How (the heck) does this bandpass work??

[ebastler]

I am trying to understand the sound generator circuits from an old arcade game -- would like to implement a version of them for a 3V supply, using rail-to-rail op-amps.

But I am struggling with the attached "Rumbler" circuit, which I have simulated in TINA-TI (a SPICE derivative). Essentially it is a narrow bandpass filter. In the game's full audio circuit, it is fed with zener-diode based noise, and the output is used for rumbling "explosion" sounds.

I am bafffled by the very steep filter response which the simulation shows for this stage. In the vicinity of the peak at ~100 Hz, it's nearly 60 dB per factor of 2 in frequency! It clearly has to do with the feedback via C8/C9/R18, but I can't figure out what is going on there.

Could anybody explain how this very sharp response is created? Or does this bandpass circuit have a name which would help me look it up? -- Or is the TINA simulation simply misleading me here and not showing the real circuit behavior? (If so, any idea why?)

Many thanks for any pointers you might have!

[Benta]

The steep cutoff around the peak is nothing unusual. It's just a question of the Q (or damping factor if you like). The asymptotes on each side will flatten out to normal cutoff rates.

[SuzyC]

I read the graph showing -30 DB from 100Hz peak to 200Hz, not 60 DB unless you are making a power measurement rather than gain.
The negative feedback is amplified by approx. 15 by the transistor above 100Hz, and above 100-Hz the filter becomes a sharp 3-element active low-pass filter with a freq response that is trending towards approx 100% negative feedback.

All the resistors and capacitors of this circuit determine this circuit response, not just those you mention, especially C10 and R21 which interact negatively with C9 though C8.


You might consider this circuit to be a high-pass filter to 100Hz and a 3-pole amplified x15 low-pass filter above this peak.

[ebastler]

The steep cutoff around the peak is nothing unusual. It's just a question of the Q (or damping factor if you like). The asymptotes on each side will flatten out to normal cutoff rates.

But where does the high Q come from? My "naive" understanding is that each order of an RC filter will add a 20 dB/decade roll-off. So which element(s) of the circuit cause that local, steeper response?

There is no LC resonance in the circuit. Is it an effect of non-linear transistor behavior? Or what else am I missing here? (Something fundamental, I guess...)

[SuzyC]

The additional high attenuation is due to this circuit being active (not passive), so with the x15 neg feedback gain of the transistor starting about 100Hz and all freq  higher, this filter circuit becomes a high Q low-pass filter past 100Hz while just being a high pass filter below 100Hz.

[ebastler]

I read the graph showing -30 DB from 100Hz peak to 200Hz, not 60 DB unless you are making a power measurement rather than gain.

I was extrapolating the slope (both positive and negative) in the direct vicinity of the maximum. It is much steeper there -- which is what I don't understand.

All the resistors and capacitors of this circuit determine this circuit response, not just those you mention, especially C10 and R21 which interact negatively with C9 though C8.

Sure, I see (and believe that I understand) the typical RC filters on the base and the emitter. But those don't account for the very sharp peak, right? So I was assuming that the feedback does that "magic".

The negative feedback is amplified by approx. 15 by the transistor above 100Hz, and above 100-Hz the filter becomes a sharp 3-element active low-pass filter with a freq response that is trending towards approx 100% negative feedback. You might consider this circuit to be a high-pass filter to 100Hz and a 3-pole amplified x15 low-pass filter above this peak.

That's the part where I am still a bit slow. How does that negative feedback behave as a function of frequency? The peak falls off very sharply also towards lower frequencies. To obtain the sharp peak, I would assume the feedback needs to be weakest at 100 Hz, and increase towards both higher and lower frequencies. Does it do that? Your description ("is amplified by approx. 15 by the transistor above 100Hz") does not fit my -- admittedly very hazy -- picture of what "should" be going on.

EDIT: The feedback does look like a T high-pass, which is in line with your explanation. But then, why the steep drop below 100 Hz?

Could you elaborate a bit more please? Many thanks!

[Benta]

Do the maths.
It''s a plain second order BP filter with relatively high Q. Whether it's done with an LC section or with only caps and an amplifier doesn't matter.
I attach the general transfer function for both cases.

[zeta] is the damping factor, 2 * [zeta] = 1/Q
[omega] = 2 * [pi] * f

[T3sl4co1l]

Put another way: it's essentially a phase-shift oscillator (notice the three RCs in the feedback loop), so has extraordinary gain at resonance, which can be far above the asymptotic "skirts", nothing wrong about that.  What is... not necessarily wrong, but tricky anyway, is the loop gain must be very slightly less than unity for this to happen; more and it simply oscillates as usual, less and the gain peak and therefore selectivity is far less exaggerated (but, that may have sound-shaping relevance too, assuming the change in overall gain itself is compensated for; that is, this is a variable one can tune to adjust the sound quality for various purposes).

Gain depends on exact resistor values, and transistor bias, so it's tricky to balance.

This is also the basis of the regenerative receiver; but not however the superregen, which harnesses an added layer/quirk on top of it.

Tim

[srb1954]

The negative feedback is amplified by approx. 15 by the transistor above 100Hz, and above 100-Hz the filter becomes a sharp 3-element active low-pass filter with a freq response that is trending towards approx 100% negative feedback. You might consider this circuit to be a high-pass filter to 100Hz and a 3-pole amplified x15 low-pass filter above this peak.

That's the part where I am still a bit slow. How does that negative feedback behave as a function of frequency? The peak falls off very sharply also towards lower frequencies. To obtain the sharp peak, I would assume the feedback needs to be weakest at 100 Hz, and increase towards both higher and lower frequencies. Does it do that? Your description ("is amplified by approx. 15 by the transistor above 100Hz") does not fit my -- admittedly very hazy -- picture of what "should" be going on.

EDIT: The feedback does look like a T high-pass, which is in line with your explanation. But then, why the steep drop below 100 Hz?

Could you elaborate a bit more please? Many thanks!

If you redraw this circuit a bit, with the input voltage source shorted out, you will see that it looks like a classic phase-shift oscillator. Feedback is passed through the 3 RC networks to produce a total phase shift of 180deg which, together with the 180deg phase inversion in the transistor, gives you overall positive feedback. With sufficient gain the circuit oscillates but here the gain is tamed down a bit with the emitter resistor so that the circuit is not quite on the verge of oscillation.

It is the positive feedback around the circuit that gives the apparent high Q. With a noise signal injected into the circuit this excites the high-Q circuit which rings vigorously at its resonant frequency giving the appearance of a narrow bandwidth BPF.

This technique is known as "Q multiplication" and was once used in RF circuits to give very high gain and selectivity with simple circuits.

[Benta]

I'm  with Teslacoil and srb1954 here, a medium-to-high Q active BP filter can be hard to tame and will easily turn into an oscillator. On top of that, it's even harder to tune (center frequency).
Still, it's a second order filter and behaves normally.

[MarkT]

R20 and R22 set the voltage gain for ac, I presume this is carefully arranged to be somewhat below the oscillation point.