What are these opamps doing?

[Sage]

This is the example mono circuit from the datasheet for the Accutronics/Belton BTDR-3 reverb module.



Can someone please explain to me what the opamps in this schematic are doing, exactly?  The first one, which feeds the BTDR-3's input, looks like a textbook example of a differentiator amplifier, but I don't understand what the purpose of that is.

The second opamp is obviously working as a summing amplifier to combine the stereo outputs from the BTDR-3 into a single mono signal... but the dry signal is being plugged into the opamp's non-inverting input, which makes it look like a differential amplifier, and I don't understand exactly what that does.

Can someone explain this circuit to an untrained electrical engineer like myself?

[Groovenut]

The first opamp is an inverting amplifier with a gain of 2.2 above 723Hz, boosting the signal into the brick so the output of the brick is at unity gain. Second opamp is as you said a summing amplifier, mixing the dry signal with the signal coming from the brick. The dry signal is 180 degrees out of phase with the signal coming from the brick because of the first inverting opamp stage.

I personally dont like this arrangement for the belton brick circuits.

[Rob Strand]

Can someone please explain to me what the opamps in this schematic are doing, exactly?  The first one, which feeds the BTDR-3's input, looks like a textbook example of a differentiator amplifier, but I don't understand what the purpose of that is.

It looks like a differentiator but it's actually a high-pass filter.  The filter only feeds the upper mids and higher frequencies into the reverb.   The idea is to stop reverb of low frequencies.

The difference between a differentiator and a high-pass filter can be subtle.  When the cut-off is above the highest frequency of interest it could be called a differentiator, when it's not, it's more correct to call it a high-pass filter.
[Since your an engineer I'll add:  a differentiator often has an upper hpf frequency to stop it boosting higher frequencies by a crazy amount. Such boosting doesn't add anything to the in-band frequecies]

I personally dont like this arrangement for the belton brick circuits.

They don't seem too switched on! They have bugs in their datasheet:
http://www.tubeampdoctor.com/images/File/BTDR-3%20DIGI-LOG%20REVERB%20MODULE.pdf

For the stereo circuit the high-pass cut off should have R1 replaced by R1/2.  The mono formula is OK.
The gain has a factor of two in the stereo formula but that is also wrong.  The "2" should be removed.

(I don't what to look at it anymore it probably has more bugs in the summer.)

[Sage]

I personally dont like this arrangement for the belton brick circuits.

What is it you don't like about it?  How would you change it?

[anotherjim]

Dry signal is always present, so it can't properly be used in a parallel FX send/return arrangement that most amps have -  it would have to be in a series insert FX loop (some amps do have this though).

Input impedance is too low to use directly with guitar. That is a line level circuit.

The use of the term "common" could be misleading. Some will assume that means 0v, but that can only work if the op-amps have bi-polar power supplies. For single supply, "common" is Vref or half the single supply voltage.

[Sage]

The first opamp is an inverting amplifier with a gain of 2.2 above 723Hz, boosting the signal into the brick so the output of the brick is at unity gain.

Why does boosting by 2.2 at the input result in unity gain at the output?  Is that just how the BTDR-3 works, or does it have to do with what's happening after the brick's output?  It looks like the feedback resistors on the second opamp are actually reducing the volume (10/39).  Am I reading that wrong?  Why is that also being applied to the dry signal?

Second opamp is as you said a summing amplifier, mixing the dry signal with the signal coming from the brick. The dry signal is 180 degrees out of phase with the signal coming from the brick because of the first inverting opamp stage.

OH!  I get it, it's set up like a differential amp because when you subtract a negative value from a positive one, it's like adding two positive values.  1 - (-1) = 2.

Dry signal is always present, so it can't properly be used in a parallel FX send/return arrangement that most amps have -  it would have to be in a series insert FX loop (some amps do have this though).

Can't you just unplug the dry signal from the non-inverting input of the second opamp to get a pure wet signal?

Input impedance is too low to use directly with guitar. That is a line level circuit.

Which is funny, because the datasheet explicitly says the example circuits are for instrument-level signals.  I'm actually planning to use this with line level signals, I'm applying the reverb to several mic preamps, so I'm good with this.

The use of the term "common" could be misleading. Some will assume that means 0v, but that can only work if the op-amps have bi-polar power supplies. For single supply, "common" is Vref or half the single supply voltage.

Yeah, the datasheet mentions that: "'Common' is 'Signal GND' in a split-supply circuit or Vcc/2 in a single-supply circuit."  I don't understand how a split-supply even applies to this, though.  Does that mean if I power the opamps with +/-15V then common is 0V?

[R.G.]

It's easiest to understand the first opamp if you remember that the order of components in a purely series string has no, none, nada, zilch effect on the voltages and currents at the ends of the string. So you can freely swap that 10K and the cap, resulting in a textbook inverting AC amplifier with a capacitor at the input to block DC. Making that swap also makes the low frequency rolloff of the input R-C much easier to see. Since the inverting input of an opamp works on currents, not voltages, the thing that matters is how much current is being shoved into the (-) pin to be cancelled by currents fed back through the feedback (duuuh!) resistor. The impedance of that cap rises linearly with frequency, so it limits current at lower and lower frequencies, and it crossover at the frequency where its impedance equals the resistor.

Differentiators work the same way, but without the resistor, and the incoming impedance is purely limited by a cap, not cap plus resistor. That view ignores the fact that all incoming signals have some internal impedance too, but the textbook view is that it's quote "small" unquote.

I don't like the Belton schemo much myself, for some of the reasons already presented. The biggest is that I prefer to do my own mixing. I've laid out a PCB that will use the BTDR as the replacements for the guts of a reverb tank, using the tank's shell and I/O jacks and producing much the same input and output characteristics, excepting that it needs DC to run. I'm a little shocked that Belton hasn't already done this, as it was on my mind some years ago. I got early samples of the BTDRs at one of the NAMM shows way back when and asked the folks at the show if they had it packaged in a reverb tank. They said they had no plans to do that, reasoning that their market was new designs. C'est la vie.

Anyway, the datasheet says that the BTDR works with signals of 1.5V peak. I made that to be 3V peak to peak, and on my BTDR designs, I put a dual pair of clipping diodes in front of it, to snip off the incoming signal at 1.4V peak and give the overload characteristic some softer clipping behavior rather than the blaaaat of an overdriven A-D converter.

The opamp on the output is a kind of mess. They're adding the two channels from inside the BTDR in a simple inverting mixer, then doing a non-inverting add of the dry signal. Good, I guess, if what you're doing is a new design for an amp, but not in general what you want. for effects and such.

[PRR]

> Why does boosting by 2.2 at the input result in unity gain at the output?

Look at one output. Full-up. Gain set by 39K and 10K. Gain is about 1/4 (0.256).

Now look at the whole thing. There's two outputs. They are summed. So the output gain is 1/4+1/4= 1/2 (0.513).

Gain of 2 in and gain of 1/2 out is unity gain. (2.2*0.513= 1.13.... they may be compensating for delay-chip loss, or aiming high because that always sounds better than coming shy of unity, or expecting you to not always turn full-up.)

> looks like a ....differentiator amplifier

As was probably said: nearly all audio amplifiers are "differentiators", up to a point. We have gain above 20Hz but zero gain at zero Hz. The range from zero to 20Hz is indeed a differentiator, but we call it a low-cut (hi-pass) or DC block.

The difference is that resistor in series with the cap. If the resistor is larger than the cap's reactance, at the frequencies you actually care about, then the differentiation is not the point.

When the corner is not "below audio" (20Hz), but "mid audio" like 750Hz here, we indeed differentiate a large slice of the lower audio frequencies. Again we call it "low cut".

We almost always cut bass into a reverb because reverb deals with short-time manipulation of sound events. Big boring bass sounds don't benefit from "small reverb" like a $10 chip. (Bass reverb is important in large rooms, but that possible benefit is hard to capture in a small box; and many performance spaces are already bass-boomy.) Also bass is most of the power in speech/music, so tends to eat-up electronic headroom faster than it gains richness. The ~~750Hz bass-cut matches traditional Fender spring-reverb drives (the 500pFd cap and 500K resistor into the reverb driver tube).

[Sage]

Look at one output. Full-up. Gain set by 39K and 10K. Gain is about 1/4 (0.256).

Now look at the whole thing. There's two outputs. They are summed. So the output gain is 1/4+1/4= 1/2 (0.513).

Gain of 2 in and gain of 1/2 out is unity gain. (2.2*0.513= 1.13.... they may be compensating for delay-chip loss, or aiming high because that always sounds better than coming shy of unity, or expecting you to not always turn full-up.)

Ah, now I get it, thanks.  I suspect the slightly-higher gain on input has more to do with using more common values for the resistors than anything else.  A 22k resistor is a little more common than a 20k resistor, I should think.  Same with 39k instead of 40k.

But why boost at all?  Why not feed unity gain into the brick and then cut the two outputs to 1/2 and then add them together?  Wouldn't there be less chance of digital clipping from the brick's internals that way?

Also, how does that math work when adding the dry signal into the non-inverting input of the second opamp?  It appears to be cut down to 1/4 as well...

We almost always cut bass into a reverb because reverb deals with short-time manipulation of sound events. Big boring bass sounds don't benefit from "small reverb" like a $10 chip. (Bass reverb is important in large rooms, but that possible benefit is hard to capture in a small box; and many performance spaces are already bass-boomy.) Also bass is most of the power in speech/music, so tends to eat-up electronic headroom faster than it gains richness. The ~~750Hz bass-cut matches traditional Fender spring-reverb drives (the 500pFd cap and 500K resistor into the reverb driver tube).

Thanks for this info, I didn't know this.  It makes sense that they'd be attempting to replicate a spring reverb sound.  I assume I would want to set the frequency cut lower if I wanted to replicate, say, more of a "hall" type sound?

[R.G.]

But why boost at all?  Why not feed unity gain into the brick and then cut the two outputs to 1/2 and then add them together?  Wouldn't there be less chance of digital clipping from the brick's internals that way?

It's at least partially because you want to use all the bits in analog to digital conversion. I don't know exactly the effective resolution of the BTDR, but let's assume they're 16 bit internally. This is almost certainly incorrect in detail, but it's a useful way to think of things.

If the signal you feed into a 16 bit converter has full scale peaks in each direction, you get to use all 16 bits. If the signal only ever peaks at 15 bits, you still have the other half of your signal conversion available, but unused. You'd get the same results with a 15 bit converter. Put another way, the residual noise from conversion is still there at the same volume it was before, but the signal is now half as loud, so you've lost 6db of whatever signal to noise you would have with a full 16 bit signal. If you only ever put in peaks at 14 bits high, you lose another 6db of signal to noise.

This thinking leads you to the conclusion that you want to feed an A-D converter a signal that just peaks at the full resolution, minus whatever you feel you have to have to avoid digital-blatting distortion from overloading the converter. companding before sampled-data processing is a time-honored way of keeping as much of the resolution in the process as possible, and expanding the inherent noise in the process downwards at the end. This is also why companding is sometimes used before bucket-brigade chips; they're a sampled-data system as well, and sampling noise applies, even if quantizing noise does not.

Belton has graciously told us where full resolution is - it's at 1.5V peak. So you get the best signal to noise ratio out of the BTDR if you feed signals that just  barely touch that as a maximum. It is possible that there's an internal compander in there, don't know. But the same reasoning applies to a compander - feed it signals that just touch its non-distorting limits on peaks. It is very likely that whomever designed the BTDR innards and the app circuits were cognizant of this line of reasoning and that the suggested gains are their best guess at what is needed to make the best simple use of it's non-distorting range.

That's also my line of reasoning with the 1.4V diodes distorter. It converts accidental overloads into more acceptable clipping than the occasional digital overload blat, and it's the kind of overload we expect from analog electronics, while staying just barely inside the max range Belton quotes.

As Freud probably said "Well, it's just a theory."   

[anotherjim]

Those PT2399's in the bricks are bit stream. It's "kinda" PWM. The delay memory per chip is something like 44k x 1bit. I'm not sure if you can directly compare to a normal digital conversion which, although the digital signal is usually handled in serial form, have 16bit or better word length.

[PRR]

> But why boost at all?

How tall is the signal? How tall is the door?

If we take R.G.'s 1.5V, then it seems Belton expected 0.7V max inputs. This is a fair number for strong guitar. (However the low input impedance sucks treble if fed straight from guitar....)

In semi-Pro audio gear, reference is -10dBV 0.3V on a VU-like meter but peaks are expected to 2V. Pro gear scales to +4dBm 1.2V on slow meter with peaks to 8V. Hi-fi is similar to semi-pro. So it's not really good for any of the common interfaces.

I am SURE Belton aimed this info at people who could adapt the gains to suit their particular use.

Also note that it may clip in the reverb module but the dry sound has far higher headroom (depending on supply voltage). A little clipping in the reverb path is no disaster. Dry should be squeaky clean.

> more to do with using more common values for the resistors

Well, yeah, those are the dead-common values any builder should have handfuls of.

But other common values could be picked and end up at gain of, say, 0.92. -0.7dB. While 1dB differences are often negligible, give guitarists a 0.7dB drop and they are disappointed. They never complain about 'unity gain' being 1.1 or +0.8dB.

[R.G.]

And that's the reason I mumbled about other conversions and not being exactly a 16 or other defined-word-size conversion. They're actually sigma-delta (or delta-sigma, both terms are used for the technique) and not exactly related to a fixed number of bits in this sense. Well, they are related, but it takes some fancy math to dig it out.

The smallest resolution a traditional A-D can do is one LSB. The quantization noise from that is 1/2 LSB, which is fixed, so each additional bit adds 6DB better signal to noise. A traditional word-wide A-D of N bits implemeted with SAR can get an N bit sample in N clock cycles, and so its fundamental sampling rate is N clock cycles. This sampling rate then has to be divided by at least two to get the fastest signal that can be reconstructed without aliasing.,

A sigma-delta has an LSB that is the smallest voltage step it can resolve as different from the varying signal, as the "sampler" looks for a difference of one LSB from the reconstructed signal to the incoming signal. It does this very quickly, as it's a single-bit A-D, but it must do it many, many times to accumulate resolution greater than its single bit. IIRC, it takes 2^N samples for the sigma-delta to resolve to N bits. the attraction for S-D is that it inherentlly is a lowpass for its own reconstructed signal and a highpass for its quantization noise, so it moves the sampling noise up where it's easier to filter out. The price is that you have to oversample much much more than with an N-bit wide converter; not 2:1 or 3:1, but 128:1 or more.

This has the same side effect, ultimately. You toss out your effective resolution by never feeding it a signal as big as it can handle.

I side stepped all this in the first post because you face the same kinds of tradeoffs, and it's far simpler to explain. SD appears to be an infinite bit stream, but it does not in fact have an unlimited resolution nor unlimited signal to noise.

[Sage]

Thanks, everyone, I think I understand this a lot better than I did before.  It seems clear that, despite the input impedance being too low, this circuit was designed with instrument level signals in mind.  Being that I intend to apply this reverb to the signal from a set of three INA217 mic preamps, I'll probably want to eliminate that gain boost at the input, and perhaps even reduce the signal into the brick.  I'll probably also want to adjust that high-pass filter since my aim is to emulate more of a hall reverb than a spring.

I must say, although you guys see it as messy, that plugging the dry signal into the non-inverting input and using it as a differential amp to combine it with the negative-polarity wet signal seems really clever to me, mainly because by giving the dry signal its own input on that second opamp, they prevent the wet signal from connecting directly back to the first opamp, which is something I'm having trouble getting around in other ways.  Can anyone suggest another way to accomplish this at the mixing stage (i.e., if I were to take the summed L/R outputs from the brick and then combine them with the dry signal via a third summing opamp)?

Anyway, the datasheet says that the BTDR works with signals of 1.5V peak. I made that to be 3V peak to peak, and on my BTDR designs, I put a dual pair of clipping diodes in front of it, to snip off the incoming signal at 1.4V peak and give the overload characteristic some softer clipping behavior rather than the blaaaat of an overdriven A-D converter.

Do you have a link to a schematic of this, or something similar to this?  I'm not at all experienced with diode clipping.

[PRR]

That type of "Virtual Earth" mixer has nominally "zero" sneak-back from one input to another. (So why did they use it for two nearly identical inputs?) In practice the sneakage should be inaudible, known only from lab tests.

Fender did fine putting a gain-stage in the dry path and then a massive attenuator to the mixer. Sneak-back through most amplifiers is very-very small.

[Sage]

Here's the circuit I came up with, incorporating R.G.'s diode clipping (what are the best diodes to use for this, btw?) along with my own idea to attenuate the reverb sends rather than the output.  I do have one question left: as you can see, while Inputs 2, 3, and 4 are all split off to be processed by the reverb brick, Input 1 is not.  The idea is that Input 1 should be unaffected by the reverb.  However, it seems to me that the only thing preventing Input 1 from leaking into Opamp 2 in this circuit is a couple of resistors.  Is that enough to isolate that signal, or do I need to be adding something else?

[Digital Larry]

The right side of R1 is at "virtual ground" and so there should be very little feedthrough going back through R2, etc to the other op amp.

All you op-amps get offa my lawn!