Hey everyone, I'm currently working on modding a friends danelectro daddy o that previously had a diode mod consisting of two leds on one leg of a spdt and two silicon diodes on the other with the middle leg of the switch wired to one side of the factory led location and the two diode pairs being joined to a wire and soldered to the other side of the factory led location with the other factory led in its original position. The switch has since broken and the friend gave me free rein on how to fix it. I've since installed a new spdt switch with a 1N34a mini diode that tested at about 690mv vf and a NTE110 with vf in the 360mv range to give two separate asymmetric clipping options with the factory led handling the other half. The issue I'm running into is that neither diode choice has a discernible difference from the other. When playing both positions of the switch sound identical with no noticeable change in the amount of clipping or character of clipping. Am I missing something here? I figured that big of a difference in vf would give a noticeable difference in tone.

If the diodes are actually clipping, one should at least give you more *volume* then the other. If you've only got a single diode in each position of the switch, you're only clipping half the signal, so the effect would be a lot less notable.

If the signal hitting it is already much larger than the Vf of either diode, I can imagine it might sound the same - past a certain point, clipping is clipping is clipping.

Thanks for the response! So from a schematic I've found it seems to be the diode to ground style of clipping setup, the two diodes on the switch are handling one half of the signal and the factory led is handling the other half.

Can you breadboard it? Would allow you to experiment a bit more.
Also sharing schematic would help.

Quote from: buford254 on July 05, 2024, 06:06:37 PMa 1N34a mini diode that tested at about 690mv vf

  

Quote from: antonis on July 06, 2024, 09:02:01 AM

Quote from: buford254 on July 05, 2024, 06:06:37 PMa 1N34a mini diode that tested at about 690mv vf

  

They should be around 300mV or less, right?

Quote from: antonis on July 06, 2024, 09:02:01 AM

Quote from: buford254 on July 05, 2024, 06:06:37 PMa 1N34a mini diode that tested at about 690mv vf

  

You know, actual constructive input would be way more beneficial to the community than just throwing shade for no reason.

Quote from: matopotato on July 06, 2024, 09:04:00 AM

Quote from: antonis on July 06, 2024, 09:02:01 AM

Quote from: buford254 on July 05, 2024, 06:06:37 PMa 1N34a mini diode that tested at about 690mv vf

  

They should be around 300mV or less, right?

Yeah, I'm aware this is wildly out of spec for what it's supposed to be, but the label on the parts bin I pulled it from said 1N34a mini so I was just going off of that. Main thing I'm interested in is why such a large difference in vf is netting no noticeable change in clipping.

Have you double-checked that they are in the right orientation? Also, is there a difference if they are out of the circuit? You might be able to hold the switch in the middle so that neither side makes contact to check.

Ge diodes have a large series resistance, meaning that the measured voltage drop depends strongly on the test current. A 1N34a should test around 300 mV at 1 mA but will be closer to 600 mV at 5mA. Mind you, it will likely only be specked at 1mA, so Vf at higher currents and indeed the I/V function in general of Ge diodes is a bit of a game of dice.

Secondly, there are plenty of documented cases of fake 1N34a diodes out there. Mostly relabeled Schottkys, it seems, but may include good ol' 1N4148 as well. Not that this necessarily matters for any particular application, as longvas it sounds good.

Thirdly, I have learned not to be so easily offended by Antonis' comments. His very short, or even emoji only, comments tend to come across a bit condescending but I have come to the conclusion over the years that this is not his intention but the result of sardonic humor not translating well in short-form text cirrespondences. YMMV, just my personal experience.

HTH,
Andy

Quote from: buford254 on July 08, 2024, 12:31:19 PMYeah, I'm aware this is wildly out of spec for what it's supposed to be, but the label on the parts bin I pulled it from said 1N34a mini so I was just going off of that. Main thing I'm interested in is why such a large difference in vf is netting no noticeable change in clipping.

This might explain how it works: Chapter labeled "Clipping of Both Half Cycles"
Not sure how big impact 600 vs 300 mV has. Also if leakage plays a role.
I built Boss BD-2 as kit by Aion FX from Das Musikding, with a mod (Galaxie Mod). I breadboarded to see the mod impact, and part of the whole effort was to see if I wanted to include a mod by Brian Wampler from his earlier mod books and also googlable. It meant swapping mainly Si diodes for Ge + some 4001 or similar combo.
It was a bit messy to make switches so I could A/B on the breadboard, but I felt there was a difference. Not so that one outtones the other, but a difference and I wanted both (of course). Quite messy to implement on one 4PDT switch (I think), but got it working.
About 2 weeks later, Wampler publish a clip (7:17 ->) about the BD-2 stating (among other) that his previous diode mod does not really have much audible impact, and even saying he tricked himself. So I fell for that, and might not be much of a moral to the story but I trusted his experience and thought I heard the difference. Tricked by proxy I guess.
Not I hear just about no difference when flicking the switch. The possibly tiny difference requires me aligning all the pots in the right directions, and then wanting to hear some difference...

I bought 1N34A's from my local shop. I fully trust him selling good stuff, but the one's I got had Vf in the 600mV region. I also tried leakage current using a battery and DMM. (https://www.youtube.com/watch?v=3ao0-gRYzBU). That did not either come out in favor.
So I kept a few for fun and returned the rest.
Recently I paid decent money from Banzaimusic.com and they came out at 300mV or a bit less and with a clear leakage current.
Still in my searches for how to test, what values to compare with other values, I find that even among people giving advise and making clips on how to test etc. there is not really as much consensus as I would have thought.
For me, I wanted the real deal. And I think I got it second time around. At least enough to keep me happy.
Can I hear the difference between a Ge clipping build vs some other diode with similar enough characteristics? Probably not. Can others? Probably.
So in the end it is up to you what you prefer. If it sounds fantastic, then it is a fantastic build for you and your playing. If you cannot shake the suspicion then you might need to continue searching.
I gave up on 1N34a's coming my way, until I found them, and hope to try them out in a Klon Clone. Just to "know they are the right ones" But that's me 

Quote from: Fancy Lime on July 08, 2024, 02:17:04 PMA 1N34a should test around 300 mV at 1 mA

Which is the test current for most of DMMs..
(there are some meters that apply an even smaller current - down to 400μA or so..)

Quote from: buford254 on July 08, 2024, 12:31:19 PMMain thing I'm interested in is why such a large difference in vf is netting no noticeable change in clipping.

Without a complete circuit schematic, we only could guess..

For a high gain hard clipping (diodes to GND) circuit, the only difference should be limited to amplitude..
(almost squared waveforms of different levels..)
Additionally, in case of output recovery stage, there should be no noticeable change related to various forward voltage drops..

IMHO, what might make the difference is a cap in parallel with clipping pair..
(it "smooths" clipping edges..)

P.S.
Diode pair inside NFB loop (soft clipping) is another kettle of fish..

Quote from: buford254 on July 05, 2024, 06:06:37 PMI've since installed a new spdt switch with a 1N34a mini diode that tested at about 690mv vf and a NTE110 with vf in the 360mv range to give two separate asymmetric clipping options with the factory led handling the other half. The issue I'm running into is that neither diode choice has a discernible difference from the other. When playing both positions of the switch sound identical with no noticeable change in the amount of clipping or character of clipping. Am I missing something here? I figured that big of a difference in vf would give a noticeable difference in tone.

Try testing a lower diode currents.

If the circuit operates at low currents then it makes more sense to compare the diodes at low currents.  In some circuits the diode current doesn't reach 1mA, or even 100uA.

I would like to throw in my experience that a hard clipping circuit usually adopts the tonality of the diode with the highest forward voltage. The reason it's so popular in the boutique world is because it's easy marketing and "Tube amps do it!".
The LED has a higher forward voltage than the modded diode pair. Also assuming the LED has a forward voltage of 2V, at saturation the 690 mv side of the waveform is 9.25 dB quieter. At 390 mv it is 14.89 dB quieter. At best, the LED side of the waveform is near twice as loud as the silicon side, at worst it is almost three times louder.
In all of my mods, if I want LED clipping, experimentation has told me I only need to swap one of the diodes to get near as makes no difference to the same effect as replacing both of them. In a quiet room the difference is subtle, in a band mix it is near inaudible. The only time I change both is if the pedal needs that small boost in output that changing both gives you.

Quote from: antonis on July 09, 2024, 04:17:23 AM

Quote from: Fancy Lime on July 08, 2024, 02:17:04 PMA 1N34a should test around 300 mV at 1 mA

Which is the test current for most of DMMs..
(there are some meters that apply an even smaller current - down to 400μA or so..)

Exactly! The important words here are "most" and "DMMs". But seeing what is being sold at Amazon or Ebay for cheap these days, I have long abandoned any expectation of "reasonable engineering practices" in these devices. I would not be surprized to find 10 mA test currents or even completely unstable ones (which totally defeats thebpurpose). Also, not all meters have a diode testing functionality, in which case you would have to set the current manually when measuring the voltage across the diode. Unless I missed it, we don't yet know how Vf was actually measured. 1 mA with a decent DMM are the most likely test conditions, but then again, this thread started because the sound of the circuit seems to be at odds with the measurement, so who knows. It is, admittedly, a bit of a pet peeve of mine but still: a measurement is only meaningful if the test conditions are known (not assumed or guesstimated).

Quote from: Christoper on July 09, 2024, 08:28:26 AMI would like to throw in my experience that a hard clipping circuit usually adopts the tonality of the diode with the highest forward voltage. The reason it's so popular in the boutique world is because it's easy marketing and "Tube amps do it!".
The LED has a higher forward voltage than the modded diode pair. Also assuming the LED has a forward voltage of 2V, at saturation the 690 mv side of the waveform is 9.25 dB quieter. At 390 mv it is 14.89 dB quieter. At best, the LED side of the waveform is near twice as loud as the silicon side, at worst it is almost three times louder.
In all of my mods, if I want LED clipping, experimentation has told me I only need to swap one of the diodes to get near as makes no difference to the same effect as replacing both of them. In a quiet room the difference is subtle, in a band mix it is near inaudible. The only time I change both is if the pedal needs that small boost in output that changing both gives you.

Another of my pet peeves! Most circuits with "diode to ground" clipping have the diodes AC coupled to the preceding and the following stage. This means that the sides of the caps connected to the "top" of the diodes settels to the middle of the clipped signal swing after only a few cycles. This makes most of the clipped signal almost completely symmetrical. Both halve clip at half the sum of both diode drops. Be aware of DC conditions throughout the circuit over the course of a real guitar signal from pluck to decay!

Quote from: Fancy Lime on July 09, 2024, 02:10:07 PMBe aware of DC conditions throughout the circuit over the course of a real guitar signal from pluck to decay!

Could you elaborate on this? Decoupling caps remove DC from the signal path so where exactly does it work it's way into it?

How would one create true assymetrical clipping?

Quote from: Christoper on July 09, 2024, 10:34:50 PM

Quote from: Fancy Lime on July 09, 2024, 02:10:07 PMBe aware of DC conditions throughout the circuit over the course of a real guitar signal from pluck to decay!

Could you elaborate on this? Decoupling caps remove DC from the signal path so where exactly does it work it's way into it?

How would one create true assymetrical clipping?

I will try to explain in a manner that is useful to as many people as possible. Please ask or correct me, if I fail to make myself clear. It is not really as easy to understand as it seems that it should be and many people are a bit confused by the marketing talk and mountains of misinformation on the internet about "asymmetrical clipping".

One common misconception is that AC coupling via a cap somehow "removes DC". While there is indeed no "direct current" (DC) flowing through the cap (at least if the cap were ideal; in a real cap there is always a very small amount of DC coming through, in bad or old electrolytics, this may be quite substantial), the real question is not one of current but one of "idle voltage" (meaning the voltage that a multimeter would read at a given point in the circuit while there is no AC signal present). The circuit after the coupling cap still has an idle voltage. Let's look at the good old Electra Distortion. Have a look at the first schematic on Jack's site here:
https://www.muzique.com/tech/electra.htm

We need to quickly remind ourselves what a capacitor (let's assume an ideal one for simplicity) does when subjected to a real signal with a DC and an AC component. Essentially, DC is blocked, meaning that the absolute voltages on both sides of the cap are independent of each other. In case of the coupling cap C2 in the Electra, we have the collector voltage of the transistor on the left, which is set by the transistor properties and resistors R1, R2, and R3, which all play a role in biasing. This should be somewhere around half supply (very roughly speaking), so about 4.5V above ground for a 9V supply. So far so good. So what is the idle voltage on the right side of C2 and therefore the top of the diodes? It is ground because all DC connections after C2 go to ground. For the purpose of this discussion, we do not need to dive into what that really means. All we need to know is that it is a very stable voltage, which we define to be 0V in this case. Crucially, the idle voltage at the top of the diodes would also be 0V if there was a second coupling cap right after the diodes, despite there seemingly being no DC path to ground. This is due to the leakage current of the diodes.

Now let's see what the AC component of the signal does. We will, for now, assume that we have taken the diodes out of the circuit. Caps pass AC, meaning that if the left side swings up by some amount from its (the left sides) idle voltage, the right side also swings up by that amount from its own (the right sides) idle voltage. If the transistors collector (and thus the left side of C2) swings from 4.5V idle to 6.5V, then the right side of C2 wants to swing from 0V to 2V, which it can do because we have taken the diodes out. If the transistors swings from 4.5V idle to 2.5V, then the right side of C2 swings from 0V to -2V. Yes, you can make a signal swing to voltages outside the supply rails with caps and AC. In fact, the vast majority of stompboxes do exactly that at their ground-referenced output.

So finally, we put the diodes back in and see what happens now. D1 will clip the negative half wave, if the right side of C2 swings below 0V-Vf1 (wherein Vf1 is the forward voltage of D1). Analogously, D2 will clip the positive half wave, if the right side of C2 swings above 0V+Vf2 (wherein Vf2 is the forward voltage of D2). So far so intuitive. The problem is that this is only true for the very first half wave (positive or negative) that is large enough to clip the signal. This is because a diode that is past its Vf conducts DC, which will drag the right side of C2 away from 0V.

Concrete example: Let's say our signal has a peak to peak voltage of 4V, the first half wave is negative, and Vf1 is 1V. D1 will start to conduct as soon as the signal has reached -1V. When the signal comes back up towards 0V, what happens next? The right side of C2 now sits at -1V, so this is where the whole positive signal swing (from negative peak towards positive peak) starts. Now let's say Vf2 is 2V. When will D2 start to clip? When the signal is at +2V, obviously. Because it starts from -1V at the negative peak voltage, that means the positive half wave clips at 3V above the negative peak. Sounds very asymmetric so far, doesn't it. Let's see what happens next. If the signal clips via D2, the right side of C2 settles to +2V, so this is where the next down swing from the positive peak starts. Now when will the negative half wave clip again? At -1V in absolute terms. But the signal now starts its journey from +2V at the peak, meaning it has 3V to travel in total before it clips. Damn, 3V from the opposite peak in both directions. Symmetrical clipping after the first full cycle. And in a real guitar signal, the first few swings are so wildly chaotic that we cannot really tell by ear if the initial pluck was clipped symmetrically or not.

So far for the theory of pure and simple diode clipping. However, what goes into the diode clipping circuit matters a lot. If the coupling cap is small and there is a fairly small resistor parallel with the diodes, diode clipping can be made asymmetrical to a certain extend with different diodes but this arrangement will also loose bass frequencies, which may or may not be what we want. In case of the Electra Distortion, the diodes load the output of the transistor, making the transistor put out a wildly asymmetric wave form. Depending on the diodes, this can result in a very asymmetric wave form after the diodes, even if the two diodes are identical! This has to do with the different output impedance of a simple common emitter stage on the positive vs. the negative swing. How this all works out in detail is not at all simple and depends on the characteristics of the diodes and the transistor, the values of the resistors and caps, and also (in simple cases with no output buffer, like the Electra) on the input impedance of the next thing in the signal chain. This makes designing such circuits on paper rather pointless. Better to just breadboard it and play with the components until you like the result. But I digress, that was not the topic of this thread, was it?

The easiest way to get real asymmetric clipping is by using an opamp and putting a pair of diodes in the negative feedback (NFB) loop, like in a tube screamer. There, we have well defined absolute voltages and low circuit impedance, which means that clipping happens where you expect it to, namely at Vf from the idle voltage. If you want this type of arrangement to clip "hard", you can put the diodes in the NFB loop of an inverting opamp.

HTH,
Andy

Edit: I amended the explanation of how a cap works based on Antonis' suggestion. I may have assumed a bit too much prior knowledge on the first attempt. Thanks, Antonis!

Andy, I'm not sure if OP is enlightened or confused more..

e.g.
>Let's say our signal has a peak to peak voltage of 4V, the first half wave is negative, and Vf1 is 1V. D1 will start to conduct as soon as the signal has reached -1V.<

In a single supply CE amp, signal can't go negative..
You'd better describe it somehow like:
When idling, C2 left side sits on Collector voltage and right side on 0V..
For negative waveform, C2 left side is dragged down followed by its right side, making D1 to conduct..

Or better, use a dual supply CE amp example with Collector ideally biased at GND..  

P.S.
No intention for insult or understate your very well established writting..

Quote from: Fancy Lime on July 10, 2024, 06:40:24 AMI amended the explanation of how a cap works based on Antonis' suggestion.

So you finally managed to make me feel hangdog...

P.S.
After OP gets familiar with all those nasty things, we'll pass to "diode's non-linearity", dynamic resistance  and Shockley equation..
(just to realize that we live in a world with no symmetry at all..)