overdrive / distortion that reacts to a Fuzz, TS-7/DS-1 ??

[Steben]

Hi there!

I'm planning te start all over with pedals!
First project would be to get small volume without a speaker load, while still being enable to get mushy responsive fuzz (meaning: with a little drive after the fuzz).
I have a TS - 7 tube screamer and a DS-1 distortion.
Can I do something with them?

Some suggestions:

-  In a normal TS the output is only limited bij de supply voltage of the opamp, because of the unity gain circuit. With an effect in front, the signal can be a multitude of the 0.7 treshold of the feedback diodes, resulting in a complete different sound than an amp. I suggest putting a pair of extra clipping diodes in front of the opamp. This prevents a way too big signal by clamping the input below a treshold. This way, the pedal gets complexer clipping above a 0.7V input (if silicon).
- swapping the DS-1 clipping diodes with diode MOSFETS.
- changing the gain transistor in the DS-1 to lower, more lineair gain.

[jasperoosthoek]

I don't get what you are saying about the TS7, actually I'm a bit lost.

[Steben]

I don't get what you are saying about the TS7, actually I'm a bit lost.

I know it sounds awkward  ...

In an opamp feedback clipper, there is unity gain once the diodes conduct fully. If the input signal is way above the forward voltage of the diodes, there is a strange sum of signals in there, with a hint of crossover distortion. Think of germanium in there instead of LED's for example. The germanium will sounds mushy with less headroom feeling.
A guitar output is never much bigger than the forward voltage.
If you want the TS to function as an gentle amp distortion with a large signal in front (dist, fuzz, boost...), it doesn't work that well. Not at all as an amp.
Because of this, I suggest a clipping pair of diodes in front of the opamp stage. This way, the input signal gets limited. Of course, first of all, there will be soft clipping in the opamp stage, but with a harder limiting with bigger signals.

[jasperoosthoek]

In an opamp feedback clipper, there is unity gain once the diodes conduct fully.

That's not true. The diodes have an IV curve which is an exponential relation between current and voltage. The input to the clipper is a resistor which applies a current to the diodes. In a stable feedback loop the inverting input is a virtual ground, i.e. always the same voltage as the non-inverting input. So the resistor R applies a current to the diodes V_in/R. The output from the diodes is reverse exponential or logarithmic.

So the gain depends on the resistor feeding the clipping stage. Make it smaller and the gain goes up. The relation between input voltage and forward voltage is completely irrelevant  . It is the input current, which depends on the input voltage and the resistor feeding the clipping stage, that matters.

[Steben]

In an opamp feedback clipper, there is unity gain once the diodes conduct fully.

That's not true. The diodes have an IV curve which is an exponential relation between current and voltage. The input to the clipper is a resistor which applies a current to the diodes. In a stable feedback loop the inverting input is a virtual ground, i.e. always the same voltage as the non-inverting input. So the resistor R applies a current to the diodes V_in/R. The output from the diodes is reverse exponential or logarithmic.

So the gain depends on the resistor feeding the clipping stage. Make it smaller and the gain goes up. The relation between input voltage and forward voltage is completely irrelevant  . It is the input current, which depends on the input voltage and the resistor feeding the clipping stage, that matters.

The resulting impedance at a given voltage is irrelevant to the impedance of the resistors in the non-inverting stage.
A = R1 + R2 / R2
R1 (with diodes conducting, around 100 - 1 ohms) <<<<<<<<<<<<< R2 (mostly range of 1k to 5k ohms)
With gain at infinity (or open loop) the result would be the sum of a square wave added to the input signal. Soft clipping.

In "clipping to ground" circuits (DS-1, rat, ....) a = R2 / R1 + R2. At max input signal R2 drops to several ohms or even less and then a = 0. Square wave. Hard clipping.

[jasperoosthoek]

The resulting impedance at a given voltage is irrelevant to the impedance of the resistors in the non-inverting stage.

Which resulting impedance of what? Which voltage, input, output, supply voltage? What do you mean by the impedance of the resistors? Don't you mean diodes and resistors?

A = R1 + R2 / R2
R1 (with diodes conducting, around 100 - 1 ohms) <<<<<<<<<<<<< R2 (mostly range of 1k to 5k ohms)

Which ones are R1 and R2? If it's a diode then please call it R_diode or something similar. What circuit do you base your analysis on? So R1 is a forward biased diode and R2 is the 4.7k resistor? R6 here: http://www.freeinfosociety.com/electronics/schematics/audio/pictures/tubescreamerts808.gif

With gain at infinity (or open loop) the result would be the sum of a square wave added to the input signal. Soft clipping.

Of which circuit? In case of a Tube Screamer there will indeed be a clean signal added to a clipped signal. That won't be soft clipping but just hard clipping mixed with a clean signal. You don't get rid of the harsh harmonics by mixing it with a clean signal.

In "clipping to ground" circuits (DS-1, rat, ....) a = R2 / R1 + R2. At max input signal R2 drops to several ohms or even less and then a = 0. Square wave. Hard clipping.

Which ones are R1 and R2? This time R2 is a diode?

Say the diode and feeding resistor (http://www.muzique.com/schem/pcrat1.gif) get a 4V amplitude signal. Just before the opamp clips. The voltage on the 1k resistor will be 3.3V as the diode takes about 0.7. 3.3mA runs through the resistor and diode meaning a diode resistance of over 200 ohms. That's not less than several ohms. At these resistances the diodes fry.

My friend, I am even more lost now. If you want feedback on your ideas then please make it more easy for the reader to understand what you mean and where you base your analysis on. Add schematics, make diagrams and don't expect me to do your homework for you.

[Steben]

The resulting impedance at a given voltage is irrelevant to the impedance of the resistors in the non-inverting stage.

Which resulting impedance of what? Which voltage, input, output, supply voltage? What do you mean by the impedance of the resistors? Don't you mean diodes and resistors?

A = R1 + R2 / R2
R1 (with diodes conducting, around 100 - 1 ohms) <<<<<<<<<<<<< R2 (mostly range of 1k to 5k ohms)

Which ones are R1 and R2? If it's a diode then please call it R_diode or something similar. What circuit do you base your analysis on? So R1 is a forward biased diode and R2 is the 4.7k resistor? R6 here: http://www.freeinfosociety.com/electronics/schematics/audio/pictures/tubescreamerts808.gif

With gain at infinity (or open loop) the result would be the sum of a square wave added to the input signal. Soft clipping.

Of which circuit? In case of a Tube Screamer there will indeed be a clean signal added to a clipped signal. That won't be soft clipping but just hard clipping mixed with a clean signal. You don't get rid of the harsh harmonics by mixing it with a clean signal.

In "clipping to ground" circuits (DS-1, rat, ....) a = R2 / R1 + R2. At max input signal R2 drops to several ohms or even less and then a = 0. Square wave. Hard clipping.

Which ones are R1 and R2? This time R2 is a diode?

Say the diode and feeding resistor (http://www.muzique.com/schem/pcrat1.gif) get a 4V amplitude signal. Just before the opamp clips. The voltage on the 1k resistor will be 3.3V as the diode takes about 0.7. 3.3mA runs through the resistor and diode meaning a diode resistance of over 200 ohms. That's not less than several ohms. At these resistances the diodes fry.

My friend, I am even more lost now. If you want feedback on your ideas then please make it more easy for the reader to understand what you mean and where you base your analysis on. Add schematics, make diagrams and don't expect me to do your homework for you.

Zeg het eens in 't nederlands dan 
On the forum, in english.

Indeed, I ment R_diode. In a non-inverting opamp stage the gain will be (R_diode + R to Vref) / R to Vref resulting in unity gain at high levels.
It's common use to call tubescreamer soft clipping and the clipper to ground hard clipping (as well as inverting stages).

[Steben]

Say the diode and feeding resistor (http://www.muzique.com/schem/pcrat1.gif) get a 4V amplitude signal. Just before the opamp clips. The voltage on the 1k resistor will be 3.3V as the diode takes about 0.7. 3.3mA runs through the resistor and diode meaning a diode resistance of over 200 ohms. That's not less than several ohms. At these resistances the diodes fry.

Now I'm a bit lost as well.

That would mean the type of diode is futile, which is not the case.

And if you have a 2k resistor? That would make in your calculation 1.65mA and a diode impedance of 400 ohms.

[jasperoosthoek]

Now I'm a bit lost as well.

That would mean the type of diode is futile, which is not the case.

That does not mean the type of diode is futile. For most diodes, small signal and large signal, the forward voltage at a reasonable current will be around 0.5 to 0.8. In that range the current can change several orders of magnitude. That means that, as the voltage doesn't change that much the current does and so does the resistance.

So loading a small signal diode or a (1A) power diode with a 1k resistance to 4.0 volts total will always yield a resistance of about 200 ohms because both devices are saturated. The small signal diode is much more saturated though. The only relevant part is the onset of conduction of both devices. Not at 4V input but around the forward voltage of the diodes (in a Rat type circuit). The small signal device will clip much earlier than the other device and provide more gain. But the 'brick wall' clipping edge will be roughly at the same voltage.

And if you have a 2k resistor? That would make in your calculation 1.65mA and a diode impedance of 400 ohms.

Yes, because that is what happens. Dividing the current by half only changes the forward voltage a small bit meaning that the resistance almost doubles.

[cpm]

Hi there!

...

lets start simple again...
you want more headroom: raise your clipping threshold (series diodes, leds, etc)
or... scale down your input

anyway, you have to make assumptions about what will be the expected input level for the circuit to work as intended.

[Steben]

Hi there!

...

lets start simple again...
you want more headroom: raise your clipping threshold (series diodes, leds, etc)
or... scale down your input

anyway, you have to make assumptions about what will be the expected input level for the circuit to work as intended.

To use very simple words: I want any of my pedals to sound a bit more as a gentle driven amp that works fine with a fuzz in front. Emphasis on: making the fuzz bloom.
Headroom issues are not that big a problem. It's all about gentle clipping.


First idea would be swapping the diodes in the DS-1 with mosfet/ge pairs. OCD-like, shaka braddah - like...
Combined with reducing distortion in the transistor stage by moving bias more towards linear region.

Another option is a clean boost between my amp and the fuzz pedal to drive the tube input a bit.

[jasperoosthoek]

To use very simple words: I want any of my pedals to sound a bit more as a gentle driven amp that works fine with a fuzz in front. Emphasis on: making the fuzz bloom.
Headroom issues are not that big a problem. It's all about gentle clipping.

The Marshall Bluesbreaker pedal has a resistor in series with the back to back diodes to soften up clipping:

Searching for the schematic I found this site: http://www.gmarts.org/index.php?go=217. It has a lot of info on different types of clipping between units. The Marshall Bluesbreaker was the first pedal I built. Made on perfboard directly from the schematic. The layout is very chaotic but it worked . Now it is in my junk box. I might start to use it again. I'm looking for a drive pedal that is less generic than the Tubescreamer.

[Steben]

To use very simple words: I want any of my pedals to sound a bit more as a gentle driven amp that works fine with a fuzz in front. Emphasis on: making the fuzz bloom.
Headroom issues are not that big a problem. It's all about gentle clipping.

The Marshall Bluesbreaker pedal has a resistor in series with the back to back diodes to soften up clipping:

Searching for the schematic I found this site: http://www.gmarts.org/index.php?go=217. It has a lot of info on different types of clipping between units. The Marshall Bluesbreaker was the first pedal I built. Made on perfboard directly from the schematic. The layout is very chaotic but it worked . Now it is in my junk box. I might start to use it again. I'm looking for a drive pedal that is less generic than the Tubescreamer.

the "GM arts overdrive circuit" below does exactly what i suggested   : putting in the front two clipping diodes. Not only do they prevent high input into the opamp, they also limit the max output of the soft clipping section.


EDIT: finally I understand comparison between response of a valve / transistor and a diode clipper...
The current on the left of a forward voltage I/V diode curve is the signal and the curve gives the output voltage. I always did it the other way around....
A resistor in series with a diode just adds a linear curve to the diode curve. With a silicon diode you get a less exponential curve. Just like in a tube (triode).
With fine tuning (pot) in series with a diode, you can in theory match a (1.5) function.
This means that double germanium with very small resistor should sound similar to a silicon diode with bigger resistor.