What does this op-amp implementation do?

Started by GGBB, October 17, 2022, 01:56:19 PM

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GGBB

I have a tube amp that uses an op-amp based reverb driver which seems unusual to me - one op-amp with some gain is fed into a second unity op-amp and the outputs of both are joined and fed to the reverb tank. Could someone explain what this does/why it's used?





I have another amp from the same family (very similar topology, single channel vs dual, 15W vs 30W) that just uses the single op-amp minus R78 and C47. It uses half of a TL072 instead of NE5532.
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PRR

It is noted as a 150 Ohm tank. Cheap opamps can drive 2,000 Ohms. The '5532 can drive 300 Ohms. Two together can drive 150r. R78 R79 enforce equal current sharing.

But you don't have to do reverb "perfect". A lot of designs just use a part of a TL07x. Just like the first Fender Reverb (standalone) used a 6F6 (a 7/8 scale 6V6) to drive, and the onboard reverbs usually used 12AT7, a much smaller bottle.
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FiveseveN

Rod Elliott goes in depth on strategies for tanks of various impedance, including this kind of driver (Figure 5A): https://sound-au.com/articles/reverb.htm
Quote from: R.G. on July 31, 2018, 10:34:30 PMDoes the circuit sound better when oriented to magnetic north under a pyramid?

Rob Strand

#3
There's actually two ideas here:

- The spring is current driven, that's the R27 part and the way the spring is wired within the feedback loop.
- The opamps are wired in parallel to boost the drive current of the opamps.

The thoughts behind the booster are:
- The output of the second opamp follows the first but current is supplied from each opamp via the 10 ohm resistors, essentially doubling the current drive.   
- The 10 ohm resistors allow the outputs to be combined, without those there's no way to wire in the second opamp.
- The 10 ohm resistor also prevent unwanted oscillations and unwanted DC flowing *between* the opamp outputs due to slight differences in opamp's DC offset.  A 10mV offset on each will produce at most 20mV/20 ohm = 1mA DC current to flow - very small in comparison to the output currents.  At the same time the voltage drop is small.
- I suspect 10 ohm is actually too small to prevent the opamps heating up at high levels.  That's where you need
  higher resistor values and more opamps in parallel.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

GGBB

Thanks folks. I wondered if it was a current thing.

Rob - you said
Quote from: Rob Strand on October 17, 2022, 06:52:33 PM
- The spring is current driven, that's the R27 part and the way the spring is wired within the feedback loop.

I would have assumed that R27 was purely for biasing and setting the input impedance. How does it affect current drive? Is it as simple as Iin * gain = Iout?


More peculiarities - in comparing the two versions of the amp which both use the same reverb tank, the reverb tank out is returned into a TL072, but the amp with the NE5532 send lacks a bias resistor R31 47k on the return op-amp. Does that seem right/wrong? Could it just be an error on the drawing?




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Rob Strand

QuoteI would have assumed that R27 was purely for biasing and setting the input impedance. How does it affect current drive? Is it as simple as Iin * gain = Iout?
My apologies for the annoying typo!  I meant R28.

Because of feedback the opamp + and - inputs are at the same voltage which is Vin.   Vin appears across R28 (and C27, which is large enough to be ignored.  A current Vin/R28 flows through R28 and that must equal the current through the reverb coil.

QuoteMore peculiarities - in comparing the two versions of the amp which both use the same reverb tank, the reverb tank out is returned into a TL072, but the amp with the NE5532 send lacks a bias resistor R31 47k on the return op-amp. Does that seem right/wrong? Could it just be an error on the drawing?
It looks OK to have it.

You will often see load resistors and capacitors across the receive coil.    The idea behind these components is to tune the HF response.   Some circuits try to maximize bandwidth without too much peaking others deliberately roll-off high end.    It's a good idea to have at least 100k there so the opamp has some bias when the reverb spring is pulled.   (It also lowers an crackle due to bad connections to the tank, which is insanely loud without a bias resistor.)
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

GGBB

Interesting - thanks again. I always felt the two reverbs sounded a little different - this would explain that.
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Rob Strand

Quote from: GGBB on October 20, 2022, 09:08:02 PM
Interesting - thanks again. I always felt the two reverbs sounded a little different - this would explain that.
Yes, it can be audible but subtle.

On quite a few amps I've found a parallel cap soldered inside of the reverb tank (1n to 2n2) , then another parallel cap on the PCB (say 100pF).   The cap soldered inside the tank *doesn't appear on the schematic* but it is much larger in value!  If the reverb receive coil is say 250mH then resonating that with a 2n2 will give a resonant frequency of 6.8kHz.  Whereas a receive coil of 450mH inductance will resonant with a 2n2 cap at 5kHz and resonate with a 1n cap at 7.5kHz.   In these cases you tune the parallel resistor to so the peak at resonance isn't too high.

If however we don't have a cap then a 250mH coil and a 47k load resistor we might expect the roll-off to be 30kHz but what normally happens is the coil and cable capacitance causes some undesirable peaking and pickup of high frequencies.   If you at least have 100pF across the coil it keeps the peaks down below 30kHz or so.

You get similar problems with phono cartridges,
https://www.hagtech.com/loading.html

There's point where the mechanical/acoustical effect limits the frequency response and that will override any electrical cut-off.  IIRC 6kHz to 10kHz would be about it.

Nothing stopping you playing with a few values.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.