Sweet Honey Overdrive - 2nd stage LED question

[Chuck D. Bones]

I have a Joyo Sweet Baby Overdrive.  Sounds pretty good, circuit looks identical to the published schematics of the MP SHOD, except for the opamp.  So I changed the RC4558 to OP275.  Not sure if it sounds any different.  I expected the LED in the 2nd stage feedback loop to add some even-order harmonics, but as best I can tell that isn't happening.  When I measure the response to a sinewave input, there are no even-order harmonics.  I ran an LTSpice simulation of the SHOD.  Again, no even-order harmonics.  Based on simulation and measurements it appears that the 2nd stage LED never conducts enough current to affect the guitar signal.  I'm wondering if the published schematics have the right resistor values for the 2nd stage feedback.  Anyone else have thoughts on this? 

[ElectricDruid]

You don't post a schematic, so I'm guessing a bit, but basing myself on this one:



From this, it'd depend on the voltage of the zoners (or whatever diodes you use for the first stage) and the forward voltage of the LED whether the LED ever does anything. The gain of that second stage is only ~x3 so while that's enough for someone situations, it's not enough for everything.

For example, if the first clipping diodes are simple silicon like 1N4148 and the LED is a fairly typical 2V forward voltage, its never going to do anything. The signal gets hard limited to around 0.6V but the first diodes, then boosted by the gain to around 1.8V, but still never really hits the LED forward voltage at 2V. So the choice of components in this circuit is pretty critical to the gain-staging.

HTH,
Tom 

[Chuck D. Bones]

Thanks, Tom.  I should have posted my trace of the Joyo Sweet Baby, here it is.  The ceramic capacitors are unmarked, so I based their values are on the Sweet Honey.  D4 & D5 are unmarked.  Their body sizes are consistent with 1N400x diodes.  In the back-to-back arrangement, it wouldn't matter if D4 & D5 were zeners because each one would go into forward conduction before the other saw enough voltage for reverse breakdown.  The maximum voltage across the LED (D6 in my schematic) is 5.6K / 2.7K * 0.6V = 1.24V, definitely not enough to crack a red LED on.  By my estimate, a "better" value for R12 would be around 1.5K.  Then D6 would conduct when D5 is driven hard.  BTW, R7 is marked "30B" in my unit, which is the SMT code for 2K.  If I ever get my hands on a real SHOD, I'll be sure to make some measurements.

- Chuck

[antonis]

The maximum voltage across the LED (D6 in my schematic) is 5.6K / 2.7K * 0.6V = 1.24V, definitely not enough to crack a red LED on.

Actually, the voltage across the LED should be (5.6/2.7 + 1) (non-inverting configuration  ) * 0.65 (more realistic voltage drop at about 2mA current) = 2V which should be OK for most red LEDs clipping purpose..

(after all, you can try to raise 5k6 resistor value to 6k8 with better result..) 

P.S.
"Cracking" an LED is only necessary to get light - clipping may occur in lower than nominal forward voltage drop..

[Chuck D. Bones]

Thanks for the feedback (pun intended).  Your formula would be correct if the LED were connected across both R12 and R13, however the LED is only connected across R13, so my formula is correct.  By "crack" I meant get the LED to conduct more than a small fraction of R13's current.  I'm looking for the LED to conduct at least 20% as much current as R13 in order to get a detectable amount of 2nd harmonic generation.  I'm thinking of replacing R13 with a 10K trimmer.

- Chuck

[antonis]

Your formula would be correct if the LED were connected across both R12 and R13, however the LED is only connected across R13, so my formula is correct. 

Plz allow me to insist on "my" formula..!! 

LED shunts R13 so it effectively takes its role..
(which role is to be the feedback item in the non-inverting Gain formula..) 

If LED was connected across both R12 & R13, we would have a heavily halfwave oscilating op-amp..

[bool]

You could also try your luck with a IR LED.

Or alternatively, and purely by speculation, a 6.8k / 2.2k combo would get you there.

[Chuck D. Bones]

I finally got around to replacing R12 (2.7K in the Sweet Baby, 2.61K on the SHOD schematic) with 1K.  Now the LED will conduct!  At low Drive settings I get some even-order harmonics before the hard clipping from the 1N4007s takes over.  The increased gain of the 2nd stage results in bigger signal at the top of the volume pot, which I ended up replacing with 50K log taper.  I am liking this pedal even more now.

- Chuck

[Chuck D. Bones]

Your formula would be correct if the LED were connected across both R12 and R13, however the LED is only connected across R13, so my formula is correct. 

Plz allow me to insist on "my" formula..!! 

Insist all you like, I invite you to measure or simulate the circuit.  I have done both.

Cheers!

- Chuck

[antonis]

Your formula would be correct if the LED were connected across both R12 and R13, however the LED is only connected across R13, so my formula is correct. 

Plz allow me to insist on "my" formula..!! 

Insist all you like, I invite you to measure or simulate the circuit.  I have done both.

Cheers!

- Chuck

Good for you.. 
(I always had the impression that two parallel connected items share the same voltage difference but your multimeter & simulator tend to make me alter it.. - dum spiro spero..)

Cheers!

- Antonis