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PostPosted: Tue Sep 18, 2018 8:57 pm 
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Hello,

I'm building my first p2p circuit and using the V2 of the Bazz Fuss circuit. I'd like to add a B10K pot which I've ordered to bias voltage to the transistor. I'm assuming I put this pot in place of the 10k resistor that goes from positive power to the transistor's collector leg. Am I correct? Link to circuit below.

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PostPosted: Wed Sep 19, 2018 5:28 am 
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That seems like the most obvious place to me, yes. But if you want to be able to adjust the bias both directions from the stock resistor, I’d recommend trying a 25k trimpot. A 10k will max out at, well, 10k.

Do you have a breadboard? This would be a pretty simple circuit to lay out and test to make sure.

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PostPosted: Wed Sep 19, 2018 10:14 am 
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I think you're going to be fairly underwhelmed by that bias control. It isn't really a bias control. It would be more of a bias control if you connected a pot (in a rheostat configuration) between the base of the transistor and ground. A B100k. Maybe a 250k or even 500k.

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PostPosted: Wed Sep 19, 2018 12:50 pm 
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Thanks for the responses! Yes, I do have a breadboard and built this circuit on a breadboard first. I think you're right about being underwhelmed, but I figured that was because I was using the wrong kind of pot. I love the fuzz with a 10k resistor limiting voltage to the transistor in the stock setup. For what it's worth, I'm going from 9V down to 1.2V after the resistor.

Now I took out the resistor and used an old B25K pot I pulled out of a dead Boss DS-1 I had. I noticed when I turned it counterclockwise, it cut the voltage and made the tone more glitchy & gated as I expected. Then when I turned it all the way clockwise, it made it more saturated and then suddenly, silence, the circuit goes dead as soon as it's fully clockwise. I'd take a reading and see voltage was still traveling to the transistor in an increasing amount, but the whole circuit went dead. I figured this was to do with the pot, but is it possible this is because it's overloading the transistor?

Also, this is a very rookie question, but is my understanding about how a B style pot works correct? Take a B10K pot for example, if it's wired up properly and turned all the way clockwise, it's acting like a 10K resistor and not letting anymore voltage through. Then as it's turned counterclockwise, it's letting through less and less voltage (or increasing resistance) until it's letting nothing through? Or is it backwards and when you turn it you're letting more and more voltage through until there's no resistance?

Thanks again.


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PostPosted: Thu Sep 20, 2018 8:39 am 
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Pots don't usually go all the way down to zero resistance or all the way up to their stated value. It's not at all unusual to have a B10k pot that actually goes from 500R to 9.5k or 11k. You're typically dealing with a 20% tolerance range, meaning a pot with a minimum resistance of 2k and a max of 8k (or 12k) would pass.

Best to grab your digital multimeter if you want to know what's actually going on in there. Some algebra would help too, if you're so inclined :)

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PostPosted: Thu Sep 20, 2018 10:26 am 
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Thanks!

So I am correct in my understanding that basically a 10k pot lets through max approximately a 20% variance of 10K and then as you turn it you increase the resistance up higher.


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PostPosted: Thu Sep 20, 2018 1:13 pm 
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Nominal range would from 0 ohms to 10k ohms, with a tolerance of 20%.

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