Re-biasing TS808 for LED clipping

[Asymmetric]

Hello everyone, I'm new here on the forum!
I have been building pedals for a while now, and recently I've built a TS808 and I want to use LED's for the clipping, but I know need to change the bias as LED's require more power to work. I have used the tone pad schematic and i'm just wondering which resistor is used for this?

http://www.tonepad.com/project.asp?id=1

Thanks

[Mark Hammer]

I'm not sure what your goal is here.  Sticking LEDs in place of the 1N914s will give you a hotter output signal with less clipping at max gain.  If  that's what you want, then you don't need to change anything else.

[R.G.]

Welcome.

I know need to change the bias as LED's require more power to work.

??Huh??

LEDs *can* handle more internal power than some other diodes, And *can* radiate some amount of it away as light, but LEDs as clipping devices don't necessarily eat significantly more power than other clipping diodes.

Beyond that, the clippers in the TS are in the feedback loop of an opamp which is biased to nearly the middle of the available power supply. The opamp's feedback setup can drive its output to big enough currents to cause clipping on the LEDs as long as there's enough power supply voltage to drive the LEDs into conduction and as long as the opamps internal current limits don't kick in.

It's quite difficult to bias an opamp to a higher current to drive greater loads. Most opamps don't work that way.

Again, welcome, and dive into how things work. It's a great swim.

[Asymmetric]

Thanks for the replies.
What I mean is, is that when I have three LED's for asymmetrical clipping, it sounds like the clipping is mainly from the OP amp with only a small contribution from the LED's due to their forward voltages - so I'm struggling to saturate the diodes. I am using a 2134 op amp with all the other caps rated for 25v and above, so would using an 18v power supply sort this issue out? I thought that there was a resistor near the diodes that controls how much voltage is distributed to the clipping section?

So basically, I want to saturate the LED's.

[R.G.]

What I mean is, is that when I have three LED's for asymmetrical clipping, it sounds like the clipping is mainly from the OP amp with only a small contribution from the LED's due to their forward voltages - so I'm struggling to saturate the diodes. I am using a 2134 op amp with all the other caps rated for 25v and above, so would using an 18v power supply sort this issue out? I thought that there was a resistor near the diodes that controls how much voltage is distributed to the clipping section?

So basically, I want to saturate the LED's.

OK, I understand what you mean now.

If you've read The Technology of the Tube Screamer, you know that the clipping opamp follows the input signal and then adds on top of it a voltage equal to the gain of the section as clipped by whatever is clipping in the feedback loop. You're setting this up to be two LEDs one way, and one LED the other way.

To avoid opamp output clipping, you have to have enough power supply voltage so that (1) your biggest input signal voltage (pk-pk) plus three LED voltages does not exceed the power supply voltage minus how closely that opamp can swing to its power supply voltages and (2) the DC bias voltage fed to the opamp places the quiescent DC voltage so the output doesn't exceed the output voltage swing on signal peaks.

LEDs, unlike ordinary junction diodes, have quite different forward voltages depending on their color and composition. Red LEDs are lowest, about 1.2V, going up to as much as 3-4V for green, blue, and violet. There are others with bigger voltage needs. You don't say what LED colors you're using. If you're using green, you could easily need as much as seven volts in one signal polarity, 3.5 in the other, and you'd have eaten up 10.5V just to get the LEDs to conduct, not even considering the signal voltage or any lack of swing on the output of the opamp.

If you were using red LEDs at about 1.2V, the LED voltages would be 2.4 in one direction, 1.2 in the other, for a signal swing of 3.6V pk-pk for the LEDs, plus whatever dry signal the opamp is following. In this case, you might get by with a 9V power supply if you picked a DC bias voltage for the clipping stage that was offset by half the LED forward voltage in the direction that gave the two-LED polarity the right amount of voltage headroom.

The 2134 (OPA2134??) can swing closer to the power supply than the venerable TL07x, depending on the current it puts out. How much current is needed is not clear, as the LED currents go up a lot and may need so much current that the opamp itself starts limiting.

Probably the best thing for you to try is to run it at 18V, bias it in the middle of the power supply  and see if you like the sound better. If so, you're probably banging into the limits of your power supply voltage - which you don't state, but I'm guessing might be 9V. If that's true, either run it with a bigger power supply voltage or choose lower forward drop diodes.

Have you tried it with asymmetrical silicon diodes? Do you know you like LEDs better in this circuit? That could be a much simpler answer than a bigger power supply.

And in the tube screamer and its ilk, remember that the output signal will always be as big as the incoming signal, so there is always an incoming signal size that will push the opamp to the limits of its power supply.

[Mark Hammer]

A lot of folks forget that, when one considers the typical input signal level from plain ordinary pickups, your average op-amp and average supply voltage can't really support the sorts of gains we normally associate with overdrives.  Not cleanly.  Take a 100mv signal and there's a limit to how much clean gain can be applied.  Apply even more gain and what you hear is not just the clipping from the diodes, but the op-amp running out of headroom.  If one wants to hear ONLY the impact of the diodes, you either have to ue diodes whose forward voltage is well below the maximum voltage swing of the op-amp, or crank up the supply voltage.

[Asymmetric]

Thanks for the replies!
I will try running it at 18v and I'll let you know how it goes - not only should it kick the led's in more, it'll also give more headroom to the op amp.

R.G, I don't understand what you mean by "Probably the best thing for you to try is to run it at 18V, bias it in the middle of the power supply"

- What is biasing in the middle of the power supply?

[thermionix]

- What is biasing in the middle of the power supply?

I could be wrong, but I think he means 9v bias (given an 18v supply).

[antonis]

- What is biasing in the middle of the power supply?

What thermionix said..

If you want to have a symmetrical output from the Amp you have to place signal's X axe (horizontal) as near as possible to the middle of power supply margin - so you have "equal" signal variations (positive & negative)(*)

If you have a symmetrical bipolar PS (i.e. +9V & -9V) then your bias should easily been taken from the middle (which is GND in this case).
If your PS is "single" (i.e. +9V & GND) then you have to bias Amp's input to somewhere close to 4.5V..
(the above also stands on "peculliar" supplying, like +15 & -5 or whatever else..)

Some guys like to call this bias point as "virtual ground" (perhaps because this point sets the zero point for signal swinging - although the term was originaly defined as a point of a circuit that is maintained at a steady reference potential, without being connected directly to the reference potential...)

(*) Considering that Op-Amp is able to equally hit the power supply rails or, at least, has equal voltage swing between Vcc & Vee..
(if your Op-Amp swings to, say, 2V lower than Vcc and 1V higher than Vee, then for a +9 & GND you ideally must set the bias point at 4V..)

[Asymmetric]

- What is biasing in the middle of the power supply?

What thermionix said..

If you want to have a symmetrical output from the Amp you have to place signal's X axe (horizontal) as near as possible to the middle of power supply margin - so you have "equal" signal variations (positive & negative)(*)

If you have a symmetrical bipolar PS (i.e. +9V & -9V) then your bias should easily been taken from the middle (which is GND in this case).
If your PS is "single" (i.e. +9V & GND) then you have to bias Amp's input to somewhere close to 4.5V..
(the above also stands on "peculliar" supplying, like +15 & -5 or whatever else..)

Some guys like to call this bias point as "virtual ground" (perhaps because this point sets the zero point for signal swinging - although the term was originaly defined as a point of a circuit that is maintained at a steady reference potential, without being connected directly to the reference potential...)

(*) Considering that Op-Amp is able to equally hit the power supply rails or, at least, has equal voltage swing between Vcc & Vee..
(if your Op-Amp swings to, say, 2V lower than Vcc and 1V higher than Vee, then for a +9 & GND you ideally must set the bias point at 4V..)

This is one thing I'm struggling to grasp. Are there any examples? How would I measure all this?

[antonis]

This is one thing I'm struggling to grasp. Are there any examples? How would I measure all this?

Now you make me struggling to grasp what you mean... 

If you trace the non-inverting (+) IN of your TS-808 circuit will find that is connected, via a 10k resistor, to a node of a resistive voltage divider, formed by two 10k resistor in series betweewn Vcc & GND.

The middle point of this divider (junction between the two resistors) is sitting on 4.5V (assuming a +9V PS) and is "pure transfered" to pin 3 (+ Input) - assuming also that there is no significant voltage drop on the connecting resistor because of negligible current through it (Input impedance tooooooo high..) 
(the above is also true for the Inverting Input (-) via direct ac coupling to the same point - there isn't any pure DC on pin 2, because of the ΗPF capacitor, but it's always there "waiting" to be added on or subtracted of the incoming signal's voltage)

Now, your Amp has a voltage point to "refer on" (something like an "offset" on pin 3) and compare it with the incoming signal at pin 2 (inverting IN) trying (via it's OUT & NFB loop) to make zero the voltage difference between + & - Inputs.

Considering your signal (almost) sinusoidal with an amplitude of, say, 2Vpp you have a voltage alternation between +1V and -1V.
If your Amp is supplied with +9V & GND, it should be able to produce on it's OUT an undistorted signal between 9V and 0V but in such a case you "loose" the half signal waveform because your Amp's OUT can't amplitude nothing "lower" than 0V.
(remeber that your signal is negative during it's half life time..)

So you have to raise the reference point somewhere in the middle of Amp's OUT swinging capability..
This point here is the notorious +4.5V .!!!

P.S.
Neither my English nor my "teaching" skills are suitable for making your life easier - maybe it should be better to google something like "Op-Amp bias" or wait for a more appropriate person to answer..

[Asymmetric]

This is one thing I'm struggling to grasp. Are there any examples? How would I measure all this?

Now you make me struggling to grasp what you mean... 

If you trace the non-inverting (+) IN of your TS-808 circuit will find that is connected, via a 10k resistor, to a node of a resistive voltage divider, formed by two 10k resistor in series betweewn Vcc & GND.

The middle point of this divider (junction between the two resistors) is sitting on 4.5V (assuming a +9V PS) and is "pure transfered" to pin 3 (+ Input) - assuming also that there is no significant voltage drop on the connecting resistor because of negligible current through it (Input impedance tooooooo high..) 
(the above is also true for the Inverting Input (-) via direct ac coupling to the same point - there isn't any pure DC on pin 2, because of the ΗPF capacitor, but it's always there "waiting" to be added on or subtracted of the incoming signal's voltage)

Now, your Amp has a voltage point to "refer on" (something like an "offset" on pin 3) and compare it with the incoming signal at pin 2 (inverting IN) trying (via it's OUT & NFB loop) to make zero the voltage difference between + & - Inputs.

Considering your signal (almost) sinusoidal with an amplitude of, say, 2Vpp you have a voltage alternation between +1V and -1V.
If your Amp is supplied with +9V & GND, it should be able to produce on it's OUT an undistorted signal between 9V and 0V but in such a case you "loose" the half signal waveform because your Amp's OUT can't amplitude nothing "lower" than 0V.
(remeber that your signal is negative during it's half life time..)

So you have to raise the reference point somewhere in the middle of Amp's OUT swinging capability..
This point here is the notorious +4.5V .!!!

P.S.
Neither my English nor my "teaching" skills are suitable for making your life easier - maybe it should be better to google something like "Op-Amp bias" or wait for a more appropriate person to answer..

I sort of get it but I still don't quite get in. I'm new to how opamps work! Are there any simple examples maybe on a picture or something? I'm confused!

[karbomusic]

I sort of get it but I still don't quite get in. I'm new to how opamps work! Are there any simple examples maybe on a picture or something? I'm confused!

If you look at an AC waveform it is swinging above and below '0 Volts'. All you are doing is moving the opamp's view of where 'zero' is to the middle of the power supply's range so that the AC wiggles up and down properly. We need to do this ourselves with a single ended supply such as a straight +9 or +18 DC power supply because there is no below zero so to speak. We just move the opamp's reference of zero to the middle - as far as the opamp's input is concerned.

If you were using the opamp to amplify DC, none of this would matter because it doesn't swing like that, it's just a positive voltage. If running the opamp as dual powered with a dual power supply such as +9/-9, we already have a zero in the middle scenario and don't need to create one.

[Asymmetric]

I sort of get it but I still don't quite get in. I'm new to how opamps work! Are there any simple examples maybe on a picture or something? I'm confused!

If you look at an AC waveform it is swinging above and below '0 Volts'. All you are doing is moving the opamp's view of where 'zero' is to the middle of the power supply's range so that the AC wiggles up and down properly. We need to do this ourselves with a single ended supply such as a straight +9 or +18 DC power supply because there is no below zero so to speak. We just move the opamp's reference of zero to the middle - as far as the opamp's input is concerned.

If you were using the opamp to amplify DC, none of this would matter because it doesn't swing like that, it's just a positive voltage. If running the opamp as dual powered with a dual power supply such as +9/-9, we already have a zero in the middle scenario and don't need to create one.

I am getting. I understand the theory I think. Lets say I was using a TS808 like this http://fuzzcentral.ssguitar.com/schematics/TS808.gif

Which parts would I change?

[thermionix]

There's a reprint out there of a Mullard tube book from 1959...

https://www.amplifiedparts.com/products/B-655

(Maybe it's available online, I don't know.)

In there is a chapter called "Sources of Distortion in Recorded Sound" or something close to that.  It was a real eye-opener for me.  Among other things it talks about bias for audio tape, which is used for similar reasons as bias on an opamp.  Basically so the signal can swing down as well as up.  Wouldn't sound too good if only half of it got through.

Lot of good vinyl stuff in that chapter too.

[antonis]

I am getting. I understand the theory I think. Lets say I was using a TS808 like this http://fuzzcentral.ssguitar.com/schematics/TS808.gif
Which parts would I change?

I'm afraid of you don't completely understand the theory if you consider that you have to change "parts" other than clipping diodes..

You can change many parts if you wish but you must have in mind the purpose of change (i.e. more/less gain, assymetrical diode clipping, assymetrical power rails clipping, IN/OUT impedance, Tone Hi/Low threasholds .....)

[Asymmetric]

I am getting. I understand the theory I think. Lets say I was using a TS808 like this http://fuzzcentral.ssguitar.com/schematics/TS808.gif
Which parts would I change?

I'm afraid of you don't completely understand the theory if you consider that you have to change "parts" other than clipping diodes..

You can change many parts if you wish but you must have in mind the purpose of change (i.e. more/less gain, assymetrical diode clipping, assymetrical power rails clipping, IN/OUT impedance, Tone Hi/Low threasholds .....)

Yes I understand that - I was aiming to up the power so the LED diodes can clip.

[merlinb]

Yes I understand that - I was aiming to up the power so the LED diodes can clip.

What colour LEDs are you using?

[GibsonGM]

If you have voltage above the LED threshold voltage, they will turn on as-is!    That's why Merlin asks what color, as the have varying threshold voltages.

Ex:  If you have a 6V peak to peak signal running thru the feedback loop...signal goes up to +3V, swings down to -3V.....you will turn on any diodes that have a threshold below 3V, and they will clip.    As they are placed in opposition to each other, one clips on the positive part of the cycle, and one on the negative.    If you radically increase the power supply, you will assure yourself of more clean headroom, but the diodes are going to clip at the same voltage all the same, wasting the headroom, unless you want to really jack it up in another stage....

You can try all this without changing anything but the diodes themselves...I love LED clipping, and use it un-modded all the time.