Author Topic: Unwanted distortion in "Shoot the Moon" tremolo. Can I tweak it to reduce gain?  (Read 2676 times)

D.C.

I just assembled a Shoot the Moon tremolo on my breadboard last night, using the TagboardEffects 2-knob layout. (Also, here's a schematic I found online. Not sure whether it's accurate, but it seems right.)

The tremolo effect works, but unfortunately I'm getting a bit of distortion when I hit low notes hard (on the low E string). The note fizzes for a bit, and then the distortion goes away. It's subtle (I thought it was just ordinary noise at first), but it's definitely there.

I see people have had this issue with the EA tremolo (see here) and the Hearthrob tremolo (see here). Some folks have tried to tweak the circuits to reduce the gain, but those are both transistor-based circuits. I'm not sure what would need to be tweaked in this circuit to reduce the gain.

I'm no analog electronics expert by any means, but I tried to do the first steps of an analysis. See here for a couple comments on the schematic. In terms of the actual signal path, it appears it's just a high-pass filter (in the form of an inverting amplifier), followed by the LDR going into a low-pass filter (also in an inverting amplifier). The cutoff frequency of the high-pass filter is given by:

f = 1/(2*pi*220k*1u) = 0.72 Hz

Essentially, everything audible goes through. In terms of the inverting amplifier, the gain is given by:

G = -Rfeedback / Rin
= - 220k / 220k
= -1

So it's pretty much just an inverting buffer; no gain.

The low-pass filter's not as easy; it can vary, because of the variable resistor (a 10k trimpot). I've tweaked my trimmer to provide a final gain of unity, but I don't have my multimeter, so I can't measure its resistance. But to make things easy, let's just assume it's at the halfway point:

f = 1/(2*pi*5k*330p) = 96.46 kHz

Again, everything audible is passed through. Ignoring the filter, the gain is given by:

G = -Rtrimmer / RLDR

This also isn't gonna be a fixed number, not only due to the trimmer, but also due to the LDR, which fluctuates. The datasheet for my LDR from Tayda says the dark resistance will be a minimum of 0.5M, and the lit resistance will be between 10k and 20k. Again, assuming the trimmer is 5k, we have:

Gdark = -5k/0.5M = -0.01
Glight = -5k/10k = -0.5

But remember that the gain from the first inverting amplifier (the first op-amp) was -1. So the final gains will actually be positive.

It seems like there's something I'm missing here. Seems like the only way to get unity gain out at the end (G = Rtrim/RLDR) is when Rtrim = RLDR. But the LDR isn't spec'd to go lower than 10k, so the trimmer would have to be at least 10k. I know the trimmer's not set to its maximum value, and even if it were, that seems like it would be bad design -- there's no wiggle room in the case of an LDR that goes no lower than 10k.

In any case, I'm not seeing any large gain in this circuit that should be causing any clipping. Anyone have any ideas? Anything I'm missing here?

Transmogrifox

You are correct, the gain isn't very high in this circuit.  With the correct LDR you should be able to adjust it to unity.  As you have already shown, its max gain is already less than unity.

The cause of the distortion may or may not be clipping.  I haven't noticed these LDR's are terribly nonlinear, at least, not enough to hear audible distortion, so I wouldn't look there first.

The oscillator might have a jagged enough waveform to give you some intermodulation that sounds like distortion at certain places in its sweep -- and it would become more audible when mixed with lower frequency audio.

Just to confirm where the distortion is coming from,  replace the LDR with a 10k resistor and see if you can hear any distortion.  It should be just a buffer with no effect on the sound.

Next replace the LDR and set the LED to full light (max gain) and play through just plain.  It should sound the same as when the LDR was replaced with a resistor.  If you hear distortion, then you know it's coming from the audio path and not getting modulated into the audio path by the oscillator.

If it's coming from the audio path, then it might be due to DC bias shifting with the resistance change on the LDR.  You can put a meter on and confirm everything is really close to 4.5V at op amp outputs.  In that case maybe you would want to trim the output of the second op amp until you have 4.5V. 

Finally (actually first), check your battery charge.  If your battery voltage is low enough you will start getting clipping at pretty low levels.  A strong humbucker might drive it into clipping on the low strings.

trans·mog·ri·fy
tr.v. trans·mog·ri·fied, trans·mog·ri·fy·ing, trans·mog·ri·fies To change into a different shape or form, especially one that is fantastic or bizarre.

D.C.

Thanks for the tips; I think we've found it.

Quote
Just to confirm where the distortion is coming from,  replace the LDR with a 10k resistor and see if you can hear any distortion.  It should be just a buffer with no effect on the sound.

Correct. However, the volume gets reduced somewhat. (Maybe about half, like I calculated...)

Quote
Next replace the LDR and set the LED to full light (max gain) and play through just plain.  It should sound the same as when the LDR was replaced with a resistor.  If you hear distortion, then you know it's coming from the audio path and not getting modulated into the audio path by the oscillator.

Found it! I wired the positive end of the LED straight to 9V and used a 1k current-limiting resistor. Sure enough, there's distortion. It's really noticeable now.

It seems to me that the solution would be to reduce the maximum brightness of the LED.

This brings me to my next question. What's the deal with the 1k resistor in parallel with D2 (in the schematic), in addition to the 1k resistor in series? Why isn't there just a single 1k resistor in series with the LED, as usual? (Of course, there's also some contribution to the resistance from the "Depth" pot.) Could I replace those two resistors with one? Maybe make it a trimpot, so I can adjust the maximum brightness?

D.C.

Got it! Thanks again!

Removing the 1k resistor in parallel with the LED and replacing the 1k with a larger resistor didn't cut it. Doing that reduced the tremolo's depth.

Instead, I replaced both 1k resistors with increasingly larger values. 2k didn't completely get rid of the distortion. Nor did 3.9k, 4.7k, or 5.1k. But 10k seems to do the trick. And no depth is lost; when the "Depth" knob is at maximum, it still cuts all the way.

This solution will probably vary with each individual build. Maybe my particular red LED needs 10k resistors to ensure that it doesn't get too bright and cause distortion. But maybe someone else's red LED has different specs and will work just fine with 1k resistors.

I think I'm starting to figure out what that resistor in parallel with the LED is doing in this circuit. Kind of hard to say what's running through my head. If I can think of a good way to explain it, I'll make another post.

duck_arse

if you decrease the value of the resistor across the led, it will shunt more current away from the led, making it dimmer. if you increase the value of the resistor below the led, it will decrease the maximum current through the led/upper resistor, reducing the led brightness.
duck a-duckka not fade away .....

Kipper who?

Transmogrifox

The parallel resistor shunts current around the LED, in a sense "tuning" the range of voltage applied across the LED, or shunting current around it.

Typically the parallel resistor with an LED is meant to soften the transition from full dark to bright.  It's normally an exponential (very sudden --  can cause a thump when it turns on) if this resistor wasn't there, but if you put it in parallel with a resistor you can get a linear + exponential shape, so the transition isn't quite so hard.

In my day job, I use this arrangement with opto isolators in isolated feedback loops in regulated switching power supplies.  Same idea, but the purpose is to prevent regulation from going unstable due to the rapid nonlinear transition from feedback to no feedback.  It often balances right on the point of the nail, or so to speak.  The parallel resistor blunts the top of the nail, by analogy.

Since the distortion is coming from the LDR, another trick would be to reduce the gain of the first stage to decrease the amplitude of the signal on the LDR then increase the gain of the second stage to make up for it.

In either case, I'm glad you found a way to fix the distortion without losing depth.
trans·mog·ri·fy
tr.v. trans·mog·ri·fied, trans·mog·ri·fy·ing, trans·mog·ri·fies To change into a different shape or form, especially one that is fantastic or bizarre.

duck_arse

The parallel resistor shunts current around the LED, in a sense "tuning" the range of voltage applied across the LED, or shunting current around it.

Typically the parallel resistor with an LED is meant to soften the transition from full dark to bright.  It's normally an exponential (very sudden --  can cause a thump when it turns on) if this resistor wasn't there, but if you put it in parallel with a resistor you can get a linear + exponential shape, so the transition isn't quite so hard.


this would explain what I see (and don't see) in the ea trem osc led when I put a parallel resistor across. nice one.
duck a-duckka not fade away .....

Kipper who?