Importance of IV knee graph of a Diode in soft clipping applications

Started by Vivek, November 10, 2020, 12:17:50 PM

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Rob Strand

Quote1 diode X 2 (scale up one diode graph 2 times,to compare with 2 diode graph)

1 diode X 3 (scale up one diode graph 3 times,to compare with 3 diode graph)
I 100% get why you want to see those.   However I feel silly doing it because the for 3 diode case I just scale up the one diode case.     If I scale it down by 1/3 I have to get what I started with.

When the diodes are in series they have the same current, so if one diode is Vd, I put three in series I get 3*Vd, but then when you scale the output of the 3*Vd case by 1/3 you *have to* get Vd again.

So you can't change the underlying shape of the diode characteristic by putting diodes of the same type in series.   You can get "in between" characteristics by combining two diodes of different types.

If you have resistor in parallel with diodes the overall non-linearity curves change a bit from the raw diode VI characteristic.  In order to keep the non-linearity the same  you would need to up the resistance by a factor of 3 for 3 diodes.   That's not only to up the gain, it's to preserve shape.  If for example you increased the gain by adjusting the non-feedback resistor then the non-linearity is subtletly different.   You can see this even for the one diode case.   For feedback diodes the feedback R is say 100k to 1M whereas  for  output clippers  the R is 1k to 10k.     Even if we ignore the fact the opamp can also clip, the change in R affects the non-linearity.      IIRC smaller R's make the effective linearity less soft.

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Vivek

Quote from: Rob Strand on November 14, 2020, 03:28:29 AM


When the diodes are in series they have the same current, so if one diode is Vd, I put three in series I get 3*Vd, but then when you scale the output of the 3*Vd case by 1/3 you *have to* get Vd again.

So you can't change the underlying shape of the diode characteristic by putting diodes of the same type in series.   You can get "in between" characteristics by combining two diodes of different types.


It means that a few of the earlier posts and graphs on this thread might need to be edited.

Rob Strand

Since you get the scaling idea, you might be interested in this.     It's a whole heap of diode measurements scaled to the 300uA diode voltage so you can see the underlying shape of the diode curves after removing the diode voltage scaling,

https://www.diystompboxes.com/smfforum/index.php?topic=121964.msg1148986#msg1148986

Just to be clear the cases that start with a "d" are when the device is operated as a forward biased diode.  For example d2N7000 is when the MOSFET is acting as a body diode whereas m2N7000 is when the gate is shorted to the drain and the MOSFET is operating as a MOSFET.   The same with the zeners  dBZX55...  means forward biased.  zBZX55... means it is operating as a zener.
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Vivek

Quote from: Rob Strand on February 21, 2019, 04:48:28 PM
FWIW, I did a stack of measurements in 2001 and 2009.


If you take a single diode and two of the same diode in series and perform the normalization you will conclude the softness is the same.



Wow Rob

I can't thank you enough for your hard work

And kindness in explaining things.

Now the diode graph matches intuition

I still need to understand that bit on how this affects mix of unclipped + clipper signals in some configurations

Steben

This all applies to the total of diodes, yes. But a noninverting feedback opamp stage works as it does, regardless of the diodes. The diodes have some influence, but beyond a certain point the resistance(s) of the diodes is / are so small the result becomes [dVo] = Vd + [dVi].

If, lets say an ideal diode behaves as a 50 ohms resistor beyond 0.6V, three diodes will behave as a 150 ohms resistor. (These are very (too) large values and the resistance in fact diminishes even further...) This is parallel to the feedback resistor. In a tube screamer at half gain you have around 250 k ohms feedback resistor.
Gain above 1kHz is (250k + 4k7) / 4k7 aka (250/4k7) + 1.

Beyond 0.6V this becomes (50 / 4k7) + 1 =  1.01
With three diodes this becomes beyond 1.8V (150 / 4k7) + 1 = 1.03, not 3 !

In reality, the knee of the diodes makes for a transition zone. But the result is the same: a relatively higher portion of distorted signal content vs clean content.
That is why germanium in a tube screamer sounds badly compressed yet less distorted: the circuit is the dominant factor, not the knee nor the resistance.

The series resistor of output clippers to ground does something, but there you have a direct voltage dividing. Vo = (Vi * Rd) / (R + Rd)



Just hear how everything with LEDs sounds MORE tightly distorted and germaniums sound dull in a TS808. Although the controls are the same!
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PRR

Quote from: Rob Strand on November 14, 2020, 03:28:29 AM...you would need to up the resistance by a factor of 3 for 3 diodes.   That's not only to up the gain, it's to preserve shape.  If for example you increased the gain by adjusting the non-feedback resistor then the non-linearity is subtlety different.   You can see this even for the one diode case. ...  IIRC smaller R's make the effective linearity less soft.

That's where I was going. Not just the diode, or number of diodes, but the circuit gain and all series and shunt resistors.

IMHO it is possible to swap nearly any diode if ALL other circuit components are adjusted, basically for the same current density in the crystal.

The "knee" is a mirage. Plot diodes on log paper and the knee is straight.



> a 1N4148  signal diode tends to be around 1.8

Thanks for the reminder. I was concerned my plot did not cross 26r@1mA per Shockley.
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Steben

With clipper to ground, smaller output Rs make for more current. This is why I asked before: is the diode current driven or voltage driven? If current driven this matches the R theory. With the R shaping the knee image .
With lots of current, the relative interval between 0 and max current position in voltage is larger, but with a steeper portion. The lower the R, the higher the current and higher the output voltage, "driving harder". The higher the R, the lower the current interval with less voltage and less steep knee. The output R behaves a bit like feedback in a tube power amp but reversed: higher R for softer action.
So the resistors (be it in signal series or in series with diodes) shape it all. A non feedback power amp style needs higher output R with perhaps some series resistance at the diodes, a tight clipping ("marshall") nees lower R and no resistors in series.

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Steben

Quote from: Rob Strand on November 14, 2020, 03:55:38 AM

https://www.diystompboxes.com/smfforum/index.php?topic=121964.msg1148986#msg1148986


In which you state the portion of clean waters down with higher amount of diodes / treshold in a non-inverting soft clipper.  ;D In other words: the response slope beyond treshold does not add up with the diodes.

Time to test a TS/SD style overdrive on 18V supply with zeners....
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Rob Strand

QuoteThe "knee" is a mirage. Plot diodes on log paper and the knee is straight.
In some ways an illusion and in some ways not.

If you plot the diode characteristic from 0 to Vmax vs 0 to Imax  then there is a knee as far as the eye is concerned.   It's the point on the VI characteristic which is closest by eye to the top left corner point V = Vmax and I = 0.  The plot is a box within some range of values on both axes.   All plots are linear vs linear.

From the eye's perspective you really need to have equal distance plots for vertically for 0 to Vmax and horizontally for 0 to Imax. If not we could stretch stretch the current axis to be the width of the screen and the vertical voltage axis to be 10 pixels, which skews the close distance by eye.   (Computing the "distances" mathematically we would divide Vd by Vmax and Id by Imax so the graphs are always normalized to 0 to 1 on each axis then find the point closest to (0,1).  That takes care of all scaling factors.)

More to the point suppose we plot the *same*diode characteristic  over a different ranges say from 0 to Vmax'  vs 0 to Imax' where Imax' is say 100th Imax in the first case.   Now Vmax' is somewhat lower than Vmax, perhaps lower by approx 230mV.   If we do this the knee seen by the eye is now lower.

In the first case we might have:       0 to 600mV and 0 to 1mA and a knee at say 500mV.
In the second case we might have 0 to 370mV and 0 to 10uA and a knee at say 260mV.

So the knee from eye's perspective is not a so much a function of the diode by more the range of current we plot the VI curve over.

If we used a diode as a current controlled resistor  we might be interested in keeping the distortion below some nominal figure and we would end-up with a completely different effective "knee", perhaps something like 10mV as the "knee" or threshold for the distortion limit.   A very different number.

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Steben

Quote from: Rob Strand on November 15, 2020, 05:29:09 AM
QuoteThe "knee" is a mirage. Plot diodes on log paper and the knee is straight.
So the knee from eye's perspective is not a so much a function of the diode by more the range of current we plot the VI curve over.

exactly
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Rob Strand

QuoteIn which you state the portion of clean waters down with higher amoutn of diodes / treshold in a non-inverting soft clipper.  ;D
I don't quite follow you.  I suspect what is being said in my post is in a non-inverting case it always mixes the clean signal with the clipped signal.  If you have more diodes  the magnitude of the diode contribution becomes higher, in relative terms, so in effect it would sound more distorted.
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According to the water analogy of electricity, transistor leakage is caused by holes.

Steben

Quote from: Rob Strand on November 15, 2020, 05:32:22 AM
QuoteIn which you state the portion of clean waters down with higher amoutn of diodes / treshold in a non-inverting soft clipper.  ;D
I don't quite follow you.  I suspect what is being said in my post is in a non-inverting case it always mixes the clean signal with the clipped signal.  If you have more diodes  the magnitude of the diode contribution becomes higher, in relative terms, so in effect it would sound more distorted.

Yes, which was what I tried to explain with the curves drawn. Vivek drew red lines which are IMHO incorrect.
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PRR

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Rob Strand

QuotePosted by: PRR
« on: Today at 02:09:13 PM »

    Insert Quote

PPR, I don't need any convincing there's no real knee.   All was doing was explaining why we see one.  I actually drew these yesterday,





[I actually drew these with different x and y sizes to show how scale causes the eye to shift the knee. It's a bit subtle perhaps easier to see in the 0.4V case.]

When we use silicon diodes it is true we see a voltage around say 500mV to 700mV.   The exact voltage depends on the current an the diode.   All that's saying is over some practical range of currents diode log(current) function, which has no knee, squashes the range of voltages we see into a smaller range which gives the illusion diodes a have a constant voltage drop and a knee.

The fact the "knee" in those two plots is not the same is actually evidence there isn't a true knee from the diode's perspective.

If we turn the problem around, the point where the effective of the diodes *starts* to be heard is more of a threshold point than any knee.  That's going to be at some low voltage, not at any perceived knee.   It will depend on the resistor in parallel with the diode feedback case, or  resistor feeding the diode in the R + D clipper.

QuoteYes, which was what I tried to explain with the curves drawn. Vivek drew red lines which are IMHO incorrect.
If you model the diode as a voltage drop with the series resistance it makes sense the resistances must add (that's where I caught the bug in my previous post).    For apples to apples comparison of distortion is make sense to scale the output of three diode case by 1/3.    So for the non-inverting opamp cases:  the output for one diode is Vclean + Vdiode  but for the three diode case it's  (1/3) (Vclean + 3 * Vdiode) = Vclean/3 + Vdiode   so it's clear the amount of Vclean is watered down by a factor of 1/3.
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According to the water analogy of electricity, transistor leakage is caused by holes.

PRR

We often draw on Linear paper. But we hear with Log ears. No knee.

In_a_circuit, we often have a constant gain/resistance and a variable impedance on the diode. If the resistance is 26k, the diode "does nothing" for up to 1uA current, diode dominates (2.6k) at 10uA.

(Yes Rob, you know all this, I'm preaching at lurkers.)
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Rob Strand

QuoteIn_a_circuit, we often have a constant gain/resistance and a variable impedance on the diode. If the resistance is 26k, the diode "does nothing" for up to 1uA current, diode dominates (2.6k) at 10uA.

The low voltage side is overlooked IMHO.   If you have a high level input the diode current is pretty much independent of the feedback resistance (ie. the pot on the TS9).     However we clearly hear a difference with different pot settings, which means the sound is being determined by the lower input signals.
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Steben

Quote from: Rob Strand on November 17, 2020, 05:35:25 PM
QuoteIn_a_circuit, we often have a constant gain/resistance and a variable impedance on the diode. If the resistance is 26k, the diode "does nothing" for up to 1uA current, diode dominates (2.6k) at 10uA.

The low voltage side is overlooked IMHO.   If you have a high level input the diode current is pretty much independent of the feedback resistance (ie. the pot on the TS9).     However we clearly hear a difference with different pot settings, which means the sound is being determined by the lower input signals.

Min gain is ca 11 x on a TS9. This means at 0.06Volts input things start to get topped off and the treble gets cut dependent on the pot setting as well.
Ild say the "knee" (yes I know this a contraption) is softer with lower R in parallel given the circuit. R = Rf // Rd With higher Rf (high gain) the R drops dramatically with changes Rd. Or in other words the higher the Rf, the less important the diode characteristics.
Where lower R in series with output clippers means harder knee.
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Rob Strand

I did this the other day but I don't have the free time to finish it off.

Here's part 1 which shows how different diodes perform.

For apples to apples comparison, the main constraints are:
- low signal gains are equal.   For any reasonable voltage this means different feedback resistors.
- all diodes have 0.6V drop at 1mA

I've created three diodes with different ideality factors; the "n" parameter in the diode equations.  So the diodes have different softness.  It is not possible to set the n higher than about 2.6 otherwise the soft clipping starts so early that even tiny signals start to get compressed.

In the figures below only the 100Hz source is active, and the level of that source is increased.

FYI if you click on the image a second time on the image site you will get a larger clearer image.

The basic set-up: (the text should read 2.6 not 2.3, I squeezed a bit more out of it at the last minute.)
[click to enlarge]



The DC characteristic for small signals.   The slope is the small signal gain.  You can see they are all well matched.
[click to enlarge]


The DC characteristic for large signals.  You can see at 1V in they all hit 0.6V output.
[click to enlarge]



Small signal AC input 100uV pk.  Shows gains are the same.
[click to enlarge]


Increasing the AC input to 1mV pk.  We can see there is already some compression of the fundamental on the softer diodes.
[click to enlarge]


Larger AC signals of 100mV peak.  More compression but softer clipping and less output for the softer diodes.   Harder for the harder diodes.
[click to enlarge]


"Full" input of 1V where all diode peak at 1mA and 600mV. Similar to prev except peaks now match.
[click to enlarge]

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Steben

That is quite intriguing.
How do the diodes have their response beyond 1.0V? Do they converge?
The "softest one" almost looks like a schottky with some extra resistance in series or a high treshold germanium.

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Rob Strand

QuoteThat is quite intriguing.
How do the diodes have their response beyond 1.0V? Do they converge?
No, the softer diodes then start to produce a higher output.   The easy way to think about it is the soft diodes squash less.   Once two diodes hit the same point on the VI curve the softer one will continue to rise whereas the hard one will stay flat.

On the 1V DC curve, you can see the soft diode has a *higher* slope (around the point 1V in 600mV out) than hard diode so by extending the line the soft diode will rise-up more.

QuoteThe "softest one" almost looks like a schottky with some extra resistance in series or a high treshold germanium.
It's definitely around that zone.

If I get time, I'll try show tweaking the slope with resistances isn't the same as having a diode which has the lower slope.   The slope the eye sees has the same problems as the knee the eye sees.    There is only a fixed slope because of the scale it is drawn on.  If you draw the curves on different scales you get different slopes.   For a reasonable upper limit in currents, the diode curves are always curving whereas with resistors the slope levels off.   The curvature is what produces the upper harmonics.    Especially the nasty harmonics which are far away in frequency from the fundamental.   A resistor will not produce harmonics.   If your guitar signal is 200Hz and the harmonics are 4kHz  that's a lot of harmonics away and only curvature can produce that.
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According to the water analogy of electricity, transistor leakage is caused by holes.