Super low volume in muff overdrive clone. Unsure of cause.

[antonis]

To me, the explanation makes sense only when there is a feedback resistor on the second Opamp.

Consider Diode incremental resistance as an almost short...
(that turns second amp into a buffer..) 

[Mark Hammer]

Antonis, your response could be a haiku if it was but one word shorter. 

Vivek, I find the way to understand op-amps is that they really really REALLY want to amplify at their maximum gain-bandwidth product.  In our case (audio frequencies) that is often in the vicinity of 60-80db.  The way to adjust gain is to feed or restrict negative feedback from the output, generally by bleeding it off.  When we put a cap in the feedback loop, we are essentially giving a "pass" to lots of negative feedback for high frequencies, but less negative feedback for whatever lies below the corner frequency set by the cap value.

The metaphor I like to use is that of a car whose gas pedal is welded and bolted to the floor, and the only way to control the speed of the vehicle is via the brake pedal.  Less brakes = more speed.

What Antonis notes is correct.  Below the forward voltage, the diodes allow for the negative feedback to be blocked.  But get near the forward voltage, and as the diodes conduct, that negative feedback begins to flow, reducing the gain, near instantly....clipping!

Conceivably, a smoother sound, coming from a more seamless transition between maximum open-loop gain and the reduced gain, could be achieved by having a resistor in parallel with those diodes.

[nocentelli]

For anyone still searching in vain for suggested values for the resistor and cap in parallel with the diodes, 100k and 470p are suggested in this thread:

https://www.diystompboxes.com/smfforum/index.php?topic=117223.0

It is bewildering that virtually every single schematic on the opamp muff fuzz is wrong: i know that analog guru posted a correct, traced (1977) version at "the other place" a number of years ago. He has since pulled every schematic of his from the web, so i'll check my pc to see if I downloaded it at the time.

[Vivek]

Vivek, I find the way to understand op-amps is that they really really REALLY want to amplify at their maximum gain-bandwidth product.  In our case (audio frequencies) that is often in the vicinity of 60-80db.  The way to adjust gain is to feed or restrict negative feedback from the output, generally by bleeding it off.  When we put a cap in the feedback loop, we are essentially giving a "pass" to lots of negative feedback for high frequencies, but less negative feedback for whatever lies below the corner frequency set by the cap value.

Still, I maintain,

If there is no feedback resistor and capacitor on OpAmp 2, then

Changing diodes = no perceptible difference except volume

Changing gain of the first IC from 470 to 47 = no perceptible difference at all

[willienillie]

I didn't read the thread, sorry if this is already solved.

[11-90-an]

Changing diodes = no perceptible difference except volume

Changing diodes affects clipping characteristic, and “tone” as some people call it...

I didn't read the thread, sorry if this is already solved.

Indeed, it is solved... 

[Vivek]

Changing diodes = no perceptible difference except volume

Changing diodes affects clipping characteristic, and “tone” as some people call it...

Generally yes/maybe

But not in the situation where there is no feedback resistor.


Please be kind to analyse the situation with no feedback resistor  and guide me in case I misunderstood.

[Vivek]

If there is a feedback resistor setting gain to 200

input of 100mv peak

without diodes, the signal would have become 20V peak

si diode will clip that at 0.65 ie at 3.25%

LED will clip that at 2.5V ie at 12.5%


Big difference in tone due to clipping the input at different places (also, more volume from LED)



Suppose no feedback resistor on 2nd Opamp

Signal would have become 9 Million Volts

Clipping that at 0.65V is 0.000000%
Clipping that at 2.5V is 0.0000000%


No difference in tone due to clipping the input at same place effectively (also, more volume from LED)

[j_flanders]

For anyone still searching in vain for suggested values for the resistor and cap in parallel with the diodes, 100k and 470p are suggested in this thread:

Should be the other way round: 100pF and 470k
Best to read this thread instead, from this post on:
https://www.diystompboxes.com/smfforum/index.php?topic=123951.msg1174407#msg1174407

[11-90-an]

So I BB’d the circuit just now and here are my observations...
the diodes used where either 1n4148s or blue LEDs

- this circuit oscillates better than all the LFOs I wanted to oscillate but never worked (prolly a bb problem)
- from wiring a 50k pot in series with the 1k, dialing it to least resistance yields the most oscillation free-sound while the other extreme oscillates like crazy. Vivek’s speculation about opamp1 gain not affecting overall distortion  was correct, however there was a teeny “distortion characteristic” change... or maybe I’m hearing things since it’s 3am and I have to lower amp volume to not disturb family...
- leds yielded a thinner sounding fuzz while the 1n4148s have a bit more bass...

[iainpunk]

i believe i read somewhere that this mini muff fuzz uses germanium transistor leakage to set the gain in the 2nd gain stage, that's why at least one of the diodes should be germanium or they have to have a large value resistor in parallel.

cheers

[antonis]

Antonis, your response could be a haiku if it was but one word shorter. 

OK, Mark..
(I revoke the word "almost"..)

:cheers:

[antonis]

If there is a feedback resistor setting gain to 200
input of 100mv peak
without diodes, the signal would have become 20V peak
si diode will clip that at 0.65 ie at 3.25%
LED will clip that at 2.5V ie at 12.5%
Big difference in tone due to clipping the input at different places (also, more volume from LED)

Suppose no feedback resistor on 2nd Opamp
Signal would have become 9 Million Volts
Clipping that at 0.65V is 0.000000%
Clipping that at 2.5V is 0.0000000%
No difference in tone due to clipping the input at same place effectively (also, more volume from LED)

Now, suppose a couple of Ohms feedback resistor (forward dynamic resistance) for Diodes and a couple of tens Ohms for LEDs..
(bot non-linear but of different slope..)

Make your calculations as above and let us know your result..

[iainpunk]

If there is a feedback resistor setting gain to 200
input of 100mv peak
without diodes, the signal would have become 20V peak
si diode will clip that at 0.65 ie at 3.25%
LED will clip that at 2.5V ie at 12.5%
Big difference in tone due to clipping the input at different places (also, more volume from LED)

Suppose no feedback resistor on 2nd Opamp
Signal would have become 9 Million Volts
Clipping that at 0.65V is 0.000000%
Clipping that at 2.5V is 0.0000000%
No difference in tone due to clipping the input at same place effectively (also, more volume from LED)

Now, suppose a couple of Ohms feedback resistor (forward dynamic resistance) for Diodes and a couple of tens Ohms for LEDs..
(bot non-linear but of different slope..)

Make your calculations as above and let us know your result..

i guess you are pointing towards the fact that after the clipping threshold is reached, the gain is 1?
that only works for non inverting gain, this unit inverts. the voltage at the inverting input node is always equal to the voltage at the non-inverting node (unless the op amp itself is clipping), this restrict the output being more than Vbias +or- Vfdiode
also the order of magnitude of that stages input resistance is so large that the dynamic resistance of diodes is negligible.

cheers, Iain

[PRR]

• -7
• -6.5
• -6.4
• -6.8
• -6.3
• -6.9
• -6.5
• -7.5

-8 at pos battery connect point

Hint for "All": when you have a 9V battery, you "expect" some parts to be spanning a large fraction of 9V. Specifically an op-amp will (in our world) have one leg on Zero, one leg on 9V, and other legs in-between.

Here EVERY pin is within about one volt span. That could be right but is very odd. All the numbers are near 9V (actually the 8V of the apparently half-dead battery). Where is the ZERO we expect??

Spoiler: JD reported that "one of the IC pins wasn't soldiered properly!" If it was one of the Power pins, that matches the wrong readings correctly.

[antonis]

the voltage at the inverting input node is always equal to the voltage at the non-inverting node (unless the op amp itself is clipping), this restrict the output being more than Vbias +or- Vfdiode

also the order of magnitude of that stages input resistance is so large that the dynamic resistance of diodes is negligible.

Never said anything different...

I'm trying to make Vivek alter his consideration from "no feedback resistance - no NFB" to "short feedback resistance - 100% NFB"..

[PRR]

...If there is no feedback resistor and capacitor on OpAmp 2, then
Changing diodes = no perceptible difference except volume
Changing gain of the first IC from 470 to 47 = no perceptible difference at all

Agree.

At least for an ideal diode. Some diodes leak *approximately like* a resistor. This is of course unpredictable and should not be relied on.

The open-loop gain falls with frequency. Highs boosted less than lows.

The signal will rise easy to a MAX which depends of the diode. A part-volt for common diodes, a couple volts for LEDs. Once it hits max it flat-tops, further increase of signal make hardly any increase of output.

- this circuit oscillates better than all the LFOs I wanted to oscillate but never worked

Amplifiers oscillate and oscillators don't.

With no gain-limit resistor this may be more gain than anybody can handle without squeal.

[Fancy Lime]

Yepp, one more thread turned into the realization that diodes are not ideal. In fact, in this arrangement, you get forward and reverse leakage currents in parallel, which can be quite substantial but is rarely specified in datasheets. The leakage current reduces gain in the same way a parallel resistor would, only less predictively, which can be fun for diy but I would not recommend it for a commercial pedal. Generally Ge diodes tend to have rather a lot of leakage, no news there. If you need high leakage Si diodes, low-voltage Zeners or high current diodes are usually leakier than dedicated switching diodes.

Cheers,
Andy

[11-90-an]

With no gain-limit resistor this may be more gain than anybody can handle without squeal.

This may be an understatement...

A strange thing that happens (I mentioned it before)... upon lowering gain of the first opamp stage from around 470 to 4.7, the squeal becomes LOUDER when using leds, but when using 1n4148s, the inverse is true..(lower gain, low squeal)

I do gotta say, when it isn't squealing, it sounds quite nice... 

[Vivek]

If there is a feedback resistor setting gain to 200
input of 100mv peak
without diodes, the signal would have become 20V peak
si diode will clip that at 0.65 ie at 3.25%
LED will clip that at 2.5V ie at 12.5%
Big difference in tone due to clipping the input at different places (also, more volume from LED)

Suppose no feedback resistor on 2nd Opamp
Signal would have become 9 Million Volts
Clipping that at 0.65V is 0.000000%
Clipping that at 2.5V is 0.0000000%
No difference in tone due to clipping the input at same place effectively (also, more volume from LED)

Now, suppose a couple of Ohms feedback resistor (forward dynamic resistance) for Diodes and a couple of tens Ohms for LEDs..
(bot non-linear but of different slope..)

Make your calculations as above and let us know your result..

Thanks,

Forward voltage of diode decides WHERE will the diode start clipping.

So all above calculations are the same

Internal resistances and capacitances will decide HOW the clipping will take place

But in the situation with no feedback resistor to limit the gain, we are either chopping the wave at 1/ 14 million with Si or 1/4 million of its peak with LED. Both are very extreme clipping.

Then minute differences in internal resistance of diodes , especially if in the 10 to 100 ohm range, will make no perceptible change at all

Basically, the clipping is so severe in both cases, we get almost pure  square waves while considering / ignoring diode resistance WHEN THERE IS NO FEEDBACK RESISTOR.

The only difference i see between different diodes is volume.

I did a LTSPICE FFT analysis comparing Harmonic content of 9 million Volt peak signals chopped at 0.65V versus 2.5V.

It's same frequency content, except the 2.5v clip has higher volume. That was with ideal diodes.

Now suppose we considered diode resistance, the 0.65 could have become something like 0.65001 maybe.

So instead of clipping the wave at 1/14 millionth of the peak
, we will clip at 1/13.999999 millionth of the peak.

No perceptible difference.