ampage Tube Amps / Music Electronics	For current discussions, please visit Music Electronics Forum.

ampage archive

Vintage threads from the first ten years

previous: MHelin Preben:One topic affecti... -- 3/6/1998 4:25 AM

Re: Fuzz Face for the people!

3/6/1998 8:36 AM
MHelin	Re: Fuzz Face for the people! I wrote: However, the Ic is 0.25 mA only when the transistor is conducting and the base voltage is then most negative (with PNP Q1). When the transistor is not conducting, there's just the base current of Q2, which is not much, and the input impedance is much more, I'd say it's close to 100k (the feedback resistor is there). So during negative phase input impedance is low and during positive phase it's high, and this is one way to see how the asymmetric distortion is created. Just one thing to add, when the input voltage at base of Q1 goes positive (towards the ground) the voltage at the collector of Q1 goes negative, as well as the emitter of Q2 thus charging the cap bypassing the 1k pot. When the cap is being charged the negative voltage affects on the base of the Q1. As the frequency of voltage is quite fast the cap just levels all the changes and changes the operation point of Q1 so that when the signal dies the asymmetric distortion disappears (I think so). So a bigger cap would change the attack/decay times longer. Now the internal re of Q2 with the cap sets the time constant RC for it, but you could connect a pot there between the cap (or actually the 1k pot) and the emitter of Q2 to limit the current charging the cap and so adjusting the time constant. Maybe with two diodes and pots you could adjust both attack and release times separately, except that the minimum 0.3 V voltage difference may create some problems with only a modest changes. Now if somebody only could find out how to get the tone with silicon transistors! -Mikko

3/6/1998 8:36 AM

MHelin

Re: Fuzz Face for the people!

I wrote:
However, the Ic is 0.25 mA only when the transistor is conducting and the base voltage is then most negative (with PNP Q1). When the transistor is not conducting, there's just the base current of Q2, which is not much, and the input impedance is much more, I'd say it's close to 100k (the feedback resistor is there). So during negative phase input impedance is low and during positive phase it's high, and this is one way to see how the asymmetric distortion is created.

Just one thing to add, when the input voltage at base of Q1 goes positive (towards the ground) the voltage at the collector of Q1 goes negative, as well as the emitter of Q2 thus charging the cap bypassing the 1k pot. When the cap is being charged the negative voltage affects on the base of the Q1. As the frequency of voltage is quite fast the cap just levels all the changes and changes the operation point of Q1 so that when the signal dies the asymmetric distortion disappears (I think so). So a bigger cap would change the attack/decay times longer. Now the internal re of Q2 with the cap sets the time constant RC for it, but you could connect a pot there between the cap (or actually the 1k pot) and the emitter of Q2 to limit the current charging the cap and so adjusting the time constant. Maybe with two diodes and pots you could adjust both attack and release times separately, except that the minimum 0.3 V voltage difference may create some problems with only a modest changes. Now if somebody only could find out how to get the tone with silicon transistors!

-Mikko

Replies:

MHelin Now if s... -- 3/6/1998 9:13 AM