"Compression Quotient" of AIAB / Distortion pedal

[iainpunk]

In small doses, like the Cornish CC1,  the clean and distorted components become one and it sounds quite natural.

another example is of course the "Pedal That Shall not be Named (or have its schematic shared on this forum)"

[Steben]

Tube screamer style pedals are almost the same. Above the diode treshold, gain is 1 (which means the original signal). When maxed out, you get the same as a square wave mixed with a clean signal.
As I said, compression ratio will give dry numerical information useful for a rig or chain of gain stages, but will tell you nothing about "dynamic response" because that is subjective.

[Vivek]

Tube screamer style pedals are almost the same. Above the diode treshold, gain is 1 (which means the original signal)

We need to properly define :

Gain
Clipping radians


Please consider two scenarios :

A) We multiple the input by 10. When output should have been more than 0.6V, we replace with 0.6V

B) We multiply the input by 200. When output should have been more than 0.6V, we replace with 0.6V


Can we say "Above the diode threshold, gain is 1 (which means the original signal)"

[Steben]

Tube screamer style pedals are almost the same. Above the diode treshold, gain is 1 (which means the original signal)

We need to properly define :

Gain
Clipping radians


Please consider two scenarios :

A) We multiple the input by 10. When output should have been more than 0.6V, we replace with 0.6V

B) We multiply the input by 200. When output should have been more than 0.6V, we replace with 0.6V


Can we say "Above the diode threshold, gain is 1 (which means the original signal)"

Yes we do in non inverting clippers, gain is 1 of the signal voltage at that point. that's what I ment with at extreme gain, we get square wave of 0.6V with original signal on top.

Vout = Vin * (1 + R feedback /R gnd)

[Vivek]

R feedback is not 0 when a diode clips

It  acts like a variable resistor, adjusting itself so that the output remains close to one diode drop, irrespective of input signal level


I understand you are saying

Vout = Vin * (1 + R feedback /R gnd)

is same as

Vout = Vin * 1 + Vin(R feedback /R gnd)

[Steben]

R feedback is not 0 when a diode clips

It  acts like a variable resistor, adjusting itself so that the output remains close to one diode drop, irrespective of input signal level


I understand you are saying

Vout = Vin * (1 + R feedback /R gnd)

is same as

Vout = Vin * 1 + Vin(R feedback /R gnd)

Yes that just math

The parts of Rf/Rgnd is variable as the diode starts conducting.
The onset region exists yes, it is not a "perfect diode", but it sets a voltage offset to unity gain.

It becomes

Vout = Vin * 1 + Vin(f(R) feedback /R gnd) + f(D Vf)

With f(D Vf) going to Vf and f(R) going to 0.



In an inverting stage this becomes

Vout = Vin(f(R) feedback /R gnd) + f(D Vf)

With f(D Vf) going to Vf and f(R) going to 0.
0 is never really reached. It is a limit

With a germanium pair in a Tube screamer, R of the diode is higher, which means f(R) is higher than silicon, while the f(DVf) is always lower. This means the result is muddier.
Tube screamer style circuits benefit from low diode R (silicon, LED) usually with higher Vf to achieve "crisp" tones. That's why a LED pair in such circuits are the "Marshall" setting since it creates sharper, less soft clipping.

On the other hand, very high input signals ruin the TS style tone, since they can clip the opamp if high enough and the ratio of signal above the square wave is high.

[Vivek]

Tube screamer style pedals are almost the same. Above the diode treshold, gain is 1 (which means the original signal)

I think it might be better to say :

Tube screamer style pedals are almost the same. Above the diode threshold, we get the original signal + a clipped signal.

If gain = output / input, We cannot say "Above the diode threshold, gain is 1"

[Steben]

Tube screamer style pedals are almost the same. Above the diode treshold, gain is 1 (which means the original signal)

I think it might be better to say :

Tube screamer style pedals are almost the same. Above the diode threshold, we get the original signal + a clipped signal.

If gain = output / input, We cannot say "Above the diode threshold, gain is 1"

Not truely, no, but it describes what is going on: there is a voltage offset with unity gain signal on top.
Because in reality we neither have a true mix of a clipped signal and a clean one that is what happens in other pedals.

[iainpunk]

Please consider two scenarios :

A) We multiple the input by 10. When output should have been more than 0.6V, we replace with 0.6V

B) We multiply the input by 200. When output should have been more than 0.6V, we replace with 0.6V

what you describe here is hard clipping. this happens in an inverting op amp gain stage (and also in an inverting transistor gain stage, like in the big muff, the signal is also hard clipping [don't believe Jason from Fuzzlord effects, he claims otherwise] [except when there are resistors in series with the diodes to up the gain after the threshold])

when the opamp is non-inverting its soft clipping. (you can also softclip to ground by using a resistor divider to the diodes)
a soft clipping scenario is when:

> We multiply the signal with 30. When output should have been more than {0.6}, we replace with {0.6}+{the original unamplified input signal}.

cheers, Iain

[PRR]

I am basically a Chemical Engineer .... .... .... .... .... ....

Oh, OK. I thought this place was for musicians, and musical engineering. I guess I'm confused.

If the "customer" is a peak meter, then compression and distortion may be measured the same.

If the customer is a Music Listener, they are not the same. Compression makes loud parts not-so-loud. Distortion makes loud parts "louder" (to the ear).

[Rob Strand]

We need to properly define :

Gain
Clipping radians

The gain from the perspective of the perceived level is Vout/Vin so that's roughly Vdiode / Vin.

It's not good measure of the "gaininess"  of the pedal.     You can can also have a gainly pedal with a a very low output but cranking the drive and backing off the level.

Gaininess is to do with what you are calling "Clipping radians".   From a practical point of view we all know more gain means making the feedback resistor larger.     For the circuit we can translate that to a clip threshold.

As a first pass approximation, a silicon diode start to conduct at say 500mV.   If there is 500mV across the 220k than means there is 2.3uA passing through the 220k.   If the diode current is zero the input voltage is at least 2.3uA * 10k = 23mV.   In simple terms all we are doing here is taking the output clip point and dividing by small signal gain.    gain = 220k/10k = 22,  clip point  relative to the input = 500mV/22 = 23mV peak.     So if we put in a 100mV peak signal the start of clipping is  at about 13 deg for a sinewave.   The angle that it will be clipping over a half cycle is (90deg - 2*13) = 64 deg.  That means it's clipping 64/90 = 71% of the time.

If we crank the gain by increasing the 220k to 1MEG the small signal gain increases to 1M/10k = 100.  The input relative clip point decreases to 0.5/100 = 5mV.    The start clipping angle is now 2.87 deg,  which is (90-2*2.87)/90 = 94% of the time.   The start of clipping angle is a much more sensitive measure for the degree of clipping or gaininess, however more gaininess means a lower start of clipping angle.

For the case where the feedback resistor is 220k we assumed 500mV as the diode clipping point.   For a silicon diode with a drop of 600mV at 1mA, the diode current at 500mV is about 120uA  which is already much higher than the 2.3uA passing through the 220k.   In fact if the input level only reaches 1V peak the diode current won't get over 1/10k = 100uA and the diode voltage won't go over 490mV.   The point where the diode current and the current through the 220k are about equal occurs at around 290mV and 1.3uA!.   What that means is the onset of clipping has a soft knee, somewhere between 290mV and 500mV.   It's unlikely we will hear the distorting effect of the diode at 290mV but it will act a bit like a resistor and reduce the effective gain.

[Rob Strand]

Quote from: Vivek on Today at 04:21:28 AM

    I am basically a Chemical Engineer .... .... .... .... .... ....

Oh, OK. I thought this place was for musicians, and musical engineering. I guess I'm confused.

It's not fair to pigeon-hole people,  there's people from all walks of life here.