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Everybody agrees that the shape of the IV knee of a diode is very important.

In all my SPICE simulations so far, I found the maximum current through the clipping diodes in the feedback loop is normally around 100 to 250uA

I drew red lines in the graph to show the region of the IV curve that we are operating in.

Seems hard to prove there is a big difference or impact of IV curve in such a small range.

(https://i.postimg.cc/QtBsSCBP/MRX-D.jpg) (https://postimg.cc/cgNVLsBM)

(https://i.postimg.cc/YqDRBSj8/AAA-124850443-3679244822133581-811767252105737916-n.jpg) (https://postimg.cc/w7JLhgXs)

What did I overlook ?

Change the scale at the right side to get at +/- 300 µA max. Then you'll see.

I take it R666 is a metal resistor?   :icon_lol:

QuoteEverybody agrees that the shape of the IV knee of a diode is very important.

i don't
in the bedroom, playing alone, sure, one might sound different than another, and there is plenty of change noticeable in the feel,
but when you are out there on a stage, in a rehearsal room, at your local bar's jam session or somewhere else where you play at decently high volume, together with more musicians... you aren't going to notice the difference. the sound difference of a different speaker is about 10x as large. placing the amp in a corner or along a straight wall has more impact on the sound than the knee of the diode. so... yes and no, there is a difference in sound and feel, but its nowhere near ''very important"

QuoteSeems hard to prove there is a big difference or impact of IV curve in such a small range.

in that bedroom situation, the difference will be biggest in the small beginning of the diode curve. as the currents are higher, the difference will be less and less noticeable, since we're going towards a square wave anyways, it doesn't matter much how we get there.

cheers, Iain

Quote from: ThermionicScott on November 10, 2020, 02:22:33 PM
I take it R666 is a metal resistor?   :icon_lol:

Thanks Scott, you just made my otherwise rotten day :)

Quote from: Steben on November 10, 2020, 12:55:45 PM
Change the scale at the right side to get at +/- 300 µA max. Then you'll see.

+1

Andy

Quote from: Vivek on November 10, 2020, 12:17:50 PM
Everybody agrees that ...

I disagree.

Unidirectional diodes are mostly an exponential (which looks the same after scaling) with shunt and series resistors (which are intended to be insignificant).

They are not THE SAME, but you can adjust drive and circuit resistances to make it Real Real Close.

Note that the wide range of music and the exponential function makes a Linear scale pretty misleading.

Quote from: ThermionicScott on November 10, 2020, 02:22:33 PM
I take it R666 is a metal resistor?   :icon_lol:

Noice !!!!!

You are my type of guy !!!!

I love puns, play on words, mixed metaphors, cryptic crossword puzzles, etymology ..............

Quote from: PRR on November 11, 2020, 12:08:51 AM

Quote from: Vivek on November 10, 2020, 12:17:50 PM
Everybody agrees that ...

I disagree.

Quote from: iainpunk on November 10, 2020, 03:50:43 PM

QuoteEverybody agrees that the shape of the IV knee of a diode is very important.

i don't

cheers, Iain

My original contention was the the role of shape of diode IV knee is overstated in clipper applications.

So would you agree that exact shape of diode knee is quite uncritical in clipper applications since we operate in very small region of the IV curve

and can change knee with compliance resistors in any case.

Quote from: Vivek on November 11, 2020, 01:20:26 AM
So would you agree that exact shape of diode knee is quite uncritical in clipper applications..

If so, then C3 should be sheer useless... :icon_wink:

Most diodes follow basically a log function with a fixed internal resistor.
Add a resistor to a classic silicon diode, double the gain and you will get close to any germanium sound.

The change happens to an extend once you use zeners and active devices as clipping diodes.

Quote..... since we operate in very small region of the IV curve
and can change knee with compliance resistors in any case.

no, it doesn't matter what part of the vi curve were in, the knee is not important however we try to change i. distortion sounds like distortion regardless of curve, especially when you are in a live mix, it all gets mushed out. what really matters is the frequency shaping before and after the distortion, this, i didn't understand the first few years of building pedals, before i joined the forum.
the only place where knee and curve and such things matter is when you play at low gain and low volume in a bedroom or so. if the volume is high enough, you'll lose that difference in feel and sound (yes, even when wearing ear protection).

cheers, Iain

Quote from: iainpunk on November 11, 2020, 11:47:22 AM

Quote..... since we operate in very small region of the IV curve
and can change knee with compliance resistors in any case.

no, it doesn't matter what part of the vi curve were in, the knee is not important however we try to change i. distortion sounds like distortion regardless of curve, especially when you are in a live mix, it all gets mushed out. what really matters is the frequency shaping before and after the distortion, this, i didn't understand the first few years of building pedals, before i joined the forum.
the only place where knee and curve and such things matter is when you play at low gain and low volume in a bedroom or so. if the volume is high enough, you'll lose that difference in feel and sound (yes, even when wearing ear protection).

cheers, Iain

I would add that what matters in the bedroom also matters on tape. Wait, that came out wrong... I mean, the differences between diode characteristics are subtle but audible under good conditions. In a live situation, all that matters is volume. Well, almost. A really loud guitar will sound awesome, no matter the diodes.

It is also true that you can bend the behavior of diodes by adjusting the circuit around them. Or you can choose the diode according to what purpose it shall serve in a given circuit. It you really want to get a specific clipping characteristic, you will have to select the diodes, as well as the voltages and currents they are operated at quite carefully. And even then, it only really matters under ideal listening conditions.

I for one always design with ideal listening conditions in mind. I think it is fun to tweak those tiny little details that almost no one will ever hear. But that is a personal choice. Or OCD. I also vacuum my apartment although I have a dog and everything looks like under Satan's sofa again five minutes after I'm done. It's a bit pointless but it feels good to know I've given it my best.

Andy

I agree on the difference in application. A low to mid gainy circuit will have lots of signal crossing the knee. 10x1N4148 in series will not sound as 2x3V Zenders and that will not sound as a 6V Zener diode. And there is capacitance as well...  Of course this only applies to the possibility of high supply voltage. The lower the headroom, the less design choices.

DO not forget transient curves of typical amps are different as well! A typical vox with non feedback cathode bias push pull ("creamy", "shimmer") has a much more curved transition into clipping than feedback fixed bias brothers of most marshalls and fenders ("crisp" "tight").

(https://i.postimg.cc/BjYBJK3f/knee.jpg) (https://postimg.cc/BjYBJK3f)


(https://i.postimg.cc/23LkGRyQ/amp-diff.jpg) (https://postimg.cc/23LkGRyQ)

Quote from: Fancy Lime on November 11, 2020, 01:24:50 PM

It you really want to get a specific clipping characteristic, you will have to select the diodes, as well as the voltages and currents they are operated at quite carefully. And even then, it only really matters under ideal listening conditions.

Andy

Almost no distortion pedal DC biases the clipping diodes to particular operating points

At low signals, the diode is working at 0 current

At high clipping, the diode in feedback loop passes somewhere between 100 to 250uA.

I never saw any circuit where the designer decided that the  clipping diode is biased around 5ma with no signal, and varies around it when signal is passed.

( Solid state diodes in feedback loop)


So I do not agree that any pedal builder choses the voltages and currents passing through the clipping diodes.

Quote from: Steben on November 11, 2020, 01:41:55 PM


(https://i.postimg.cc/BjYBJK3f/knee.jpg) (https://postimg.cc/BjYBJK3f)

Thanks

Would it be possible to mark the values of current on the y axis

And also to show only 0 to 350 uA on the y axis since that is the range of diodes in the feedback loop of an opamp

Quote from: Vivek on November 11, 2020, 02:28:26 PM

Quote from: Steben on November 11, 2020, 01:41:55 PM


(https://i.postimg.cc/BjYBJK3f/knee.jpg) (https://postimg.cc/BjYBJK3f)

Thanks

Would it be possible to mark the values of current on the y axis

And also to show only 0 to 350 uA on the y axis since that is the range of diodes in the feedback loop of an opamp

(https://lh3.googleusercontent.com/proxy/yw-ZaQnI-J9CtI5ppg-7UxABJjKOnUxZtoI9x5hMNeGUaZuC3DlQ4kaTnavaPX471EsZ30euH4l6AiRMs0Yo1sfkIHh8jnw)
(https://usermanual.wiki/Document/bzt52c2v0bzt52c51r372diodes.57529292-User-Guide-Page-3.png)
(https://i.stack.imgur.com/2AdwD.png)

And of course already posted:
(https://therepaircafe.files.wordpress.com/2019/10/ihz3udv.png)

Quote from: iainpunk on November 11, 2020, 11:47:22 AM

Quote..... since we operate in very small region of the IV curve
and can change knee with compliance resistors in any case.

what really matters is the frequency shaping before and after the distortion, this, i didn't understand the first few years of building pedals, before i joined the forum.

On top of that, those EQ details are very important in combination with the guitar. A strat usually has a flat to almost scooped character (in between pos) and in that way loves mid hump stuff (like most overdrives, brown sounds, rats...) more than tele's or humbucker glued necks. I always found strats suck with overdrive circuits that are cheered for their "neutral sound".

Quotewhat matters in the bedroom also matters on tape

hahahaha, thats a good one. :icon_lol: :icon_lol: :icon_lol: :icon_lol:
but on that note, if i'd be in a studio recording "good sounding guitar", i don't think i'd use an overdrive pedal. ill rent or loan a good tube amp, and maybe put an overdrive pedal in front of the already overdriven amp, but in that situation, the diodes won't matter either.

cheers, Iain

QuoteI always found strats suck with overdrive circuits that are cheered for their "neutral sound".

i always found that those circuits suck for every guitar you put through it. the only way i like a Klon is everything on max with a rangemaster in front of it.

cheers, Iain

Quote from: Vivek on November 11, 2020, 02:24:15 PM

Quote from: Fancy Lime on November 11, 2020, 01:24:50 PM

It you really want to get a specific clipping characteristic, you will have to select the diodes, as well as the voltages and currents they are operated at quite carefully. And even then, it only really matters under ideal listening conditions.

Andy

Almost no distortion pedal DC biases the clipping diodes to particular operating points

At low signals, the diode is working at 0 current

At high clipping, the diode in feedback loop passes somewhere between 100 to 250uA.

I never saw any circuit where the designer decided that the  clipping diode is biased around 5ma with no signal, and varies around it when signal is passed.

( Solid state diodes in feedback loop)


So I do not agree that any pedal builder choses the voltages and currents passing through the clipping diodes.

Oh, but I think we do agree! :) I said "If you want to...". I did not say anyone does that, and for good reason. In fact, I played around with DC-biased clipping diodes years ago and abandoned that because I found it was just not worth the bother. There were always easier solutions to get the same sound. However, controlling the currents through diodes by adjusting the impedance of the surrounding circuit, can be quite useful for tuning the clipping threshold and clipping onset behavior. The series resistor before the diodes in an MXR distortion+ or DOD250 does that, for example. The effect is, as has been pointed out, subtle, especially for "normal" Si diodes. For reverse biased low voltage Zeners and some Ge diodes this can be more pronounced, see Steben's posts. Still not something you could not also simulate with a different diode and a resistor or two. Weirdly, MOSFET body diodes, which are often praised for they're "tube-like soft clipping characteristics" in marketing brochures, are exactly the opposite: super hard with almost no knee. At least the commonly used ones. Make of that fact what you will but I say it proves that diode choice is a rather small part in a rather big puzzle. Pre and post clipping EQing is the big one, indeed. Diodes are super important for the mojo, though. Track down a pair of 1968 soviet Ge diodes salvaged from a radio receiver in Baikonur and you got yourself one hell of a fancy upgrade for a Distortion+. It won't sound any better but, oh, the mojo!

Andy

Quote from: iainpunk on November 11, 2020, 03:54:24 PM

Quotewhat matters in the bedroom also matters on tape

hahahaha, thats a good one. :icon_lol: :icon_lol: :icon_lol: :icon_lol:
but on that note, if i'd be in a studio recording "good sounding guitar", i don't think i'd use an overdrive pedal. ill rent or loan a good tube amp, and maybe put an overdrive pedal in front of the already overdriven amp, but in that situation, the diodes won't matter either.

cheers, Iain

Really? I'll take my breadboard straight into the mixer over any tube amp and cab combination any day of the week. Way mode flexible ;) Also, I never got the fuss about tube amps. A good distortion pedal and a basic cab sim just sounds so much better to me. You can always EQ it in post. Ah well, it takes all kinds...

Andy

i am not a fan of good sounding guitar to be honest, i get why people use good tube amps cranked, the speakers behave in a way that just makes overdrive so boring ideal. i personally just took 3 small (all with 6 or 8 inch single speakers, all under 20W) practice amps to the studio, all on clean channels cranked to 10 and a breadboarded fuzz to drive those clean channels in to distortion and it sounded perfectly bad. we did some cool recordings, but the douche who wanted/payed to put me and my sound on his record thought he could sing... he didn't... it was worse than the shaggs.

if you have clipping diodes in the feedback loop, you can schale all resistors and caps by 10 up or down and keep the gain the same, but it will sound somewhat different. if you want to test this for yourself, try this out on a breadboard: scaling the resistors up 10x won't change the gain but it will change the clipping characteristic.
(https://i.postimg.cc/T5Y3NVf9/simle-example.png) (https://postimg.cc/T5Y3NVf9)

cheers, Iain

Quote from: iainpunk on November 11, 2020, 04:03:45 PM

QuoteI always found strats suck with overdrive circuits that are cheered for their "neutral sound".

i always found that those circuits suck for every guitar you put through it. the only way i like a Klon is everything on max with a rangemaster in front of it.

cheers, Iain

Aha, so you are advocating the Klon as surrogate for amptone?  :icon_mrgreen:

Quote from: iainpunk on November 11, 2020, 05:33:47 PM
i am not a fan of good sounding guitar to be honest, i get why people use good tube amps cranked, the speakers behave in a way that just makes overdrive so boring ideal. i personally just took 3 small (all with 6 or 8 inch single speakers, all under 20W) practice amps to the studio, all on clean channels cranked to 10 and a breadboarded fuzz to drive those clean channels in to distortion and it sounded perfectly bad. we did some cool recordings, but the douche who wanted/payed to put me and my sound on his record thought he could sing... he didn't... it was worse than the shaggs.

if you have clipping diodes in the feedback loop, you can schale all resistors and caps by 10 up or down and keep the gain the same, but it will sound somewhat different. if you want to test this for yourself, try this out on a breadboard: scaling the resistors up 10x won't change the gain but it will change the clipping characteristic.
(https://i.postimg.cc/T5Y3NVf9/simle-example.png) (https://postimg.cc/T5Y3NVf9)

cheers, Iain

Yes thats it.
Because the knee characteristics of most diodes are comparable and the circuit works as it does, the treshold voltage is the main difference indeed. At max gain (think infinite) you simply get a square wave with the original signal on top. The larger resistance of the summed diodes is nihil compared to the R of the leg to ground, hence unity gain remains. The higher the treshold, the higher the ratio square vs clean. In other words: the overdrive goes towards distortion like character with every voltage treshold you add.
The little detail that can ruin it all is the input signal: if this is an amplified signal, you go well beyond the classic guitar amplitudes. This means the original signal can be dominant even clipping the opamp. That's why some pedals add a clipper pair to ground before the stage.


(https://i.postimg.cc/PvpnMm8s/soft-clipping-Vd.jpg) (https://postimg.cc/PvpnMm8s)

yes, exactly, i have a guitar with fairly high output and it never sounds good with such soft clipping circuits, there is always the nasty fizz mixed with clean effect. (the guitar's transient easily makes red led's clip if you strike it semi-hard)

the knee actually changes when you schale the resistances up or down, this is something rarely used in overdrive designs.

cheers, Iain

Quote from: Steben on November 12, 2020, 01:04:07 PM

(https://i.postimg.cc/PvpnMm8s/soft-clipping-Vd.jpg) (https://postimg.cc/PvpnMm8s)

Brother Steben, please help me understand this graphic.

Does it imply that if we place many diodes in series, the clipping threshold changes but the total effective resistance of multiple series diodes when conducting is same as one diode ?

Quote from: Vivek on November 13, 2020, 12:17:47 AM

Quote from: Steben on November 12, 2020, 01:04:07 PM

(https://i.postimg.cc/PvpnMm8s/soft-clipping-Vd.jpg) (https://postimg.cc/PvpnMm8s)

Brother Steben, please help me understand this graphic.

Does it imply that if we place many diodes in series, the clipping threshold changes but the total effective resistance of multiple series diodes when conducting is same as one diode ?

Mmmm, yes and no.... normally overdrive circuits tend to have several tens of kohms at max gain on the feedback loop and a couple of kohms in the leg to ground. The internal resistance of diodes is couple of ohms max. Since the impedance above the treshold is those couple of ohms you easily see that even some of those in series still means almost nothing compared to the leg resistor.

Quote from: Steben on November 13, 2020, 12:31:45 PM

Mmmm, yes and no.... normally overdrive circuits tend to have several tens of kohms at max gain on the feedback loop and a couple of kohms in the leg to ground. The internal resistance of diodes is couple of ohms max. Since the impedance above the treshold is those couple of ohms you easily see that even some of those in series still means almost nothing compared to the leg resistor.

But this implies that the exact IV curve of a diode is not important

Which was what I was trying to prove

Quote from: Vivek on November 13, 2020, 01:10:56 PM

Quote from: Steben on November 13, 2020, 12:31:45 PM

Mmmm, yes and no.... normally overdrive circuits tend to have several tens of kohms at max gain on the feedback loop and a couple of kohms in the leg to ground. The internal resistance of diodes is couple of ohms max. Since the impedance above the treshold is those couple of ohms you easily see that even some of those in series still means almost nothing compared to the leg resistor.

But this implies that the exact IV curve of a diode is not important

Which was what I was trying to prove

It is rather unimportant in that circuit yes.
yet, again, some "diodes" are not standard silicons or germaniums, which means applying some might give different results. But those can be simulated with standard diodes and resistors.
A simple dist + for example is called "slightly soft clipping" because it has germanium diodes instead of silicons. Why? Because of the resistance. Double the gain and use a silicon with resistor in series and you'll get about the same result.  Or use 18V, make the gain really high and use 2V7 Zeners. About the same.
The thing is, it is always a question of "yes and no". The benefits of different diode types is combining them in "ladders" with resistors, shaping your own "knee".

Quote from: Steben on November 13, 2020, 01:19:04 PM
The benefits of different diode types is combining them in "ladders" with resistors, shaping your own "knee".

But almost no commercial guitar distortion pedal has a DFG.

At the most, one series resistor for compliance in some designs

and very rarely, a parallel resistor.

Very interesting article

https://2n3904blog.com/1n4148-diode-forward-biased-i-v-curve/

Quote from: Vivek on November 13, 2020, 02:16:15 PM

Quote from: Steben on November 13, 2020, 01:19:04 PM
The benefits of different diode types is combining them in "ladders" with resistors, shaping your own "knee".

But almost no commercial guitar distortion pedal has a DFG.

At the most, one series resistor for compliance in some designs

and very rarely, a parallel resistor.

Commerce is about sellibg at the least effort. Usually copying.

Quote from: Steben on November 13, 2020, 12:31:45 PM

Quote from: Vivek on November 13, 2020, 12:17:47 AM...if we place many diodes in series, the clipping threshold changes but the total effective resistance of multiple series diodes when conducting is same as one diode?

Mmmm, yes and no.... normally overdrive circuits tend to have several tens of kohms at max gain on the feedback loop and a couple of kohms in the leg to ground. The internal resistance of diodes is couple of ohms max. ...

The slope does add-up with series diodes. As you say, with limited voltage the current is less and this too increases diode impedance. Diode impedance can be many-K, higher than typical circuit impedance. Shockley's Law predicts 26r@1mA and 26k@1uA.

Quote
But almost no commercial guitar distortion pedal has a DFG.

At the most, one series resistor for compliance in some designs

and very rarely, a parallel resistor.

funny how you assume that pedal companies give $0.02 about the diodes, the knee and the response, especially if you keep in mind that most players don't care as long as it does the job, it need not be perfect. especially at high volume levels, you won't hear the difference anyways, so who cares? making a pedal cheaper and faster to design saves a lot of money on a large scale.

that's where boutique pedal builders and DIY hobbyists come in, we take our time to perfect the sound of a design, giving that rare guitarist who isn't satisfied with comercial pedals a place to get good, 'tuned' and specialised pedals. but that is a small market, so small scale is the only way, so prices get higher... simple supply and demand... God, i love capitalism

cheers, Iain

Quote from: PRR on November 13, 2020, 05:23:48 PM

Quote from: Steben on November 13, 2020, 12:31:45 PM

Quote from: Vivek on November 13, 2020, 12:17:47 AM...if we place many diodes in series, the clipping threshold changes but the total effective resistance of multiple series diodes when conducting is same as one diode?

Mmmm, yes and no.... normally overdrive circuits tend to have several tens of kohms at max gain on the feedback loop and a couple of kohms in the leg to ground. The internal resistance of diodes is couple of ohms max. ...

The slope does add-up with series diodes. As you say, with limited voltage the current is less and this too increases diode impedance. Diode impedance can be many-K, higher than typical circuit impedance. Shockley's Law predicts 26r@1mA and 26k@1uA.

Is the slope of conducting part of 2 diode and 3 diode drawn correctly ?



(https://i.postimg.cc/PvpnMm8s/soft-clipping-Vd.jpg) (https://postimg.cc/PvpnMm8s)

QuoteQuote from: PRR on Today at 05:23:48 PM

    Quote from: Steben on Today at 12:31:45 PM

        Quote from: Vivek on Today at 12:17:47 AM

            ...if we place many diodes in series, the clipping threshold changes but the total effective resistance of multiple series diodes when conducting is same as one diode?

        Mmmm, yes and no.... normally overdrive circuits tend to have several tens of kohms at max gain on the feedback loop and a couple of kohms in the leg to ground. The internal resistance of diodes is couple of ohms max. ...


    The slope does add-up with series diodes. As you say, with limited voltage the current is less and this too increases diode impedance. Diode impedance can be many-K, higher than typical circuit impedance. Shockley's Law predicts 26r@1mA and 26k@1uA.



I was expecting higher slopes

The slope Vd vs Id for a single diode is rd = n VT / Id,
where VT = is the thermal voltage which is about 26mV at room temperature, Id is the diode current, and n is the ideality factor.

The slope is derived from the exponential diode relationship.

iD  = I0 exp  ( vD / (nVT) )

So,
vD  = n.VT ln(iD / I0) = nVT ( ln(iD) - ln(I0))

Take the derivative,

rd = d vD / d iD    =     nVT/iD

The result is independent of vD, so if you put three diode in series to make a compound diode with voltage uD = 3 *vD, the VI equation is now
   
iD  = I0 exp  ( uD / (3 nVT) )

If you derive the slope it's the same maths but  now get rd_3diodes = 3 * rd_1diode.

The difference between diodes is in the ideality factor n.   A BJT junction has an n just above 1 whereas a 1N4148  signal diode tends to be around 1.8,  +/- 0.2 depending on how it's derived.

EDIT:  Dumb arsed mistakes fixed.

Quote from: Rob Strand on November 13, 2020, 11:40:00 PM

QuoteQuote from: PRR on Today at 05:23:48 PM

    Quote from: Steben on Today at 12:31:45 PM

        Quote from: Vivek on Today at 12:17:47 AM

            ...if we place many diodes in series, the clipping threshold changes but the total effective resistance of multiple series diodes when conducting is same as one diode?

        Mmmm, yes and no.... normally overdrive circuits tend to have several tens of kohms at max gain on the feedback loop and a couple of kohms in the leg to ground. The internal resistance of diodes is couple of ohms max. ...


    The slope does add-up with series diodes. As you say, with limited voltage the current is less and this too increases diode impedance. Diode impedance can be many-K, higher than typical circuit impedance. Shockley's Law predicts 26r@1mA and 26k@1uA.



I was expecting higher slopes

The slope Vd vs Id for a single diode is rd = n VT / Id,
where VT = is the thermal voltage which is about 26mV at room temperature, Id is the diode current, and n is the ideality factor.

The slope is derived from the exponential diode relationship.

iD  = I0 exp  ( vD / (nVT) )

So,
vD  = n.VT ln(iD / I0) = nVT ( ln(iD) - ln(I0))

Take the derivative,

rd = d vD / d iD    =     nVT/iD

The result is independent of vD, so if you put three diode in series to make a compound diode with voltage uD = 3 *vD, the VI equation is now
   
iD  = I0 exp  ( uD / (nVT) )

If you derive the slope it's the same maths and you get the same rd.

The difference between diodes is in the ideality factor n.   A BJT junction has an n just above 1 whereas a 1N4148  signal diode tends to be around 1.8,  +/- 0.2 depending on how it's derived.

Brother Rob

(https://i.postimg.cc/626st1MM/2020-11-14-08-20-40.jpg) (https://postimg.cc/626st1MM)

Are the red lines correct or wrong ?

QuoteBrother Rob
Are the red lines correct or wrong ?

Lucky you posted,  I came back and looked at what I wrote before and my intuition tells me something isn't right.

The bug is,

iD  = I0 exp  ( uD / (nVT) )

For three diodes it should be,

iD  = I0 exp  ( uD / (3 * nVT) )

Because the three diodes needs three times the voltage.   That means the compound diode has three times the slope.   Which is far more agreeable to my common sense.

So... the red lines are correct in principle.  They are a bit wonky though.

More like, [click to enlarge]

Shown here with n=1.9.

(https://i.postimg.cc/gxJMn9pp/1diode-2diode-3diode-series.png) (https://postimg.cc/gxJMn9pp)


Here's n = 3.8,
[click to enlarge]
(https://i.postimg.cc/2bLHxg8W/1diode-2diode-3diode-series-n-3-8.png) (https://postimg.cc/2bLHxg8W)

It's unlikely to get a diode with n=3.8.


In both examples the diodes have 600mV drop with 1mA current.

I felt
3 rd in series = 3.rd

Or very slightly different due to nonlinearities


May I request you to plot

1 diode

2 diodes in series

3 diodes in series

1 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)

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.

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.

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.

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

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!

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.

(https://i.postimg.cc/MXLWP8D6/Diode-Z-log-42.gif) (https://postimg.cc/MXLWP8D6)

> 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.

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.
(https://live.staticflickr.com/65535/50603670403_f0aaf52019_c.jpg)

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....

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.

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

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.

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.

(https://i.postimg.cc/MXLWP8D6/Diode-Z-log-42.gif) (https://postimg.cc/MXLWP8D6)

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,

(https://i.postimg.cc/ThjCkGm5/diode-knee-Vmax-0-6-V.png) (https://postimg.cc/ThjCkGm5)

(https://i.postimg.cc/xXZRkdrh/diode-knee-Vmax-0-4-V.png) (https://postimg.cc/xXZRkdrh)

[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.

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.)

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.

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.

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]
(https://i.postimg.cc/7bFq6zqm/Dclip00-Schematic.png) (https://postimg.cc/7bFq6zqm)


The DC characteristic for small signals.   The slope is the small signal gain.  You can see they are all well matched.
[click to enlarge]
(https://i.postimg.cc/BjMWxPGW/Dclip01-DC-1m-V-1u-A.png) (https://postimg.cc/BjMWxPGW)

The DC characteristic for large signals.  You can see at 1V in they all hit 0.6V output.
[click to enlarge]
(https://i.postimg.cc/rKN7kRsd/Dclip02-DC-0-to-1-V-1m-A.png) (https://postimg.cc/rKN7kRsd)


Small signal AC input 100uV pk.  Shows gains are the same.
[click to enlarge]
(https://i.postimg.cc/4Km0V85t/Dclip03-100-Hz-100u-V-100n-A-pk.png) (https://postimg.cc/4Km0V85t)

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]
(https://i.postimg.cc/sG9HsbjF/Dclip04-100-Hz-1m-V-1u-A.png) (https://postimg.cc/sG9HsbjF)

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]
(https://i.postimg.cc/QFBPW4S7/Dclip05-100-Hz-100m-V-100u-A-pk.png) (https://postimg.cc/QFBPW4S7)

"Full" input of 1V where all diode peak at 1mA and 600mV. Similar to prev except peaks now match.
[click to enlarge]
(https://i.postimg.cc/gX4478Nd/Dclip06-100-Hz-1-V-1m-A-pk.png) (https://postimg.cc/gX4478Nd)

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.

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.

Quote from: Rob Strand on November 18, 2020, 06:22:35 PM

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.

I understand. Although on a given scale (voltage span) the effect of crossing curves comes close to added resistance, hence the higher output, dynamics and lower harmonic content.
That is why germanium and silicons sound different in the SAME circuit, but can sound alike in different circuits.

 :icon_mrgreen: 2015

(https://i.postimg.cc/VrR13nQH/Naamloos-X.jpg) (https://postimg.cc/VrR13nQH)

Effect of "Compliance" in clipping


(https://scontent.ffjr1-4.fna.fbcdn.net/v/t1.0-9/120231692_3667478889930350_354066600508861264_o.jpg?_nc_cat=107&ccb=2&_nc_sid=b9115d&_nc_ohc=ocR7C4TE1kAAX-SIpla&_nc_ht=scontent.ffjr1-4.fna&oh=58b28974e616dec36e127b69b961796c&oe=5FDEF046)


(https://scontent.ffjr1-6.fna.fbcdn.net/v/t1.0-9/120134982_3667479143263658_3087619767076465419_o.jpg?_nc_cat=104&ccb=2&_nc_sid=b9115d&_nc_ohc=ZWZUVvSGmUEAX9IdwG2&_nc_ht=scontent.ffjr1-6.fna&oh=b38ee7428899094c82452c253f2930ce&oe=5FDFBC8C)


(https://scontent.ffjr1-5.fna.fbcdn.net/v/t1.0-9/120025333_3667479549930284_6941035803522767364_o.jpg?_nc_cat=100&ccb=2&_nc_sid=b9115d&_nc_ohc=vrIO2DpQ8tUAX8hNEBm&_nc_ht=scontent.ffjr1-5.fna&oh=3d18501d9548f1d1d612170c97839722&oe=5FDF8498)



Based on plotting following equation

V=(abs(V(Out1))< clip)*V(Out1)
+(V(Out1)>= clip)*(clip+((V(Out1)-clip)*compl))
+(V(Out1)<=-clip)*(-clip-((-clip-V(Out1))*compl))

Where V is the output, Out1 is the signal that has to be processed, clip is clipping voltage, compl is compliance after clip

Compliance = 0.5 means

If there is a signal with 3V peak
and its clipped at 0.65V

then it means that the input signal has 3-0.65 volts more to go after clipping

But 0.5 compliance means that the output signal will have a peak of 0.65 + 0.5(3 - 0.65) V

(https://i.postimg.cc/xCNTgPvV/Boneyard.jpg) (https://postimg.cc/R69BZKpg)

Boneyard, Carl Martin Plexitone

Why did designer make 5-7 mA go through LED in the feedback loop ?

What extra benefit did he get instead of only allowing 100uA into the LEDs ?

QuoteV=(abs(V(Out1))< clip)*V(Out1)
+(V(Out1)>= clip)*(clip+((V(Out1)-clip)*compl))
+(V(Out1)<=-clip)*(-clip-((-clip-V(Out1))*compl))

So the way I look at this these days would be,

   V1 =  abs(V(Out1))< clip)*V(Out1)

which is just the clipped waveform. 
ie.  If V(out1) > +clip then V1 = +clip, if V(out1) < -clip then V1 = -clip  otherwise V1 = V(out1)

Then make the combination,

   V  = V1 *(1-comp)  + V(Out1) * comp

which is just a mixer or blender:
- when comp = 1 only the clean signal comes through.
- when comp = 0 only the clipped signal comes through.
- when comp = 0.5 equal amounts of clipped and clean signal are mixed together.

The important point here is this V is *identical* to the one you gave.   All I'm doing is writing it differently.

The second view makes it very clear what you are listening to - a blend of a hard clipped signal and the clean signal.  The character of clipped signal is nasty.  And adjusting comp does not change that nasty character.   So comp appears to soften but it's actually not.  All comp does is make the nasty signal softer.

A softer transition for the "clipping" would change the character of the clipped signal.    That would appear as a more rapid tapering of higher harmonics (in the same way that a triangle wave sounds less nasty than a square wave).  In the blend case the tapering of the harmonics is fixed and is determined by the hard clipping.

QuoteThat is why germanium and silicons sound different in the SAME circuit, but can sound alike in different circuits.

And it's pretty much why this holds.    The parallel resistor is the only thing you can play with and it has limited scope to change things.

QuoteWhat extra benefit did he get instead of only allowing 100uA into the LEDs ?

Smaller R's make the clipping a little harder.

Respect and Thanks, Sir Rob !!!!!


PS : Why have 2 LEDS in series in PLEXITONE ?

QuotePA : Why have 2 LEDS in series in PLEXITONE ?

My guess is since the non-inverting stage blends a bit of clean putting in two LEDs waters down that clean component.   The whole clipping stage is setting itself up for  a harder-clipped sound (but not necessarily nasty, maybe less hair at low drive).     The reason they can get away with two LEDs in series is because they have up'ed the supply voltage.    Two LEDs on a 9V set-up will clip the opamp in most cases.

When this thread has run its course, I wish we had a synopsis post :

A) Adding more diodes in series does this :

B) Changing parallel resistor does this :

C) Increasing current through the diodes does this :

Quote from: Rob Strand on November 21, 2020, 06:44:23 AM

QuotePA : Why have 2 LEDS in series in PLEXITONE ?

My guess is since the non-inverting stage blends a bit of clean putting in two LEDs waters down that clean component.   The whole clipping stage is setting itself up for  a harder-clipped sound (but not necessarily nasty, maybe less hair at low drive).     The reason they can get away with two LEDs in series is because they have up'ed the supply voltage.    Two LEDs on a 9V set-up will clip the opamp in most cases.

Maybe 2 LEDS is the cause and higher voltage requirements is the effect

They decided on 2 LEDS in series because of ...........

But the Opamp will clip with 9V supply and 2 series LEDS in the loop

Hence they needed to up the supply voltage

Quote from: Vivek on November 21, 2020, 06:51:24 AM
When this thread has run its course, I wish we had a synopsis post :

A) Adding more diodes in series does this :

make the amount of clean, compared with dirty, smaller (harder clipper)

Quote
B) Changing parallel resistor does this :

change the gain, add more clean input to the output

Quote
C) Increasing current through the diodes does this :

make the knee harder

Quote from: iainpunk on November 21, 2020, 11:01:08 AM

Quote from: Vivek on November 21, 2020, 06:51:24 AM
When this thread has run its course, I wish we had a synopsis post :


B) Changing parallel resistor does this :

Quotechange the gain, add more clean input to the output

I should have said :

B) increasing the parallel resistor does this :

Quote from: Rob Strand on November 21, 2020, 06:44:23 AM

QuotePA : Why have 2 LEDS in series in PLEXITONE ?

My guess is since the non-inverting stage blends a bit of clean putting in two LEDs waters down that clean component.   The whole clipping stage is setting itself up for  a harder-clipped sound (but not necessarily nasty, maybe less hair at low drive).     The reason they can get away with two LEDs in series is because they have up'ed the supply voltage.    Two LEDs on a 9V set-up will clip the opamp in most cases.

Higher treshold means less clean component and indeed this reflects in harder clipping character.
Less "softer than amp drive" character of for example the Tube Screamer. let's say IMHO low clipping tresholds in overdrives only work as opamp clipping protection in a pedal that is used as booster and not as main drive tone. Most valued overdrives have one thing in common : higher treshold. The Plexitone is in fact a really good example and sounds great with single coils to get that snappy strat-in-amp break-up timbre without clipping the opamp (which happens almost all the time in diode to ground clippers).

Thanks !!!

I know somebody had posted the math behind "More diodes means less of clean signal is added with the clipped signal"

Could you please post it again.

Quote from: Vivek on November 21, 2020, 12:10:27 PM

Quote from: iainpunk on November 21, 2020, 11:01:08 AM

Quote from: Vivek on November 21, 2020, 06:51:24 AM
When this thread has run its course, I wish we had a synopsis post :


B) Changing parallel resistor does this :

Quotechange the gain, add more clean input to the output

I should have said :

B) increasing the parallel resistor does this :

higher gain, the knee doesn't change, but the amount of clean sticking out the top is more.

cheers, Iain

Quote from: Vivek on November 21, 2020, 01:34:27 PM
Thanks !!!

I know somebody had posted the math behind "More diodes means less of clean signal is added with the clipped signal"

Could you please post it again.

can i ''visualize'' it for you, without the numbers?
if the clipping threshold is surpassed, the gain is 1, AKA you add clean signal
if the clipping threshold is 0,6V and the gain is extremely high, there is basically a addition of the 0,6V square wave to the clean signal.
if the clipping threshold is 2V and the gain is extremely high, there is basically a addition of the 2V square wave to the clean signal.
in comparison, there is less clean on the 2V square than the 0,6V square because the gain after the threshold is 1. that's why i don't recommend putting a signal larger than 1/3 the clipping threshold into soft clipping high gain, because it will sound like clean with a bit of distortion going on in the background, instead of real soft clipping like in tubes. this is also why i don't think its wise to use compliance resistors in soft clipping, it just makes it worse.

cheers, Iain

Quote from: iainpunk on November 21, 2020, 04:19:40 PM

Quote from: Vivek on November 21, 2020, 01:34:27 PM
Thanks !!!

I know somebody had posted the math behind "More diodes means less of clean signal is added with the clipped signal"

Could you please post it again.

can i ''visualize'' it for you, without the numbers?
if the clipping threshold is surpassed, the gain is 1, AKA you add clean signal
if the clipping threshold is 0,6V and the gain is extremely high, there is basically a addition of the 0,6V square wave to the clean signal.
if the clipping threshold is 2V and the gain is extremely high, there is basically a addition of the 2V square wave to the clean signal.
in comparison, there is less clean on the 2V square than the 0,6V square because the gain after the threshold is 1. that's why i don't recommend putting a signal larger than 1/3 the clipping threshold into soft clipping high gain, because it will sound like clean with a bit of distortion going on in the background, instead of real soft clipping like in tubes. this is also why i don't think its wise to use compliance resistors in soft clipping, it just makes it worse.

cheers, Iain

Unless you add clipping in front of the opamp stage, which sets the maximum amplitude added on top of the treshold and effecticely acting as a top clipping element. it makes the stage "immune" to heavy signals

So, given all the discussion here

What is the best way of making an Amp in a Box with 3 or 4 clipping stages ?

what should each stage be ?

Quote from: Vivek on November 22, 2020, 06:00:19 AM
So, given all the discussion here

What is the best way of making an Amp in a Box with 3 or 4 clipping stages ?

what should each stage be ?

Probably you refer to gainy type? not shiny fender in a box?
I suggest 2 asymmetrical clipping stages with capacitive coupling followed by tone control and finally a symmetrical clipping stage.

Quote from: Vivek on November 22, 2020, 06:00:19 AM
So, given all the discussion here

What is the best way of making an Amp in a Box with 3 or 4 clipping stages ?

what should each stage be ?

2 simple jfet stages (biased at 6V for asymetry) and one jfet/mosfet cascode at the end (with high feedback and biased to be symmetrical).
(i dislike feedback clipping opamps, a lot)

cheers, Iain

Quote from: Vivek on November 22, 2020, 06:00:19 AM
So, given all the discussion here

What is the best way of making an Amp in a Box with 3 or 4 clipping stages ?

what should each stage be ?

If there was a single best way to build any specific type of pedal, this forum would be a lot less interesting. It totally depends on what sound you are after. And even then there are probably several equally good and equally simple/complicated topologies that get you there. I for one like diodes in the NFB path of an inverting opamps better than a non inverting because I find it easier to tune the clipping characteristics. Matter of taste and personal journey, though. I also rarely find more than two clipping stages necessary, unless I am looking for extreme or experimental sound, which is probably not the case for an amp in a box type of pedal.

Well, that wasn't helpful at all, was it? Sorry. Anyhow, the best way for you is your way, grasshopper.

Cheers,
Andy

From all above

and from that famous old article

https://web.archive.org/web/20190831161652/http://www.bteaudio.com/articles/TSS/TSS.html

I understood : The output waveform of the noninverting clipper consists of two components:
(1) the amplified and clipped version of the input waveform
(2) plus the unamplified input waveform.


I also feel compliance resistors allow part of original signal to go through


My question is

What is the difference between :

A) noninverting clipper with no compliance resistors
and
B) Inverting clipper with compliance resistors in series with the diodes

QuoteMy question is

What is the difference between :

A) noninverting clipper with no compliance resistors
and
B) Inverting clipper with compliance resistors in series with the diodes

they can sound exactly the same, but the inverting stage can clip hard, and the non inverting stage can't.
the systems both respond differently to capacitor tone shaping,
the inverting one can go down to -inf dB where the non inverting can only go down to 0dB
i prefer noninverting due to easier math... you can throw almost anything in the nfb loop, calculate the response and mirror it around the 0db axis. the math is harder with inverting stages imho.

cheers, Iain

Quote from: Fancy Lime on November 22, 2020, 02:17:06 PM

I for one like diodes in the NFB path of an inverting opamps better than a non inverting because I find it easier to tune the clipping characteristics.

Quote from: iainpunk on November 23, 2020, 11:57:00 AM
i prefer noninverting due to easier math. imho.

cheers, Iain

I'm lost !!!!!!

its both valid, and up to personal preference,
he prefers noninv. for tuning the specific clipping sounds, like knee and hardness.
i prefer inv. for easier tuning the freq. response, and i care less (not at all) about the specifics of clipping, like knee and hardness.
i think his method is more usefull for you, since it allows for really fine tuning and perfecting the clipping itself.

cheers

Hmm, funny. I find inverting stages easier math-wise. Well, neither kind is rocket surgery. But for me, the ability to go below 0db is often useful when cramming filter functionality into clipping stages. One big caveat, though: if you want now noise and high input impedance, you need a non-inverting opamps stage. So the very first stage should usually be non-inverting to lift the signal out of the noise floor. For an amp in a box, I would make the first stage something similar to a tube screamer but with a higher corner frequency, maybe 1.5kHz, LEDs, possibly asymmetrical, and probably a bit less gain. Next a bass control and maybe mid control, then an inverting clipping stage and after that the treble control. At the very end I would stick a very primitive pseudo-speakersim in the form of a two to four pole Butterworth(ish) low pass filter at somewhere around 6kHz ( adjust to taste). Remember, this is just where I personally would start tinkering but there is more than one way to pickle a herring. You do you! We'll TRY to help, of course, but reading my own text here I wonder if that is really all that helpful...

Cheers,
Andy

Quote from: Vivek on November 23, 2020, 01:31:05 PM

Quote from: Fancy Lime on November 22, 2020, 02:17:06 PM

I for one like diodes in the NFB path of an inverting opamps better than a non inverting because I find it easier to tune the clipping characteristics.

Quote from: iainpunk on November 23, 2020, 11:57:00 AM
i prefer noninverting due to easier math. imho.

cheers, Iain

I'm lost !!!!!!

There is no spoon mate ;)

Quote from: Steben on November 27, 2020, 02:59:51 PM

Quote from: Vivek on November 23, 2020, 01:31:05 PM

Quote from: Fancy Lime on November 22, 2020, 02:17:06 PM

I for one like diodes in the NFB path of an inverting opamps better than a non inverting because I find it easier to tune the clipping characteristics.

Quote from: iainpunk on November 23, 2020, 11:57:00 AM
i prefer noninverting due to easier math. imho.

cheers, Iain

I'm lost !!!!!!

There is no spoon mate ;)

True!

I thought some more about my and Ian's different perceptions on this.  I think, which is easier depends on whether you think in volts or in amperes. Non-inverting is easier to reason about in voltages whereas inverting is easy in current. As long as there are only resistors involved, that does not make a real difference. But with diodes, one or the other may be easier to wrap your head around, depending on how you are wired.

Cheers,
Andy

QuoteHmm, funny. I find inverting stages easier math-wise.

well, i think when you use 'hard number math' its kinda equal, but when you are in to 'freestyle-interpretive math' non-inverting is easier because you can just ''flip over'' the response of what you put in the feedback loop.
put a mid scoop in the NFB? you get a mid boost and the frequency's above and below that still have a gain of 1!
put in crossover diodes? you get clipping!
put in clipping? you get gating! *
put in a mid boost? you get a mid scoop!
dislike the big muff midscoop, but like the functionality? put it in a NFB!!!!!!! **
the only thing you have to keep in mind is phase margins and such, but phase margins also apply to inverting stages.

QuoteBut for me, the ability to go below 0db is often useful when cramming filter functionality into clipping stages.

you can as well in a non inverting gain stage, but it needs another active element to get that to work, and the phase margins may become a bit sketchy...

QuoteRemember, this is just where I personally would start tinkering but there is more than one way to pickle a herring. You do you! We'll TRY to help, of course, but reading my own text here I wonder if that is really all that helpful...

this is exactly the message Vivek needs to hear.

cheers, Iain

* not clean gating, but what i'd like to call ''soft dirty gating'' the gain under the threshold is 1 and there is crossover distortion at play.
** make sure that the pot's range is limited on the treble boost side, use a 5k6 resistor with a 100k pot. this keeps the treble boost shelved. omitting the resistor makes the op amp try to boost the high frequency's WAY to much (near infinite)

Thanks everyone

Learnt a lot from this thread, and still learning !!!

Quote from: Vivek on November 28, 2020, 01:39:15 PM
Thanks everyone

Learnt a lot from this thread, and still learning !!!

you're welcome! i enjoy these kinds of threads the most tbh. low pressure and just throwing up some ideas, opinions and thoughts.

cheers, Iain

Quote from: Vivek on November 23, 2020, 11:41:59 AM
and from that famous old article
https://web.archive.org/web/20190831161652/http://www.bteaudio.com/articles/TSS/TSS.html

Sorry to resurrect; I felt I'd get some value out of this but I always worry when I see the only working reference being on webarchive; And given that the article is "famous", I took a moment to make 3 PDF documents out of the link/references for my own use... (and please let me know if this is an inappropriate offer on the forum) ...would be happy to make them available.

Hi.  My memory is a bit rubbish...but 40 years of experience says...
YES the knee is similar in shape (I vs V) for EVERYTHING from a Schottky diode clipping at 0.3 V to an LED clipping at 3V.
The difference in tone is 99% due to the V threshold for conduction.  At a nominal gain (say 10), a 0.1V input signal is lipped to blazes with a Schottky, not at all with an LED. 
End of story....fairly much...
I put up a post with data about this about 15 years ago...
There was a TINY exception with a 1N914/1N4148, with some early conduction, possibly due to the metal (gold?) doping of the junction.
Sorry if I'm repeating what was said above.  I'm not up for wading through tonnes of muddy thinking to find a few pearls of wisdom.  Lazy old me!
Cheers, and happy building!  Never give up.

TLDR

I am trying to say that

If two different clippers are

A) Standardised for clipping percentage
( for example, 6 Vp signal clipped with diode of Vf 0.6V and 25Vp signal clipped with diode of Vf 2.5V)

and

B) The output is standardised for volume
(both above clippers adjusted to have 1Vp final output)


Then the differences in most diodes is negligible.

I recently saw a pedal that called the switch between 1N4148 and LED clipping "mid boost" and almost wet myself laughing while watching the YouTube reviews waxing poetic about how effectively it boosts the mids. So I say, the only thing that really matters, is how you sell or label it.
Kidding aside, yes when normalized to gain, most diodes sound the same. You have to design for "riding the knee" by scaling current, if that is what you want. One exception worth mentioning, apart from the slightly softer knee on 1N4148, is reverse breakdown of low voltage Zeners. These are much softer, which can be useful.

Andy

For those who are looking for a diode drop somewhere between silicon and red LED, there are infrared LED's available and they are included in any optoisolator.  This raises the possibility that you could use the LED in an optoisolator as a circuit element as the diode and use the phototransistor output to drive a LED to show when clipping starts.  LED's are sensitive to ambient light - if you mount one on the outside of the enclosure, it will have different characteristics in high and low ambient light and it will be modulated by AC lighting, causing powerline hum even if you are running from a regulated supply.

Quote from: brett on December 06, 2022, 05:19:46 AM
Hi.  My memory is a bit rubbish...but 40 years of experience says...
YES the knee is similar in shape (I vs V) for EVERYTHING from a Schottky diode clipping at 0.3 V to an LED clipping at 3V.
The difference in tone is 99% due to the V threshold for conduction.  At a nominal gain (say 10), a 0.1V input signal is lipped to blazes with a Schottky, not at all with an LED. 
End of story....fairly much...
I put up a post with data about this about 15 years ago...
There was a TINY exception with a 1N914/1N4148, with some early conduction, possibly due to the metal (gold?) doping of the junction.
Sorry if I'm repeating what was said above.  I'm not up for wading through tonnes of muddy thinking to find a few pearls of wisdom.  Lazy old me!
Cheers, and happy building!  Never give up.

How they sound is a bit personal. I mean what some call a difference some call a small difference and some no difference.
Both are true: most diode behaviour follows the same Shockley law. But, the internal resistance is neglected in this law, like comparing 2 germanium diodes vs a silicon diode.
And stacking diodes raises the combined resistance. The lower the impedance of the signal, the more important this is.
But since a diode is an element following shockley's law with an internal resistance, diode graphs can be mutually mimicked with combinations of .... diodes and resistors.

If you connect a diode in a feedback loop of an op amp, you can get an ideal diode that can be used as a clamp circuit that stops the circuit from going in a selected direction, either positive or negative.  The diode goes from the output to the inverting input and the clamp voltage is set by the voltage at the non-inverting input.  You can put a resistor in series with this to simulate diode resistance.  The circuit input comes through a resistor connected to the inverting input and the output is taken from the inverting input, making it a clamp circuit that prevents the voltage from exceeding the clamp voltage.  By varying the clamp voltage, you can set the forward voltage and the resistor simulates the diode resistance.

(https://i.postimg.cc/kMvxgRKb/ideal-clamp.jpg)

The schematic shows specific devices and voltages, but the op amp, diode and operating voltages can be changed.

Quote from: iainpunk on November 23, 2020, 11:57:00 AM
but the inverting stage can clip hard, and the non inverting stage can't.

It can when the opamp hits the rails ;) And it is one of the pitfalls of "soft" clippers. Running large signals into non-inverting clipper circuits ruin a lot. While hard clippers retain their character.
The ratio of input signal to soft clipping treshold makes for a great sounding or meh sounding device.

But as far as diodes go you are right of course.
Inverting stages can "tailor" the clipping points, just as a passive clipper to ground can. It is like making a clipper "active" just as in tone control circuits.
Non-inverting stages can bring in the unity gain clip limitation. Which can be useful. But as I said, adding hard clipping before a soft clipper helps to retain the character.

Do not forget the plexi holy grail of (hard) rock sound is mainly a hard clipped negative feedback power amp. It is why it is compared to other sounds so easy to simulate.