to A.

What is more, when we solve for the peak midband frequency we get the same frequency as for the bandpass filter itself, which tells us that the center frequency is the same as the BP center frequency.

If we reduce that gain equation we can get more insight also...
G=(R4*(AAAA*R23*rBP+R23+R22)+R22*R23)/(R4*(AAAA*R22*rBP+R23+R22)+R22*R23)

where rBP is the BP gain (with C2=C1). Now we can set either R22 or R23 to zero and just leave the other as the total resistance and see what happens. Setting R22=0 we get:
G=AAAA*rBP+1

and since in your example rBP=1 we again get A+1 as the boost gain.
Likewise setting R23=0 we get:
G=1/(AAAA*rBP+1)

and since rBP=1 again we get again 1/(A+1) as the cut gain.

Lastly, setting R23=R22 we get:
G=1

which is what we get with the pot in the center.

[LATER]
I forgot to mention that the last equation for the gain G is for C2=C1. When both caps are the same value we get a simple gain solution for the BP section and apparently that is what you were using anyway.

 

Hello again,

My transfer function was in post #4 but i did not make an attempt to get it into a more manageable form because all i wanted to do was check the validity of the theory that R6/R5 had a direct effect on the midband gain of a BP stage. After numerous tests now it looks like it does.
Yes i get whatever gain i set for.

For the pot setting at mid range (equal upper and lower resistances of 50k each) i see a tiny variation in gain over frequency, but it's so little it's not worth mentioning being less than 0.1db. I've never seen it vary more than that and probably not even that much, but again i am using ideal op amps which i will change soon in the simulation.

I'll see if i can get a transfer function in a more manageable form. The first time i did it i threw a very general method at it and so it came up a little more complicated. Viewing the circuit and doing each piece separately it should come out simpler. I'll try to get to this later today.

Are you going to use LT Spice next? I hope so, for one i think it is better and secondly if you create the schematic you can post it here and i can download and begin simulating right away without having to draw the circuit all over again. Would be nice

Oh one more thing, in your last post your transfer function contains H(s). Is that the BP transfer function or what?

Oh yes H(s) is the BP transfer function, I forgot to mention that.
Yes i'll be using spice now, I downloaded it yesterday. So I'll see to rebuild it in Spice and i'll upload it later. Multisim gives you some wacky responses tbh...and they vary ALOT once you use different op-amps geeze (Even with virtual op-amps the response was wacky)

Hello again,

Ok i got some usable results finally. Note also there are component labels that repeat
so i had to change the first R3 to R33. If you use the very first op amp stage you will
also have to adjust other labels too because the BP section uses R1 R2 R3.

Using one BP section, the midband gain is:
G= ((AAAA*R23*R3+2*R1*R23+2*R1*R22)*R4+2*R1*R22*R23)/((AAAA*R22*R3+2*R1*R23+2*R1*R22)*R4+2*R1*R22*R23)

where R23=100k-R22 as before as those two make up the pot, and R4=10k (as are all the other 10k resistors), and
where "AAAA" same as A is the ratio R6/R5 (not -R6/R5) and i used four A's to make it easy to see where that gain is located in the equation.
What this says is that the gain is not exactly R6/R5 but it is close.
For your 100Hz example, using R1 and R3, we get in the pot center:
G=1
and for the pot at one extreme (boost) we get:
G=A+1
and for the other pot extreme (cut) we get:
G=1/(A+1)

so it's not related to A it's really related to A+1.

There was a short cut used to get to this highly theoretical result, and that was to make the two 10 ohm resistors equal to zero.
With the two being 10 ohms we'd get a much messier result, which is hard to comprehend. Also, the value of the last coupling capacitor was allowed to go to infinity in order to simplify the solution. This is valid because the value in real life is chosen to pass all frequencies of interest with little or no attenuation.
This solution makes it extremely clear how the midband gain is related to A.

What is more, when we solve for the peak midband frequency we get the same frequency as for the bandpass filter itself, which tells us that the center frequency is the same as the BP center frequency.

If we reduce that gain equation we can get more insight also...
G=(R4*(AAAA*R23*rBP+R23+R22)+R22*R23)/(R4*(AAAA*R22*rBP+R23+R22)+R22*R23)

where rBP is the BP gain (with C2=C1). Now we can set either R22 or R23 to zero and just leave the other as the total resistance and see what happens. Setting R22=0 we get:
G=AAAA*rBP+1

and since in your example rBP=1 we again get A+1 as the boost gain.
Likewise setting R23=0 we get:
G=1/(AAAA*rBP+1)

and since rBP=1 again we get again 1/(A+1) as the cut gain.

Lastly, setting R23=R22 we get:
G=1

which is what we get with the pot in the center.

[LATER]
I forgot to mention that the last equation for the gain G is for C2=C1. When both caps are the same value we get a simple gain solution for the BP section and apparently that is what you were using anyway.

Woa this makes perfect sense, this what explains why a value 10K for R6 gives 6dB, its because of the (A+1). I'll expand out my transfer function some more and see if I can get this relationship:

G= ((AAAA*R23*R3+2*R1*R23+2*R1*R22)*R4+2*R1*R22*R23)/((AAAA*R22*R3+2*R1*R23+2*R1*R22)*R4+2*R1*R22*R23).

which was obtained by setting the 10R resistor to 0. Should this be done when simulating or in real life? To get the behaviour as close to the theoretical model as possible?