Bass Fuzz Circuit Analysis Question(s)

[bushidov]

Hi all,

I am here again trying to understand some more fundamentals of some simple circuits, this time being the Bazz Fuss:



I get changing out the Q1 to higher and lower gain transistors changes the overall effect in terms of gain, simple enough. I also understand that C1 and C2 blocks DC voltages creeping into the circuit and the volume knob is pretty self explanatory.

I am guessing that changing R1 doesn't matter from a sound perspective but values of 10K to 100K seem to be all over the place. I wouldn't mind understanding what the differences would be for?

I am guessing that D1 is just a soft clipping diode, and being that there is only 1 of them instead of 2 (like in the clipping stage of a Big Muff Pi), it would be asymmetrical clipping?

Lastly, I am trying to see what is exactly going on with C1. I know that the smaller the capacitance of C1, the more lows are being filtered out. This would indicate to me a passive filter being involved, which if it is an RC passive high pass filter, I guess C1 is the "C" in the RC filter, but where is the "R"? Is it the R1 or a property of Q1?

Sorry again if I sound like a beginner, but that's probably because I am, in regards to audio circuitry.

[Fancy Lime]

Hi bushidov,

never apologize for being a beginner. You won't find many members in this forum who have not, at one point, wondered "how the hell does a bazz fuss work?" or something equally small. I'm saying small instead of simple because the bazz fuss is actually a bit more complicated than it seems at first. Lemme xplain:

D1 is not a really softclipping diode in the same way as the diodes in a Big Muff. What it does is bias the base of Q1 to one voltage drop (0.7V for silicon diodes, 1.2-1.4V for red LEDs, 2.1V for blue LEDs, around 0.2-0.5V for germanium diodes) below the collector voltage. Now if there is an input signal, the diode tries to keep the base at the same voltage but can only do so to a limited degree because R1 limits how much current is available for that. So R1 does matter quite a bit. It serves mostly as the drain resistor, which also influences the bias. Installing a log 250k (or so) pot for R1 and playing around with it should be interesting and possibly enlightening.

So let's say the input swings to the negative side. What happens? As the input becomes more negative, more current flows through D1, effectively killing the negative swing. However, this only works to the degree that R1 allows, because as a resistor it limits the current flow. At the same time the current sucked through D1 is also upsetting the collector bias. The result is a lot of distortion.
When the swing goes to the positive side, on the other hand, D1 cannot funnel current to keep the base where it is, so you get a lot of amplification, limited only by the reverse leakage of D1. The amplification is so great that it exceeds the abilities of Q1, meaning the transistor itself is driven into clipping.

So in short: Yes, the clipping is asymmetric. One half clips via the diode, the other has the natural transistor clipping. On top of that, the biasing is different for both halves of the signal, making a proper analysis not quite as simple as the low parts count of the humble bazz fuss makes it seem. Nifty little gadget, isn't it?

Hope that helps,
Andy

[bushidov]

Thanks, Fancy Lime!

That's a lot going on there with that diode, transistor and that drain resistor, but I do actually understand what you are saying. I didn't see it like that, but that makes a lot of sense.

How about C1? Am I correct in assuming there is a high pass filter going on right there?

[Fancy Lime]

Hi bushidov,

correct, C1 forms a first order high pass filter together with the input impedance of Q1. As you may by now suspect, the input impedance is also different for negative and positive swings because it is partially controlled by D1. For the negative swing, the input impedance is very low and changes with the amplitude of the signal. That means only the highert frequancies make it through. For the positive swing, the input impedance is mainly determined by the hfe of the transistor and can be quite high. That means lower frequencies come through on that half.

BTW: thanks for asking the question(s)! I had never really thought about the bazz fuss since building a few myself many, many (many!) moons ago. Was my first circuit. Now, all these years later, it is really fun to come back to it equipped with a bit more understanding (mostly through asking questions on this very forum and reading the materials I've been pointed to). I should build one of these again some time...

Andy

[bushidov]

Thanks again. My background is more on computer networking and software development, with a little trial by fire work experience in industrial electronics (pumps, motor controllers, GPS interfacing, embedded Linux environments, etc)

That makes sense on the high pass filter and input impedance. What I am curious to know though, is if I were to calculate the HP Filter fcut, I know the value of the capacitor, but I am not sure what the value of R is or where I'd get that from?

[Fancy Lime]

Figuring out the actual value of the input impedance is far from simple, mostly because it partially relies on the internal resistance of the diode, which you wont find acurate numbers for in a data sheet. The theory is explained in quite some detail here:
https://www.electronics-tutorials.ws/amplifier/input-impedance-of-an-amplifier.html
Although some of the resistances talked about there are either shortet (R=0) or open (R=inf) in the bazz fuss.

In practice you will fare better with trial and error for determining the cap value on this particular circuit. 1uF sounds like a good start for full frequency response. You can also add an emitter resistor (I suggest 1k) to Q1 and bypass that with a cap (22uF or 47uF or so) to increase the input impedance. That will alter the overall sound though.

Andy

[mac]

When I was experimenting with Arsenio Novo NPN-PNP distortion, posted decades ago, I noticed that it has a sweet spot when the colletor-to-base voltage is zero at DC.
The Bazz Fuss is a similar circuit, except it uses a diode instead of a PNP-as-a-diode.

The diode feeds the base, and clips at the same time. The more current through the diode, the more the diviation from Vc=Vb.
Darlingtons need a tiny amount of current to work, so the voltage across the diode is almost zero.
Single transistors need more current, so you'll notice larger resistors at the collector to force the base to draw less current.

You can place a big pot at the base to fine tune, Vc=VB.
This way the diode just clips, and the base resistor feeds the base. Team work.
This trick works fine with germanium transistors and diodes, but you have to adjust constantly because temparature and leakage are a nightmare.
Not a must with silicon darlingtons.

If you want symmetry, put another diode in the other direction.
Similar to a Big Muff stage with less parts.

mac