SOLDANO SLO preamp stage analysis in question

[ltournell]

Hello,​

While reading Richard Kuehnel's site AmpBooks I was interested in his analysis of the Soldano SLO in the classic circuits section. As I had already exchanged with him , I tried to contact him again with the following remark and I would like to have your opinion two, you all bearded techs )

So, to sum up this is what I wrote to the [email protected] :

With the hope that my following question can reach Richard Kuehnel !

In your analysis of stage 3 of the Soldano Super Lead Overdrive on the AmpBooks website I encounter a difficulty in understanding your point here :

From the diagram we can see that even with 400 V plate-to-cathode voltage the maximum headroom peak-to-peak that you can reach with a 12AX7 is 5 V.
So how can we speak of a 4.2 V peak headroom considerably greater etc.... ?

In my humble opinion the maximum headroom ( for clean signals by definition) is 2.5 V peak.

What I understand , studying this stage, is that it is designed to unbalance the signal limiting it to the positive swing , the negative swing being truncated with the 0.3 V left to the cut-off limit. The effect, for me, is to restore a good part of the overtones with EVEN harmonics that the previous stages have considerably reduced. All the other preamp stages show a 1k8 cathode resistor which produces a symmetric amplification : with a gain pushed hard we get square waves whose symmetry considerably attenuates the EVEN harmonics , giving way to the (also desired) ODD harmonics. It is the subtle combination of those two types of harmonics that give this creamy overdrive and make the Soldano a really super amp .

Yours truly.

 

[2L man]

That 5V is grid voltage sweep. You can read anode voltage sweep on horizontal line.

This link tool is great when testing loadlines and good for studying as well. When loadline get screwed bad just start again but for that another tube type needs to be selected and then going back to actual tube.

Headrom what you can input is anode voltage sweep and comes to the left of Operating (bias) Point. What comes to right side is a result of default linear grid drive so you immediately see is it linear or is there 2nd harmonic distortion which shows as numetic value as well.

Bacically grid voltage sweep change anode current which change the voltage loss on anode resistor and there comes the amplifying.

Universal Loadline Calculator

This calculator allows you to trace the anode characteristics graph, and to compute and tune loadline and operating point of various vacuum tubes.

www.vtadiy.com

 

[YellowBoots]

That looks like the good 'ol cold-clipper to me. Your question is a good one. On first glance this result does seem to defy logic.

First thing, you need to consider the difference of an stage without a cathode bypass capacitor (unbypassed) versus a stage with a cathode bypass capacitor (bypassed). A sufficiently bypassed cathode will stay at a constant voltage as the signal is being amplified. An unbypassed cathode will change voltage as signal is being amplified. In fact, the unbypassed cathode will go from 0V at cut-off (because no current it flowing) to a pretty high voltage at the grid-current limit (take the current and multiply it by the cathode resistor to get the voltage). This cold-clipper is unbypassed.

Second thing, imagine the stage is at cut-off. We know the voltage on the cathode at cut-off of an unbypassed stage is 0V. (No current passing through the cathode resistor means no voltage drop by the cathode resistor.) But looking at the load line, we also know that Vgk (voltage measured from grid to cathode) is -4.2V. It's that simple--the signal has to be -4.2V at cut-off. For an unbypassed stage, it doesn't matter where the bias lands, the negative signal voltage it takes to drive the tube to cut-off clipping is always the Vgk where the AC load line hits the x-axis (zero current).

Third, you can estimate the positive signal voltage that will get you to grid-current clipping by also examining the load line. Assuming grid-current starts at Vgk = 0V, we see the current should be about 2.8mA. This current is passing through the 39k cathode resistor, so by Ohm's Law we know the cathode voltage is 0.0028A x 39,000Ω = 109V! If the cathode is at 109V and the grid-to-cathode voltage is 0V, then the grid voltage must also be 109V! That is a ton of positive side headroom, and that's the result of a large value (unbypassed) cathode resistor. Obviously, you hit the -4.2V negative swing limit before you hit the 109V positive swing limit.

Kuehnel is right. The clean input headroom of the stage is theoretically about 4.2Vpeak.

 

[ltournell]

That 5V is grid voltage sweep. You can read anode voltage sweep on horizontal line.

This link tool is great when testing loadlines and good for studying as well. When loadline get screwed bad just start again but for that another tube type needs to be selected and then going back to actual tube.

Headrom what you can input is anode voltage sweep and comes to the left of Operating (bias) Point. What comes to right side is a result of default linear grid drive so you immediately see is it linear or is there 2nd harmonic distortion which shows as numetic value as well.

Bacically grid voltage sweep change anode current which change the voltage loss on anode resistor and there comes the amplifying.

Universal Loadline Calculator

This calculator allows you to trace the anode characteristics graph, and to compute and tune loadline and operating point of various vacuum tubes.

www.vtadiy.com

Thank you for the sympathetic tool but to draw loadlines my favorite tool is ... a pencil ! (color ones preferably)

On this simple diagram you can immediately check the clean headroom you get in the output , depending on the cathode resistor. You can see that the clean headroom with a 39k resistor does not go beyond 0.3 V.

 

[ltournell]

That looks like the good 'ol cold-clipper to me. Your question is a good one. On first glance this result does seem to defy logic.

First thing, you need to consider the difference of an stage without a cathode bypass capacitor (unbypassed) versus a stage with a cathode bypass capacitor (bypassed). A sufficiently bypassed cathode will stay at a constant voltage as the signal is being amplified. An unbypassed cathode will change voltage as signal is being amplified. In fact, the unbypassed cathode will go from 0V at cut-off (because no current it flowing) to a pretty high voltage at the grid-current limit (take the current and multiply it by the cathode resistor to get the voltage). This cold-clipper is unbypassed.

Second thing, imagine the stage is at cut-off. We know the voltage on the cathode at cut-off of an unbypassed stage is 0V. (No current passing through the cathode resistor means no voltage drop by the cathode resistor.) But looking at the load line, we also know that Vgk (voltage measured from grid to cathode) is -4.2V. It's that simple--the signal has to be -4.2V at cut-off. For an unbypassed stage, it doesn't matter where the bias lands, the negative signal voltage it takes to drive the tube to cut-off clipping is always the Vgk where the AC load line hits the x-axis (zero current).

Third, you can estimate the positive signal voltage that will get you to grid-current clipping by also examining the load line. Assuming grid-current starts at Vgk = 0V, we see the current should be about 2.8mA. This current is passing through the 39k cathode resistor, so by Ohm's Law we know the cathode voltage is 0.0028A x 39,000Ω = 109V! If the cathode is at 109V and the grid-to-cathode voltage is 0V, then the grid voltage must also be 109V! That is a ton of positive side headroom, and that's the result of a large value (unbypassed) cathode resistor. Obviously, you hit the -4.2V negative swing limit before you hit the 109V positive swing limit.

Thank you for your brushing up of the good old clipper.

But my question to R.Kuehnel lies beyond :

Kuehnel is right. The clean input headroom of the stage is theoretically about 4.2Vpeak.

No. I disagree. If we stick to the headroom definition it is the maximum (clean) input voltage that produces a clean output. What else can it be ? I understand there is a margin of 4.2 V on the input but what the use of it ? When you speak about 4.2 V peak headroom , you are supposed to treat a 8.4 V peak-to-peak signal, which is obviously not the case here.

Besides I invite you to read carefully what Ric.Kuehnel writes (up here) :

Input headroom is therefore 4.2V peak (+9.5dBV). This is considerably greater than for the first two stages, despite a DC operating point very close to cutoff.

All other stages use 1k8 cathode resistors ! Those values with 1k5 offer the highest headroom for clean output ! What is the point in comparing this true headroom with the supposed 4.2 V peak ? He adds " despite the DC operating point very close to cut off" ! as if both biassing methods lead to comparable results ! To me it is as if a famous biology scientist undertook to compare a rabbit with a goat and were satisfied with concluding that the goat is bigger than the rabbit .

Obviously the explanation should extend to the production of dissymmetric signals designed to produce overtones with harmonics of even order combined to the ones of odd order produced by overdriving the "symmetrical" stages. To me if one fails to understand that , one fails to understand the philosophy of the SOLDANO preamp and the reason of its success.

 

[YellowBoots]

Are we talking about input headroom or the type of clipping this stage produces?

Kuehnel’s point in comparing this stage to the previous ones is that you cannot determine input headroom just by looking at the bias point for an unbypassed gain stage. It is biased extremely cold, yet it has more input headroom than a fully-bypassed center-biased gain stage. Don’t you find that fascinating?

Yes, the clean input headroom is 4.2Vp or 8.4Vp-p however you want to slice it.

I direct you to this link:
http://valvewizard.co.uk/Common_Gain_Stage.pdf
Skip right to section 1.18 (pg 25) and understand the first paragraph there.

 

[2L man]

This attachment has 12AX7, 300V, 100k, 2k,. 0.90mA loadline which took less than a minute to set. I have drawn many loadlines to copies but don't have functioning printer so I use ULLC and store them.

 

[ltournell]

@andrewRneumann

Are we talking about input headroom or the type of clipping this stage produces?

I think we are talking about Richard Kuehnel's analysis of the 3rd stage of the SOLDANO SLO. So we speak of both, knowing that both aspects are technically linked of course.

Kuehnel’s point in comparing this stage to the previous ones is that you cannot determine input headroom just by looking at the bias point

We will not understand each other because we do not mean the same thing with input headroom. From my point of view, YES we can see the input headroom just by looking at the bias point.

the bias point for an unbypassed gain stage.

Bypassed or unbypassed the bias point IS THE SAME ! The bypass capacitor with a correct dimension only cancels the negative feedback effect of the cathode resistor ( more exactly there is a new feedback network with Rk // Ck)

See below : at the guitar input of the SOLDANO SLO100 you input a 50 mV peak ( ) sine that is 100 mV pp. Set the overdrive gain to 5 (on a log pot it means 10 % of the signal)

At the grid of the cold clipper you get 20 V p-p (LTSpice simulation) :

A the output after the coupling cap you get this :

Now if I bypass the 39k resistor with say a 10 µF capacitor, I will only increase the gain and get this :

On the SOLDANO there is no bypass capacitor only because it is useless, the gain being sufficient for clipping even with low level signals at the input.

I have made a few preamp prototypes with various versions of this clipper : unbypassed, then partly and fully bypassed : the choice was always made according to the gain I expected at the output knowing that the bias point was always doing the job.

It is biased extremely cold, yet it has more input headroom than a fully-bypassed center-biased gain stage. Don’t you find that fascinating?

Not at all. It does not make sense to me.

I direct you to this link:
http://valvewizard.co.uk/Common_Gain_Stage.pdf
Skip right to section 1.18 (pg 25) and understand the first paragraph there.

I see you have good readings. I have exchanged with the author ( Merlin Blencowe) a lot between 2005 and 2007 together with Randall Aiken. At the time of the first edition that he had released too fast I think there was about one mistake every two pages ! This second edition is far, far better. You can be sure that Merlin Blencowe and I have the same approach to guitar amps, except perhaps that he favors mainly EVEN harmonics in overdriven signals, and that I like them to be mixed.

By the way read again the paragraph you point out to me to find the reason why " bypassed or unbypassed the bias point is the same" but explained by Merlin Blencowe himself.

 

[YellowBoots]

We will not understand each other because we do not mean the same thing with input headroom.

I think we agree on what input headroom is. It's the maximum input signal (expressed in volts peak, peak-peak, or even dBV) that still results in clean output from the stage.

Bypassed or unbypassed the bias point IS THE SAME !

I never said it wasn't. The bias point in this circuit is Vgk = -4.0V with a Ia of about 0.1mA. That is the same whether there is a bypass cap or not. What changes with the addition or subtraction of the bypass cap is the headroom, not the bias point.

You seem to be very passionate about your position on the input headroom of this stage. You started this discussing why you thought Kuehnel is wrong. You did not discuss his math or where he went wrong in his formulation. I gave you another explanation of why Kuehnel is right and you insist that I am wrong, but again no discussion about where I went wrong in my reasoning. Are you just baiting the forum for an argument?

Here's a table of the math behind my reasoning. Would you be kind enough to point out where I am wrong?

	Idle	Cut-Off
Vgk (V)	-4.0	-4.2
Ia (mA)	0.1	0
Vk (to ground) (39kΩ) (V)	-4.0	0
Vg (to ground) (V)	0	-4.2

In the last row, we see that the signal on the grid swings from 0V at idle to -4.2V at cut-off. The input headroom is 4.2Vp, not 0.2Vp (or 0.3Vp). I'm not sure what else I can say to prove my point. I believe you are mistaking the Vgk swing for the input headroom. This would be correct for a fully bypassed gain stage but not for unbypassed.

I also see you have a sophisticated model of the SLO. Why don't you turn down the gain until the output of the 3rd stage is just starting to clip and then plot the input signal that caused it? Is it around 0.3Vp or around 4.2Vp?

By the way read again the paragraph you point out to me to find the reason why " bypassed or unbypassed the bias point is the same" but explained by Merlin Blencowe himself.

In the Blencowe paragraph, I was hoping you noted the final sentence: "This effect is called cathode current feedback or cathode degeneration, and as well as reducing the overall gain it also reduces distortion and increases headroom." Neither Kuehnel, Blencowe, or I have stated that the bias point changed. (It's silly to put myself in the same sentence as these two ridiculously smart people.) It is the headroom that changes, not the bias point. Possibly you are thinking that the only way to change the input headroom is to change the bias point?

Hopefully this helps you understand my point of view. I'm prepared to change my opinion if you can give me evidence that I am wrong, but I haven't seen any yet.

 

[ltournell]

Ok Andy, I certainly did not want to have an argument )

I think we agree on what input headroom is. It's the maximum input signal (expressed in volts peak, peak-peak, or even dBV) that still results in clean output from the stage.

The problem of our misunderstanding may just be a problem of language. You are American, Blencowe is English , and I am French. In the French language we use the English words to name those technical concepts because we do not have defined a precise vocabulary for that field. For instance we use the word "headroom" only with this meaning (it gives a clean output) . Reading all of you I am forced to admit that I am wrong : headroom can mean any kind of margin, gap , range or whatever lies between two limits. You are the only one who can tell me if I am right till there ! This said , I cannot still maintain that Kuehnel or you are wrong when using this word.
Now as to Blencowe it is different . In the paragraph that you point out he uses headroom with the meaning I expected , that is why I did not react : what increases headroom is creating cathode feedback by unbypassing the cathode resistor , thus producing a reduced gain and decreasing the relation between output and input so you can input a higher level without overdriving the output.

Now I suggest we start it all again step by step. Be sure that I totally agree with the math behind it all , this is why I did not contest Kuehnel's figures but only his interpretation of these and especially the use of this particular stage that he surprisingly does not deal with.

Now please , give me time to study more carefully your own point of view and make the most relevant comments if any.

 

[ltournell]

@andrewRneumann :

I have had the time to think all this over and the overall conclusion is you are right. Nevertheless my question to Richard Kuehnel remains open but should be amended.

Kuehnel’s point in comparing this stage to the previous ones is that you cannot determine input headroom just by looking at the bias point for an unbypassed gain stage

I explain now how I could misinterpret your statement :
logically if you underline unbypassed I can understand this : you CAN determine input headroom just by looking at the bias point for a bypassed gain stage. So I had two solutions : either you mean the input headroom is different or the bias point is different . As in my mind I see the load-line diagram with its fixed Vgk curves I thought you meant the bias point is different. And there I admit that I was wrong due to a bad habit of mine : I make very often use of simulators that I build myself (using LTSpice software) and thus reason too much from left to right , that is from input to output . I must say that your approach is a better one, it is the one of the amp designer who thinks from output to input and obviously you are right , it is the headroom that is different and not the bias point. Thanks for correcting in me that bad habit !

Now the math behind it all :

Here's a table of the math behind my reasoning. Would you be kind enough to point out where I am wrong?

	Idle	Cut-Off
Vgk (V)	-4.0 OK	-4.2 OK
Ia (mA)	0.1 OK	0 OK
Vk (to ground) (39kΩ) (V)	-4.0 NOT OK : + 4.0	0 OK
Vg (to ground) (V)	0 OK	-4.2 OK

There is a slight notation mistake but you are right all the same . The input headroom is actually 4.2 Vp (8.4 Vpp) and not 0.3 V as I thought .

I also see you have a sophisticated model of the SLO. Why don't you turn down the gain until the output of the 3rd stage is just starting to clip and then plot the input signal that caused it? Is it around 0.3Vp or around 4.2Vp?

Excellent idea now . As a "sophisticated" model I simply use the GNU free source LTSpice simulator and draw the OD channel like this :

I input a low level signal of 50 mVp ( to give an idea it is the attack level of the 1st string of a low level single coil PU - from a Strat for example) and we draw the response between input (the grid) and output of stage 3 after C6

1 . First we start with a minimum of gain : one half of division on the OD gain (0.5) :

As expected we get a perfect sine at that output well balanced around 0 V : ( input = green / output = yellow)

2. We push the OD gain to 1 to reach 8 Vpp at the input : we are not yet reaching the headroom as shown above by your calculations but drawing near , the output begins to appear unbalanced. If we refer to the load-line diagram we see that the Vgk curves under the bias point pile up and a strong compression occurs progressively so the real cut-off can be expected a little beyond the theoretical value of 4,2Vp.

3 Pushing the gain to 2.25 then to 2.5 we pass the cut-off limit with respectively 4.5 Vp and 5 Vp :

At 5V p the clipping is now clearly visible.

The spectrum analysis of these signals (a little fussy to represent here) show that harmonics of overtones progressively appear as we draw near the headroom limit.

So thank you for your relevant remarks which helped me compensate my own lack of perspective.

Now I think that my (awkward) question to Richard Kuehnel remains open. I still do not understand why he stopped his analysis there with the simple statement that there was in this stage a large headroom. What about the function of this clipping stage generating even harmonics among all other stages producing odd harmonics when overdriven (which is the general situation in this OD channel) ?

Strange all the more so that he is a reliable author (when compared to others) . I have bought his book "Guitar amplifier power amps" that I keep as a constant reference.

 

[YellowBoots]

headroom can mean any kind of margin, gap , range or whatever lies between two limits. You are the only one who can tell me if I am right till there !

I think what you wrote is a good way to describe headroom. When I think of input headroom, I think of the maximum clean input signal that results in a clean output signal. Maybe there is a technical definition that specifies a maximum distortion that can still be considered "clean", but I don't know what it is. (5%?) There is also output headroom which is the maximum clean output swing of the stage. This is just the input headroom multiplied by gain, taking into account the loading of the stage.

In the paragraph that you point out he uses headroom with the meaning I expected , that is why I did not react : what increases headroom is creating cathode feedback by unbypassing the cathode resistor , thus producing a reduced gain and decreasing the relation between output and input so you can input a higher level without overdriving the output.

I prefer to keep headroom and gain as separate technical terms. Take for instance a standard 12AX7 gain stage (bypassed) and change the plate resistor and the cathode bias resistor. Let's reduce the plate resistor to 50k and decrease the bias resistor so that the bias voltage is the same. What we are left with is a stage that has the same input headroom, but less gain. I give this as an example of how the headroom can stay the same, but the gain can change.

Many times, they are changing together though. If we increase the loading (steepen the AC load line), you will find the gain reduced and a small reduction in headroom. Less headroom and less gain, at the same time. So to sum up, what I think you've realized, is that "less headroom" does not mean "more gain."

Possibly the confusion lies in the word "gain"? There are some who think "gain" means the "amount of overdrive". To them, "more gain" means "more clipped." I'm not sure that is what you were thinking, but to me gain is simply the ratio of the output signal to the input signal.

Thanks for correcting in me that bad habit !

You're welcome. I'm glad this conversation has been fruitful for you.

There is a slight notation mistake but you are right all the same .

Good catch! You are correct. It shows you understand what is going on by noticing that negative sign didn't make sense. Thanks for pointing that out.

the output begins to appear unbalanced. If we refer to the load-line diagram we see that the Vgk curves under the bias point pile up and a strong compression occurs progressively so the real cut-off can be expected a little beyond the theoretical value of 4,2Vp.

Just a little side note here--all that cathode degeneration is negative feedback. This has the effect of reducing distortion (until clipping is reached.) So the final affect of the "bunching" of Vgk curves should actually be less for an unbypassed stage than a bypassed stage. I'm not 100% certain on this because you plots do show a large difference in peak voltages. If you can do a harmonic analysis you might be able to see what I am talking about. What is the harmonic content with this stage on the verge of clipping vs. the same stage with a bypass cap on the verge of clipping? (Obviously the bypassed stage will need a much smaller signal, but we are now interested in the harmonics, not the headroom.) The theory says the unbypassed stage should be more linear.

I still do not understand why he stopped his analysis there with the simple statement that there was in this stage a large headroom. What about the function of this clipping stage generating even harmonics among all other stages producing odd harmonics when overdriven (which is the general situation in this OD channel) ?

I guess that is author's discretion. I believe it is the 4th stage (cathode follower) that clips first, so the effect of this stage comes in after that one. The effect of a clipped stage can be complicated by a further stage that is already clipping. What do your models show? What is the 4th stage doing when the 3rd stage starts to clip? Can you even see the 3rd stage clipping in the output of the 4th stage?

I have found Kuehnel's writing to be valuable. Fundamentals of Guitar Amplifier System Design is very helpful. He does seem to be more focused on the electrical aspects than the tone aspects. I suspect that is because he's a technical writer and a lot of "tone talk" is subjective. The other harsh reality is that it is very hard to "see" tone in a model or an oscilloscope. There are just so many non-linearities and dynamics going on in a real amp played by with a real guitar that making guesses about tone from a model is subject to much uncertainty. He is probably sticking to the facts he can say with near certainty... and leaving the "tone talk" to us!

 

[peteb]

It is biased extremely cold, yet it has more input headroom than a fully-bypassed center-biased gain stage. Don’t you find that fascinating?

class B is the coldest bias, cutoff at idle, yet it supplies the most clean power.

it’s because the high bias allows the signal more room to operate cleany.

 

[ltournell]

@andrewRneumann

Thank you for your reply . It is very rich and interesting and I will take the time in particular to make snapshots of the spectral analysis to answer your questions. For the time being from what I have seen the harmonics appear relatively early in the clipping process while the shape is still an apparent sine but the average value clearly sinks below 0V but I will confirm precisely.

Gain means for me strictly the same thing as for you. I think the confusion may arise from the fact that amps frontfaces often show GAIN pots for overdrive channels while VOLUME pots are rather dedicated to clean channels (?)

As to Richard Kuehnel, you are probably right. I am currently replying to his reply ( because, as a Gentleman, he has replied to me today !) and I ask the question again from the "tone" point of view. I come back here as soon as things are clear with him ( to discuss the SOLDANO OD channel too ).

 

[2L man]

class B is the coldest bias, cutoff at idle, yet it supplies the most clean power.

In theory and only in a push pull circuit! And in theory PWM driven D-class.

You haven't bought Amato book yet

 

[peteb]

In theory and only in a push pull circuit!

it’s true If comparing to class A and class AB as we see on fender amps.

class AB lies between class A and class B. Class AB is bordered on both ends of the bias spectrum by classes A and B.

class A is less efficient than the entire class AB range, and class B is able to achieve an efficiency higher than the entire class AB range.

 

[ltournell]

@peteb & @2L man

This discussion about classes if off-topic for me since classes define power amps only. The main characteristic of the SOLDANO is that the "tone" is really built at the level of the preamp , and , as Richard Kuehnel puts it, the power part of the SOLDANO is supposed to act only as a HI-FI amplifier and thus not supposed to add anything by cranking it up. SOLDANO users find the same pleasure whether playing the 100 , the 50 or the 30 ( check on youtube).

 

[2L man]

Operating class differences can be seen in simple triode circuits and cascade triode circuits which kind of are PP but yes the side step me and Pete made does not belong here. Itournell, I am sorry and quit that path here!

"Cold clipper" is technically a B-class. There is no inductance but there is a resistor. It is the Operating Point (bias) which make the difference between A-, B- and C-class. You must have seen a SPDT switch on cathode which change bias adding a parallel resistor. Or a potentiometer/resistor parallel with actual cathode resistor.

Some guitar amp A-class stage has diode or LED clipper. I recall seeing also germanium transistor circuit which clip and obviously hoping to add interesting germanium distortion. Failed germanium transistor "junction diodes" have been used building effects.

Itournell, I believe you already know these and I admire how you like to study tube operation

 

[YellowBoots]

@peteb & @2L man

This discussion about classes if off-topic for me since classes define power amps only. The main characteristic of the SOLDANO is that the "tone" is really built at the level of the preamp , and , as Richard Kuehnel puts it, the power part of the SOLDANO is supposed to act only as a HI-FI amplifier and thus not supposed to add anything by cranking it up. SOLDANO users find the same pleasure whether playing the 100 , the 50 or the 30 ( check on youtube).

Perhaps the 39K cold-clipper’s main purpose shouldn’t be thought of a clipping the negative side of the wave. As we have discussed, any bypassed stage can be biased cold to induce negative side clipping. We also know that the negative side input headroom of an unbypassed stage will be similar no matter what the bias point is (this does depend on load). So what is the difference between 1.5K, 2.2K, 10K, and 39K? I propose that the main reason to bias it with a 39K unbypassed is to ensure the positive side never clips—even at very large input signals. (I’m sure there are tonal differences with the sound of the clipping too, but can’t comment on that due to lack of hands on experience.)

 

[GearGeek01]

As a side note, I have been drooling over the new Soldano SLO pedal... I gotta have one, LOL... ($229.99 MAP)

Soldano SLO Overdrive Pedal

Get the sound of a legendary amp on your board! The versatile Soldano SLO pedal has a three-band EQ, plus, volume and gain controls, and a Deep switch.

www.zzounds.com