Tuesday, September 25, 2012

Big Muff Tone Stack analyshenanigans.

So, I did a few crude experiments and have a few problems, but they've gotten me thinking about how difficult it to manage impedance-related losses in passive circuitry. Here is a more technical diagram of what I'm actually doing.

This is just a diagram of models. Basically I know that if a circuit fits a model, and the models work together a certain way, then I know that the circuits work together that same way. Every wire you see Zout in one stage, and Zin in the next stage, attenuation is occuring, and this time its unwanted. For a "perfect" transmission of voltage, Zout must be zero and Zin must be infinity. The problems with these impedance is that they change when you twiddle the knobs. If it was predictable it wouldn't be a big deal, but there are some variable resistors that have to be taken into account, so instead of saying Zin is a, we have to say Zin varies from b to c. It might take differentiation to find the min or max, depending on how you do it. Here's the schematic for the version of the tone stack I'm using:
To find Zin, we just see what the equivalent resistance to ground of the whole thing is, assuming no load. To keep life simple, we're going to just do a DC analysis, and assume that capacitors are open circuits.
There's only one path for current to go through, so the equivalent input impedance is trivial to find: just add up the three resistors. To find Zout, you see what the equivalent resistance going in BACKWARDS from the load is, assuming that the source is grounded. For this circuit, this means two branches which are effectively in parallel, and each one of them contains half of a potentiometer. bigmuffstack3 To really analyze this, we're going to have to split up the potentiometer. Let's call one half R and the other half (1M-R), since they have to equal 1M. So this means that the two parallel branches consist of R + 39k and (1M-R) + 8.8k. The formula for parallel resistance tells us that in terms of R, Zout is (R+39k)*(1008.8k-R)/1047.8k, a quadratic. It's easy to see that when R is 1M, Zout is approximately 8.8k, and when it's 0, Zout is approximately 39k, but when R is 484.9k, about halfway, Zout becomes its maximum amount: 261.95k. So there you have it. When I design the attenuator, it must be able to deal with this large change in output impedance by having an input impedance much large enough to be considered equally "larger" to both values. When I tested the circuit I plugged it directly into the 22k impedance effect return, and the result was that in the middle of the knob barely any sound came out, and it was crazy loud at the outside. Looking at the numbers, this is the result one would expect! More on attenuator design coming soon.

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