OK, I give up. Can someone explain this bass boost and possibly a formula for it

[Derringer]

this is nicked from ROG's new Thunderbird - Thank you B Tremblay

I tried my own research but just haven't had much luck finding what I want. In short, I have a fuzz on the breadboard and it sounds great. It sounds even better when I run an EQ after it and boost 100Hz a few Db's. So I though, hey why don't I try to make an active boost? I haven't done that before.

I want to use an opamp because half of that opamp is already being used as an input buffer. I found this gem and it's perfect ... says it's ready to go at 150Hz which should be fine.

But I'd love to know what it is and how it works please.
It looks a lot like the "Dual Version" of a "Twin T to Bridged-T Notch Filter" located at the bottom of R.G.'s essay on EQ's. http://www.geofex.com/article_folders/eqs/paramet.htm
Thank you Mr. Keen.

But why does this version, the picture above, have two different caps? How does the pot affect the boost? Does this circuit also cut 150Hz?
Is there a nice math formula I could use to design something like this for other frequencies? Something I could put into a spreadsheet and make my own boxed up calculator for future use?

thanks!


edit- dangint ... yep, in my haste i left a connection out, thanks for the info guys

[dschwartz]

Something is missing there, the output should be connected to the feedback.

This is a very simple and effective peak filter, the net at the feedback is a T style notch filter, removing a range of freq. since it is at the feedback, the result is the opposite, boosting that range..

The caps are mismatched to give a lower Q on a broader range of freqs, you dont want a narrow boost, unless you want a wah type of sound..

The pot works as a "mix" control. At max, the signal fed to the inverting input is 100% passing thru the T filter, at min, the signal comes from the 47k resistor, where the filter is not effective..

I recommend you to learn how to simulate circuits..with ltSpice this would be easy to simulate and play with the values..

[Mark Hammer]

First, note that the .01uf cap and 47k resistor are also tied to the output of the op-amp.

It's a non-inverting op-amp configuration.  If there was to be any gain in such a circuit, there would need to be negative feedback from the output to the inverting input.  Do we have that?  Yes.  It can range from 47k, all the way up to 97k, depending on where the pot wiper is set.

Also, for gain, we would need a path to ground from the inverting output.  Do we have that?  Yes, via the 22k resistor.  It's weird, but it's there.

The gain of a non-inverting stage is determined by the feedback resistance, and ground-leg resistance, and their relative values.  Remember that an op-amp "wants" to function at open-loop gain of several thousand,  The negative feedback essentially "steps on the brakes" to arrive at gains less than MAX.  The more negative feedback, the lower the gain, and the more you bleed off or prevent negative feedback, the higher the gain.

Let's ignore the .01uf cap for the moment, and pretend as if we only have the two fixed resistors (47k, and 22k), the pot and the .15uf cap in there.  Look atthe schematic for a DOD250 or MXR Distortion+ and you'll see pretty much the same thing: a variable resistance on the ground leg, plus a fixed resistor (to set max gain) and a cap to block DC.  Unlike the DOD and MXR pedals, though, this pot is wired to reduce the ground-leg resistance at the same time as increasing the feedback resistance.

When the wiper is moved all the way to the .015uf cap, we have 97k feedback resistance, 22k ground leg, and .15u.  That yields a gain of 5.4x with a 6db/oct rolloff at 48hz.  If we rotate the pot the other way, so that the wiper is at the 47k resistor end, we have a feedback resistance of 47k, ground leg of 72k, and .15uf, giving a gain of  1.65x and a rolloff at 15hz.

BUT...we also have that .01uf cap.  And it plays a rather oddball role by sort of being in series with .15uf, and sort of providing a path to ground via the 22k, from the output.

And at this point, my expertise runs out.  TAG, somebody else is it.

[Derringer]

that definitely helps gentlemen, thank you.

LT Spice. Probably something I should look into when I have a moment or two. Good advice.

But a formula? It's got to be out there somewhere. Like what if I wanted 300Hz?
Is simming a circuit and trial and error playing with RC networks the best way to go?

[merlinb]

It's a bridged-T, often used to create a mid hump, although presumably in this case a bass hump.
The pot allows the feedback to be varied from the output of the bridged-T (max boost or humpiness) to a point only part way along the bridge, leading to less boost. If the 47k were shorted out then it would vary all the way from full boost to perfectly flat response...

[PBE6]

The clever bit about this circuit is that by connecting the wiper to v- you keep the centre frequency constant as you boost. I've seen other implementations of this circuit (a proposed modification to ROG's Thor pedal actually, that replaced the 100k resistor in/out switch with a 100k pot) that connect the outer lug to v- instead. Same boost, but the frequency gets lower as you boost more. Both sound good.

The centre frequency formula for this type of circuit is:

Freq = 1/(2*pi*SQRT(Rf*Rg*C1*C2))

For more info see:
http://johnhearfield.com/RC/RC4.htm

Note: funny enough, swapping the resistors and capacitors gives you the same centre frequency but with different Q and gain characteristics.

[Ripdivot]

I calculate a freq of 89hz with the pot maxed and 127hz with the pot at minimum. How much gain does the circuit have in the two extreme positions of the pot and how does one calculate the gain in this setup?

[midwayfair]

I calculate a freq of 89hz with the pot maxed and 127hz with the pot at minimum. How much gain does the circuit have in the two extreme positions of the pot and how does one calculate the gain in this setup?

Standard non-inverting op amp gain.

Gain is the resistance between the negative input and ground divided into the resistance between the negative input and the output pin, plus 1 (the plus 1 is the inherent unity gain of the non-inverting op amp).

You should be able to figure out the numbers on your own for any setting on the pot. Ignore the pot for a moment and assume it's 0Ohms between all lugs. (a) What's the resistance to ground directly from pin 2? (b) What's the resistance between pin 2 and pin 1? (c) Divide a into b, add 1, what's your gain? That's your baseline.

The pot's wiper is connected to pin 2, and it increases the resistance between pins 1 and 2 at the same time as it decreases the resistance between pin 2 and ground. Put it in the exact center so that it's 25K between the wiper and the other lugs, and repeat steps a-c above. What's your number this time?

If you turn the pot fully CCW, what are the resistances for a and b? Now recalculate steps a-c.
If you turn the pot full CW, what are the resistances for a and b? Now recalculate steps a-c.

Now you know the extremes. Want to find out what any particular setting on the pot is? As long as you end up with a total of 50K for the division between the wiper and the outside lugs of the pot, you can figure out the exact gain of any resistance.

[PBE6]

Funny, I just ran a little simulation on CircuitLab and the centre frequency appears to be 89 Hz for the whole sweep of the pot. The max gain was about 14.2 dB, which is pretty close to the calculated gain:

Av = 1 + (47k + 50k)/22k = 5.409091

20*log(5.409091) = 14.7 dB

The minimum gain was 4.2 dB, quite a bit lower than I expected from a (probably incorrect) calculation:

Av = 1 + 47k/22k = 3.316364

20*log(3.316364) = 9.9 dB

I must be missing something obvious...

[PBE6]

Aha! I think I found something. If you treat the remainder of the pot as a resistance in series with the ground resistor, you calculate something quite close to the low gain value:

Av = 1 + 47k/(50k + 22k) = 1.652778

20*log(1.652778 ) = 4.4 dB

Seems to work reasonably well for the midpoint on a linear pot too:

Av = 1 + (25k + 47k)/(25k + 22k) = 2.531915

20*log(2.531915) = 8.6 dB

The simulation has the gain as 7.9 dB, not right on but not far off. Will have to try it with other resistor values.

EDIT: (Didn't see midwayfair's post above)

[Ripdivot]

Funny, I just ran a little simulation on CircuitLab and the centre frequency appears to be 89 Hz for the whole sweep of the pot.

You are correct, I miscalculated.

[Derringer]

SWEET ..... formulas and some in depth discussion. Thanks!

Funny, I just ran a little simulation on CircuitLab and the centre frequency appears to be 89 Hz for the whole sweep of the pot.

You are correct, I miscalculated.

I'm getting the same values you did, 88 and 128. What were you calculating wrong?

When I tried adjusting the resistor values like PBE6 did for the gain calculations, my new frequency was about 70 ... which seems wrong

[PBE6]

In the simulation, it appears that the frequency doesn't depend on the wiper position but only the total feedback loop resistance (47k + 50k) while the gain depends on the wiper position. This kinda makes sense for the filter, since current (ideally) only flows from Vout to ground through Rg (22k) so the filter is the same no matter where the wiper is set. It also kinda makes sense for the gain, since the opamp only cares about the feedback/ground resistance ratio. But it kinda doesn't make sense when thinking about them both at the same time, mostly because of the errors between the calculations and the simulation results (maybe due to neglecting the capacitors?)

Hats off to anyone trying to develop an algebraic expression for the transfer function, I found that I lost concentration after about 8 lines 😳 I will have to try it again with some coffee on hand...

[bool]

It will probably make it easier this way - or not: The EQ boost "action" will be in fact a reversal of a "notch" that a bridged-t circuit produces when used as an inline filter (the opamp will try to compensate for the "notch" in the NFB with a "boost" at the output). The boost added to the signal bacause of the non-inverting configuration (!) will be proportional to a pot-position, like it would be if it was a notch filter - but reversed as a boost.

Think in reversal, and apply in a given context.

[Derringer]

definitely understand the reversal "notch" concept ... good there, thank you

I'm with you on the wiper position concept too PBE6, both with regards to gain and the frequency calculation. Thanks!

So then the center frequency here is really about 89Hz and not 150Hz as advertized?
I have a spreadsheet to make 

[jatalahd]

Here is my two cents:

I calculated the transfer function for the circuit as shown in the image above. When using the given transfer function, one needs to add the remaining part of the pot (noted as Rx) to R1, for example when pot is at "minimum", R1 = 47k + 50k and Rx = 1, when pot is at maximum, R1 = 47k and Rx = 50k.  When I plot the results in Octave I get the shown bandpass plot (for max, min and in between). When Rx is very small, I got a gain of 15 dB and when Rx is 50k the gain is about 4 dB at the center frequency of 89 Hz.

Theoretically, the center frequency f is actually shifted by the factor 1 + Rx/R1, but with these component values the center frequency stays the same 89 Hz, because the factor has a maximum value of 2 in this case and does not have much weight inside the square root. The quality factor can be calculated from here if someone has interest to it.

Hopefully someone had the same results from simulations? If so, then this transfer function is adequate to model the Bridged T op-amp circuit.

[PBE6]

Nice work jatalahd! Very elegant. The simulation I ran on CircuitLab looks just like your calculated results.

[Derringer]

Nice work jatalahd! Very elegant.

+1

[jatalahd]

It looks like such a useful filter that I wanted to continue the calculations. Simplifications of earlier equations and evaluation of Q-factor lead to equations:


I have just rewritten the earlier equations into different form.

The obvious result is now that the mid-frequency is always constant and it just adds R₁ and R_x as one resistance value.

The transfer function can be written in the form of non-inverting op-amp gain, 1 + Av. Now it is so clear that the gain function is a band-pass biquad function, whose gain is controlled with R₁. Remember that I defined my analysis model so that R_x models the pot such that the remainder of R_x is added to R₁. Therefore when the pot is at minimum (Rx is very small), R₁ has the maximum value and gives the maximum gain.

I also calculated the Q. It is somewhat scaled by R_x/R₂, which was a bit surprising. Using the given values, the maximum Q is 0.5 and minimum 0.16.

Then if someone wonders how the input filter (10uF and 47k) affects the results, I added the related scaling factor into the transfer function. When the product of Cin and Rin is kept large enough (condition given in image), it will not affect the transfer function at all. That is why I neglected it in the first place.

Before this I had never seen this kind of filter before, but it seems so amazingly useful that I really want to use it in many circuits! Thats also the reason I wanted to calculate and simplify the equations, because now in the simplified form they provide a much help to the design process.

[PBE6]

+1 great work jatalahd, will definitely be referring to this many times for future builds!