Bazz fuzz trouble?

[DocAmplify]

I'm not sure if this is working as it' supposed to. 

I made this for my bass.  I destroyed my first set of components when I attempted soldering.  I had to resort to some alternate components (and maybe that's where my problem is).

My input is through a 4.7 uF capacitor; negative pole on the side of the input. 

My output is through a .1 uF capacitor; negative pole in the side of the output.

I ruined my transistor, so I got a handful of NPN switching transistors "designed for high speed medium power switching and general purpose amplifier".

I ruined my diode, so I replaced it with IN4007; stripe side to the middle pin of the transistor.


Here's what I'm experiencing.  The box seems to be dropping volume.  If I turn my bass up and really attack the strings, I get fuzz, but if I set my bass pots anywhere but max, and don't attack, it clips.  I even tried putting two transistors into darlington configuration to see if I could get more amplification, but it didn't seem to.

I initially wired this with an audio pot, but after not getting much, decided to take the pot out and run it straight, but even with no pot that's what I get.

Help!

[DocAmplify]

And if this helps, I get the most fuzz from low strings (open E).  As I go higher (D or G open strings) the sound is clipped.

[RandomGlitch]

I had similar happen to me when I first breadboarded mine.

It turned out that I had the transistor in the wrong way.

Then I got that right, but had the diode wrong, you get a horrible farty fuzz (if you can call it that!) if you do.

I recommend getting it to work on breadboard first, then note which way round the components go. Pen or paper or photo!

Transistors aren't critical. I used 2n2222s MPS13A darlington and 2N3904's, all good, just slightly different-sounding. I think diodes aren't critical either.

Your caps are correct, negative to the "outside" of the circuit is how I remember it.

[Earthscum]

I think RM is most likely correct. I still stick transistors in upside down regularly. 

However, check a couple things... look at the transistor, and it should have numbers and letters on it, most likely starting with 2N or BC or 2SC. Use this "part number" and google for "(part number) pinout". Emitter on NPN transistors goes to ground, Collector goes to V+. PNP is the other way around.

Now, use a DMM (if you have one) and confirm that you have about 0.6V from Emitter (ground) to base (input). If you use a Darlington configuration, this should be about 0.9-1.2V. Next, measure from ground to the Collector (output) and it should read whatever you had before, +0.6V.

No offense meant if you know any of this already.

[DocAmplify]

No offense  at all.  It is NPN, but the box had the pin layout labeled, so I should have it correctly placed, but I had it wrong at times. 

.98 v from emitter to ground.  (darlington setup)
1.03 v on the collector side,

I had 7.3 v across the resistor (just before the transistor), and if I'm doing this right, from the resistor to the battery ground gives me 1.4v. 

As I inspect the breadboard to answer these questions, I noticed that I have two different transistors.  The box of NPNs were apparently assorted.  I have a 2n3904 and a 2n2222.

[LucifersTrip]

.98 v from emitter to ground.  (darlington setup)
1.03 v on the collector side,

.98, you mean base to ground, hopefully...
that's not quite the difference you're looking for...you may want to keep it as a single transistor in the beginning just to keep it simple. also, remember when you use darlington, the 100K is switched to 10K


did you actually try adjusting the 100K/10K to see if you could tune in a better sound?

[DocAmplify]

I hadn't noticed that the resistor value should change; thanks.

I was hopeful that swapping the resistor would work, but nothing.  So, following your advice, I stripped it back it one transistor.  Still nothing.  Then I noticed it.  I had been plugging my amp into the input and my guitar into the output.

Pretty much one whole weekend to troubleshoot that one issue.  I can't say it's a waste of time though.  It's unbelievable how much I learned looking for a solution.

[Earthscum]

,

I hadn't noticed that the resistor value should change; thanks.

I was hopeful that swapping the resistor would work, but nothing.  So, following your advice, I stripped it back it one transistor.  Still nothing.  Then I noticed it.  I had been plugging my amp into the input and my guitar into the output.

Pretty much one whole weekend to troubleshoot that one issue.  I can't say it's a waste of time though.  It's unbelievable how much I learned looking for a solution.

High Five for that one... Lol. I think we've all done it at least once. Glad to hear it worked out for you.,,,,,,,,,,,,

[DocAmplify]

I'm also trying to learn the circuit.  The working part of this circuit is the transistor and the diode, but I can't quite figure out how it works. 

If I look from the guitar input side, the signal comes from the tip of the 1/4 inch jack and goes to the transistor base.  It is connected to a diode, but the current would hit the diode in its blocking direction. 

I believe the transistor allows current to pass from collector to emitter when current is applied to the base (right?).  That would mean the battery's current would drain off to ground when the guitar gives power to the base.  When it's not, the current would flow to the output jack. 

Is that right? If so, that means the transistor is acting as an amplifier because it's substituting the battery's current for the guitar's current (which is much smaller). 

I think I understand it, but I can't figure out what the diode is for.  It allows some output current from the collector to enter the base, but why? 

[DavenPaget]

I'm also trying to learn the circuit.  The working part of this circuit is the transistor and the diode, but I can't quite figure out how it works. 

If I look from the guitar input side, the signal comes from the tip of the 1/4 inch jack and goes to the transistor base.  It is connected to a diode, but the current would hit the diode in its blocking direction. 

I believe the transistor allows current to pass from collector to emitter when current is applied to the base (right?).  That would mean the battery's current would drain off to ground when the guitar gives power to the base.  When it's not, the current would flow to the output jack. 

Is that right? If so, that means the transistor is acting as an amplifier because it's substituting the battery's current for the guitar's current (which is much smaller). 

I think I understand it, but I can't figure out what the diode is for.  It allows some output current from the collector to enter the base, but why? 

It's a "feedback loop" type diode clipper .
It limits the maximum signal to prevent the splatty sounding power rail banging clipping induced .
The guitar's signal induces a voltage in the base and therefore the transistor switches on and off according to the signal , but , the output is taken from the collector and emitter goes to ground .
The reason why output is taken from collector is that everytime the transistor pulls high and starts to conduct the signal , it will have a waveform on the collector that resembles the base's signal and therefore the output is taken from the collector and passed via a DC blocking cap so you get the amplified signal .
Totally forgot about the inverted theory , somebody help me out on that , not sure if emitter's inverted or the collector is inverted .

[Earthscum]

Basically the diode is feeding the same amount of current to the base as what it's collector sees, so it biases itself in diode drops from the collector to the ground. This determines your output swing, too. If you have an LED, your swing is allowed to the BE voltage plus the CE voltage.
When your guitar signal goes positive, it feeds current into the base. The collector goes negative, reverse biasing the diode and clipping to ground. This is why high gain is needed. You have to reverse bias the diode to get negative clipping at the output.
When it goes negative, it tries to steal current from the base. the collector goes high and tries to rebalance itself. So basically you get half wave recti with clipping on both sides.
Remember, BJT base works on current not voltage, and a diode passes current until it fails. The resistor converts voltage to current, and the variation in current allowed to pass to ground or out is what we see as voltage swing at the collector.

[DocAmplify]

Basically the diode is feeding the same amount of current to the base as what it's collector sees, so it biases itself in diode drops from the collector to the ground. This determines your output swing, too. If you have an LED, your swing is allowed to the BE voltage plus the CE voltage.
When your guitar signal goes positive, it feeds current into the base. The collector goes negative, reverse biasing the diode and clipping to ground. This is why high gain is needed. You have to reverse bias the diode to get negative clipping at the output.
When it goes negative, it tries to steal current from the base. the collector goes high and tries to rebalance itself. So basically you get half wave recti with clipping on both sides.
Remember, BJT base works on current not voltage, and a diode passes current until it fails. The resistor converts voltage to current, and the variation in current allowed to pass to ground or out is what we see as voltage swing at the collector.

Thanks;  I don't quite get it yet, but I don't absolutely need to at this point.  Some day I'll read something else and this post will make it all crystal clear. 

[Earthscum]

Go dig through geofex.com, RG has a ton of info. Even articles like "when good op amps go bad" can create a good spark to understanding transistors.

[DavenPaget]

Now that i remember , the feedback loop in the bazz fuss is in the diode , the diode carries the inverted output to bias the non-inverted input to invert the output back again , but the diode also limits max gain to about 1.7V i think .
The diode is essential , and it is a feedback loop clipping style diode .

[LucifersTrip]

,

  Then I noticed it.  I had been plugging my amp into the input and my guitar into the output.

Pretty much one whole weekend to troubleshoot that one issue.  I can't say it's a waste of time though.  It's unbelievable how much I learned looking for a solution.

High Five for that one... Lol. I think we've all done it at least once. Glad to hear it worked out for you.,,,,,,,,,,,,

i've named it Big Muff syndrome

[DocAmplify]

I've got the Bazz Fuzz working, but now I want to modify it. 

I found this link here ( http://www.seymourduncan.com/tonefiend/wp-content/uploads/DIY-Club-Project-2.pdf).  It's a PDF, so I can't link to the specific picture, but it's the graphic on page 17, "Version 3: What's What".

On the input side of the circuit, he suggest a tone pot and two different sized caps.  Both caps in this diagram are much smaller than that used for the bass circuit (I used a 4.7). 

What I'd like to do is use both circuits so I can put a switch to throw between a bass set up and a guitar set up.  My thought was to add a SPDT switch at the input.  The single end  of the switch would be between R2 and the tone pot.  Pole 1 would complete the circuit as it is in the diagram.  Pole 2 would be to a 4.7uF cap that would ultimately connect at the cathode end of the diode. 

My thought is that I could throw the switch one way and have the input go through only the 4.7 or throw it the other way and have it go through the tone pot that trims between the .22 and .01 caps. 

I know the 4.7 could be turned off with this switch, but if the 4.7 arm of the circuit is turned on, the arm with two caps is actually still a circuit, but the caps are polarized.  Will this work?

[DocAmplify]

I have another thought (I should just be bread boarding this).  What if I put a diode in the two capacitor side?  could I ensure that the circuit was one way and this arm would only be accessed if the switch was thrown to this side?

[RandomGlitch]

I'd just be using a DPDT switch here.

I did a sketch..

[DocAmplify]

Thanks Random.  Elegant solution.