Author Topic: Where is the clipping stage in the BD-2?  (Read 14728 times)

ianmgull

Where is the clipping stage in the BD-2?
« on: March 29, 2008, 01:40:49 PM »
I'm thinking about adding a little more "bite" to my BD-2 (slightly less compression). I was going to play around with the clipping diodes and see what I could come up with. Looking at the schematic, it looks like there is clipping going on at Diodes 7-10. Are diodes 1 and 3 also for clipping? are there any other clipping stages I missed? Thanks!

Schematic:

http://www.freeinfosociety.com/electronics/schematics/audio/pictures/bossbd2.gif


ian

ayayay!

Re: Where is the clipping stage in the BD-2?
« Reply #1 on: March 29, 2008, 02:29:27 PM »
D7 & D8, D9 & D10, and you can also horse around with D1 & D3. 

Your link doesn't work.  Try this schem instead hosted @ Indyguitarist.com:

http://www.indyguitarist.com/schematics/boss/BOSS%20BD-2.gif

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aab0mb

Re: Where is the clipping stage in the BD-2?
« Reply #2 on: March 29, 2008, 04:03:07 PM »
Stick to D7,D8,D9 and D10 for diode changes.  D1 and D3 are NOT clippers.  Look for posts from Jay Doyle in the archives for more in the BD2.  He wrote a great and lengthy article describing the BD2 in great detail.  There is a modified fender tone stack consisting of C26, R37, r38, C34, r50, r51, and C35.  Download the Duncan's tone stack calculator and play w/ the values to change the frequency response to something you might like a little more.  That is, if the diode changes don't do all that you had hoped for.

Good luck!

ayayay!

Re: Where is the clipping stage in the BD-2?
« Reply #3 on: March 29, 2008, 04:37:01 PM »
Quote
Stick to D7,D8,D9 and D10 for diode changes.  D1 and D3 are NOT clippers.

OK thank you.  So I looked it up and here's what Jay Doyle said about those diodes: 

Quote
First off, ignore the diodes, they are there to protect the opamp and nothing more, no clipping. The basic job of it is to filter the output, that transistor on it's negative input and the parts surrounding it are a simulated inductor which along with C9 form a resonant frequency and in this case create a "dip" in the frequency response (I think, still need to simulate it). The easiest way to mod simulated inductors and the frequency response is to change the caps, in this case C9 and C16.

Thanks for educating me.  But does anyone know why they're necessary?  Will the Opamp blow or become damaged without those?  I guess I'm just not familiar with this scenario. 



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wampcat1

Re: Where is the clipping stage in the BD-2?
« Reply #4 on: March 30, 2008, 03:59:53 PM »
Quote
Stick to D7,D8,D9 and D10 for diode changes.  D1 and D3 are NOT clippers.

OK thank you.  So I looked it up and here's what Jay Doyle said about those diodes: 

Quote
First off, ignore the diodes, they are there to protect the opamp and nothing more, no clipping. The basic job of it is to filter the output, that transistor on it's negative input and the parts surrounding it are a simulated inductor which along with C9 form a resonant frequency and in this case create a "dip" in the frequency response (I think, still need to simulate it). The easiest way to mod simulated inductors and the frequency response is to change the caps, in this case C9 and C16.

Thanks for educating me.  But does anyone know why they're necessary?  Will the Opamp blow or become damaged without those?  I guess I'm just not familiar with this scenario. 





I know Jay (and traditional elec. knowledge) says that those diodes in front of the opamp stage aren't doing anything but my ears say differently. Sounds like they are indeed compressing the signal at least a little bit. Put the diodes on a switch and see what you think. Keeley has a bd-2 mod that puts germanium trannies there (as diodes) so I'm guessing his ears say the same.

bw


ayayay!

Re: Where is the clipping stage in the BD-2?
« Reply #5 on: March 30, 2008, 04:43:46 PM »
Quote
know Jay (and traditional elec. knowledge) says that those diodes in front of the opamp stage aren't doing anything but my ears say differently.

Well that's what I thought too.  I've put Ge's in there and I personally *thought* it to be more pleasing.  What I need to do is exactly what you said Brian, and throw a switch in there to see for myself. 

Here's my thinking:  OK, so it's probably not clipping as Jay pointed out, but filtering the output right?  We know different diodes produce different tones (clipping wise,) so even though it's not clipping it might still affect the tone right?  Especially considering where that output is leading to on the schem. 

I'm not arguing, I'm asking.  So feel free to learn me on this here one.   ;D
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aab0mb

Re: Where is the clipping stage in the BD-2?
« Reply #6 on: March 30, 2008, 05:55:54 PM »
I like the idea about the different sounds of the filtering.  Can't wait for the final verdict on this one... 

ayayay!

Re: Where is the clipping stage in the BD-2?
« Reply #7 on: March 30, 2008, 06:24:03 PM »
Uh oh, did I just get volunteered?   :D

Seriously though, if I get time tomorrow night I'll try it out and post results.  :)
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aab0mb

Re: Where is the clipping stage in the BD-2?
« Reply #8 on: March 30, 2008, 09:16:28 PM »
If I hadn't loaned mine out, I'd be in there w/ the soldering iron right now.  :icon_twisted: 


blanik

Re: Where is the clipping stage in the BD-2?
« Reply #9 on: March 30, 2008, 11:20:32 PM »
i changed mine to 3mm LEDs (D3, D9, D10) and put two 1N4002 at D7, D8... really changed the sound for the best... i was thinking about also putting two LEDs at D7, D8 to remove what's left of the fuzziness...

the BD-2 doesn't only use the clipping diodes to create distortion, the IC also does and i suspect some trannies in there does the same... it's really a multi layered distortion

R.

JDoyle

Re: Where is the clipping stage in the BD-2?
« Reply #10 on: March 31, 2008, 12:25:09 AM »
Well now, I must come and defend myself, no?  :D

OK - here is the deal with those diodes - there aren't SUPPOSED to do anything other than protect the op amp. Those diodes ensure that the differential voltage between the inputs doesn't get too large and fry the transistors that make up the differential amplifier at the front end of the op amp's circuit. Many, many, many op amps have these integrated into the internal circuit for just that purpose. R.G. has a really good write up on this issue at his site but I can't remember the link or what he called it, making this sentence near useless...

Anyway, if you connect the emitter and collector together on any transistor you can use them as the Cathode of a zener diode, with the base being the anode, the zener voltage is around 7 volts (it is actually the reverse breakdown voltage of the emitter-base junction and can be anywhere from 4 to 7 volts depending on the transistor with actual zener diodes being constructed for specific breakdown voltages). BUT, once you do this, you degrade the REGULAR operation of the transistor in terms of gain and/or noise...

Basically, a large differential voltage between the inputs of an op amp can force one or more of the input transistors into a zener state and degrade the performance. The nasty bit about all this is that it doesn't completely kill the transistor so the op amp can start out really nice and happy but end up quite pissed off and sounding like it...

In the BD-2, those diodes are on the output opamp that does several duties, it buffers the tone control, outputs a low impedance signal to drive whatever is next, and lastly, it is a filter. That transistor and the associated components in the feedback loop make up a simulated inductor that creates a mid-hump with the cap in series with it (from the output to the neg. input). When power is removed from an inductor a large voltage spike can occur and those diodes are there to bulletproof the op amp from such an event.

Now - First off those diodes SHOULDN'T ever have to be used as an op amp is supposed to keep it's inputs at the same voltage (if it can't, it becomes a switch) but when you turn off the power to the circuit, you no longer have an op amp, just a mass of semiconductor p-n junction just waiting to be diodes both the regular kind (collector to base and base to emitter) or the zener kind (as mentioned above, though the collector can be ommitted from the connection, though it isn't in ICs due to parasitic issues that occur). So, with an inductor there is the chance.

Second, if those diodes WERE conducting in reality, while I don't know for sure, my guess is that it would sound pretty bad when they did - the input signal would now be dropping itself across the negative input in a very non-linear way, with the diode creating a voltage divider with the resistance on the negative input, while the positive input would be a diode drop away, this would make the IC quite unhappy and the output would try to fix all of that - but because the distortion is outside the feedback loop, is dependent on the input signal, and the inputs are a diode drop different, there isn't all that much the op amp can do but wait for the input to correct itself...

SO - How can we both be right (though I'm a bit weary about anyone taking the previous paragraph in without a large helping of salt)?

Easy - the diodes have a parasitic capacitance that would be different for different diode types (explaining why changing the diode TYPE changes the sound, but this is also process dependent so would vary between diodes of the same type as well) and also varies with the input signal. The latter is a bit hard to explain without going into a lot of math and theory, but because a swing one way causes the paracitic capacitance to increase and with the other polarity the capacitance decreases, you have asymmetric distortion.

But in my opinion the parasitic capacitance is what you are hearing. Seeing as the circuit is beating the absolute hell out of the input signal by clipping it in the first discrete op amp, then with diodes, then again with the last discrete op amp, I really don't see how adding a tiny, tiny bit of even harmonics would make all that big of a difference.

And as we all know, a couple of pF here or there can really make a difference in the overall sound of any circuit.

So while those diodes protect the op amp the price paid is their parasitic action in series with the signal. It's either that or risk a lot of dead op amps. I wouldn't want that risk with one, forget millions for a company like Boss, thus the diodes. But just because they aren't SUPPOSED to do anything to the tone, doesn't mean they don't - anything that touches the signal has an effect.

In the end, EVERYTHING, from the power supply caps to the way everything is connected to ground, matters in the final tone - the question is figuring out what matters a lot and what matters a little. So if changing those diodes improves the tone in your opinion, great, I'm sure it does, but if you are looking to mod the circuit, I'd look elsewhere first and come back to that to fine tune.

Regards,

Jay Doyle

PS - Personally, I think a really easy mod that could do cool things to this circuit is to just swap the off board leads on one of the two ganged pots in the drive control - then as you turn it one way you get more clipping, less volume; more volume, less clipping the other... With the middle position being the same as before the mod... I don't own this pedal and I don't build commercial circuits so I don't know, but it always hit me as an easy and easily reversed mod to try...
« Last Edit: March 31, 2008, 12:48:43 AM by JDoyle »

ayayay!

Re: Where is the clipping stage in the BD-2?
« Reply #11 on: March 31, 2008, 10:40:12 AM »
Thank you for that explanation Jay.  Both for explaining what the purpose is, and explaining the inherent capacitance factor.  Now the $64,000 question is:  Does that parasitic capacitance make a noticeable enough difference to our ears to shape the tone!  :D

I'm really eager to bust open my BD-2 to throw a switch in there to try it out, but Mein Fuhrer has me assigned to a few tasks over the next couple of nights. 

Also, I read your PS and also saw it in the archive post.  It's tempting to say the least, so maybe I'll try that out too, but I'd probably have to strip back the ribbon cable on the stacked pot, or maybe wire it up new.  If anyone else wants to take a stab at it be my guest! 

...Man this is a good thread. 
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blanik

Re: Where is the clipping stage in the BD-2?
« Reply #12 on: March 31, 2008, 11:56:54 AM »
Second, if those diodes WERE conducting in reality, while I don't know for sure, my guess is that it would sound pretty bad when they did...

the stock BD-2 actually sounds pretty bad when you turn up the drive, fizzy, gritty, high pitched fuzziness...

changing the diodes for LEDs got rid of that completely (and changing D3 which isn't in the clipping diodes pattern also made a big difference...) the pedal sounds more like an overdrive but can still go in DS-1 territory when you turn up the drive  :icon_twisted:

JDoyle

Re: Where is the clipping stage in the BD-2?
« Reply #13 on: March 31, 2008, 12:55:36 PM »
Second, if those diodes WERE conducting in reality, while I don't know for sure, my guess is that it would sound pretty bad when they did...

the stock BD-2 actually sounds pretty bad when you turn up the drive, fizzy, gritty, high pitched fuzziness...

changing the diodes for LEDs got rid of that completely (and changing D3 which isn't in the clipping diodes pattern also made a big difference...) the pedal sounds more like an overdrive but can still go in DS-1 territory when you turn up the drive  :icon_twisted:

Ok, then this may be starting to make sense, a simulated inductor saturates a little over a volt, the LEDs could be taking something going on with the simulated inductor and the input and moving it out of some kind of range that makes it sound better...

Thing is, if that is true, then Keeley's mod of changing them to germanium diodes wouldn't work, UNLESS he is connecting them as zeners, then we are getting the same thing as with the LEDs. Does anyone know the actual way he connects the transistors? Sounds to me like he had a ton of sub-par germanium transistors he needed to use...

Anyway, I have no idea if they are actually clipping or not, if the input signal changes their impedance which causes some kind of distortion, or in all honesty what the hell is going on, but as long as you use diodes with thresholds under 4V, they will still protect the op amp, so if changing them improves the tone to your ears, then by all means DO IT!!!

A wild guess would be that when the output op amp with the diodes is forced into clipping by the magnitude of the input signal, which is very possible with the drive up, and when the op amp is clipping, there is no longer negative feedback and those diodes could be conducting as the output signal (and therefore the signal on the negative input) is pinned to a rail and the input signal is not, if there is a diode drop's difference between the two, the signal clips. Something to check (I would but I'm sitting in Pittsburgh at the airport praying to get on a flight to bring me home, I was supposed to be there 24 hours ago, I would KILL for a fresh pair of socks) would be the max output swing of that chip. The discrete's can swing almost to the power rails (to within the collector saturation voltage and the drop across the emitter resistor at max input signal/current) and it is doubtful that Boss used a rail-to-rail opamp, too expensive, so if the IC can't get within a diode drop of either rail, the signal will clip at the input when the output is clipping, but if it can get to within an LED's diode drop, we may have found the answer...

Regards,

Jay Doyle

kurtlives

Re: Where is the clipping stage in the BD-2?
« Reply #14 on: August 28, 2008, 03:13:52 PM »
Hate to bump an old topic but my question relates and I dont think a new topic is nessicary.

I replaced D7 D8 D9 and D10 with 3mm red LEDs... The sound doesn't seem like it changed and the LEDs don't light up as I play...

Also I can remove those four diodes and I still get distortion...

John Lyons

Re: Where is the clipping stage in the BD-2?
« Reply #15 on: August 28, 2008, 05:30:42 PM »
I've not twealed or even played though this circuit but...
The LEDs do not have to light to be working.
Unless they are fully conducting you might not even notices them lighting at all.
As mentioned above, there will be some clipping even without the diodes.
The diodes add clipping to transistors which depending on how hard they are driven clip themselves.
Take the diodes out a a tube screamer and you get a good amount of clipping as well.
Diodes to ground or in a feedback loop often smooth out harsh clipping adding harmonics and nuances of there own.
Also adding compression and reducing ldynamics. This is why adding LEDs helps bring out the touch sensitivity often times.

john

Basic Audio Pedals
www.basicaudio.net/

kurtlives

Re: Where is the clipping stage in the BD-2?
« Reply #16 on: August 28, 2008, 05:35:50 PM »
Ya I figured I would get clipping without the main diodes...

I was just surprised when I installed the LEDs cause I am used to them giving me a big fat compressed sound and I couldn't tell if they were doing anything.

slacker

Re: Where is the clipping stage in the BD-2?
« Reply #17 on: August 28, 2008, 05:54:59 PM »
If you stuck 2 LEDs in series then you would need at least 2-3 volts of signal to make them clip, the circuit probably isn't that loud at that point.

kurtlives

Re: Where is the clipping stage in the BD-2?
« Reply #18 on: August 28, 2008, 06:44:58 PM »
So basically my LEDs ain't doing squat?

comfortably_numb

Re: Where is the clipping stage in the BD-2?
« Reply #19 on: August 28, 2008, 07:15:57 PM »
You got it man.  I replaced mine with 1n4001s and they sound pretty nice.  I really like the sound of this circuit.  I use it with the gain dimed.  I'd like to get a bit more out of it actually.  It sounds sweet with a booster in front too.