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

[Fancy Lime]

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

[Fancy Lime]

what matters in the bedroom also matters on tape

hahahaha, thats a good one.
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

[iainpunk]

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.


cheers, Iain

[Steben]

I 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? 

[Steben]

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.


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.

[iainpunk]

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

[Vivek]

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 ?

[Steben]

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.

[Vivek]

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

[Steben]

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".

[Vivek]

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.

[Vivek]

Very interesting article

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

[Steben]

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.

[PRR]

...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.

[iainpunk]

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

[Vivek]

...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 ?

[Rob Strand]

Quote 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.

[Vivek]

Quote 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



Are the red lines correct or wrong ?

[Rob Strand]

Brother 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.




Here's n = 3.8,
[click to enlarge]


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


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

[Vivek]

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)