Hey Everyone,

Ok first off I guess it should be said that I have only been doing circuitry for about 2 weeks now (Yes I know ohms law and power). So I can understand if you want to tell me to go read a book on it or something (I have borrowed a basic electrical engineering book from my physics teacher).

UPDATE:

-Thanks very much.

 

Not sure about the op amps: been away from them too long, but Q4 won't work very well as shown. I'd try it as a common emitter amp if the pickup mike input Z is no higher than 10K. If it's over about 100K I'd use a FET or matching transformer. I'd also make the "in/bypass" switch a 3PDT type and wire the hot side of the output jack, J1, in the same manner as the input jack, J2, and use the third pole on this switch to turn off the power to Q4 in bypass mode. If you still want the LED, connect an 820 ohm resistor in series with it and the switched 9 V source. Also, arre you getting the power for the op amps from the main amp supply? Good luck.

 

Well the idea is that I want to create distortion on the guitar signal and return it back through J1 all with only a 9 volt battery.

J2 (input) -> Pre-Amp?? -> Distortion OR Overdrive -> J1 (output)

Thats what I would like it to do.

(Ive been reading up a lot on on the allaboutcircuits.com tutorials and its starting to make a little sense.)

(thanks for your help so far)

 

The opamps aren´t connected in the right way, you have to add resistor before the upper one, and a second resistor from the lower one´s - pole to ground.
The gain of the upper one will be Pot resistance / the added resistor, and for the lower one 1+ (Pot/resistor)

Also the master volume is wrong, you have to connect the input to one pole, ground to the other and take signal out from the viper.

And probably the input transistor can be ommitedd, or replaced by a buffer.
If it stays there, it needs resistor from emitter to ground to create some load for it.

 

This is the kind of clipping I'm getting now:

But why isn't it on the bottom as well.

 

This is the kind of clipping I'm getting now:

But why isn't it on the bottom as well.

That is hard to tell, you should measure the dc voltages in the circuit,the signal path should have 4.5V DC component, do you understand that the ground point should be in the middle of supply voltage?

 

Ok it has the 4.5V DC component...but what do you mean about the ground in the middle of the supply voltage?

 

Ok it has the 4.5V DC component...but what do you mean about the ground in the middle of the supply voltage?

That - pole of battery goes to -supply of the opamp and + pole goes to +supply. And the ground should be in the middle between + and -, which here is 4.5V.