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Rat biasing

Started by scaesic, November 07, 2006, 05:51:32 PM

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PenPen

Quote from: scaesic on November 09, 2006, 12:13:23 PM
just thoguht i'd point this out for clarification

in jacks corrected rat schem pins 2 and 3 arethe other way around from the multirat schem: http://www.diystompboxes.com/pedals/PCRAT1.GIF , which is how i assumed the schem was, and also makes more sense to me as pin 2 is the non-inverting input for the lm308.

correct.

scaesic

Quote from: PenPen on November 09, 2006, 12:16:44 PM
Quote from: scaesic on November 09, 2006, 12:13:23 PM
just thoguht i'd point this out for clarification

in jacks corrected rat schem pins 2 and 3 arethe other way around from the multirat schem: http://www.diystompboxes.com/pedals/PCRAT1.GIF , which is how i assumed the schem was, and also makes more sense to me as pin 2 is the non-inverting input for the lm308.

correct.
so the multirat schem is wrong? or is it using a different pin out?

PenPen

Many schems are wrong on this. The Rat had a few slight variations.

Compare the datasheet for the LM308 to get the exact pin outs, but the basics are pin 1 is compensation, pin 2 is inverting, ping 3 non-inverting, 4 is ground, 5 not used, 6 is output, 7 is your V+, and 8 is your other compensation pin.

Now, go to that schem you linked above, the pin MARKED 2 should be pin 3 (non-inverting), pin MARKED 3 should be pin 2 (inverting).

PenPen

#23
I'll repeat what I said before, but using the schem linked above to illustrate. See the 560/47 ohm resistor pair coming off of the inverting input (pin 2)? Let us call those R2. The pot we will then call R1.

The gain formula for any opamp set up like this is gain == R1 / R2. I am completely ignoring the caps on R2 and the 100pf bypass cap around R1 for simplicity's sake.

In that schem, R1 = 150k, R2 = (560+47)/2, or R2 ~= 300 ohm. Therefore, the stock gain with the pot dimed is 150k/300 = 500. Lets say you change one of the two resistors in R2. You make them both 1k, thus R2 = 500 ohm. 150k/500 = 300. You just dropped out your max gain, but you changed the frequency response also, so it is going to sound different. HOWEVER, we can instead parallel a 47k with the pot, R1. Now, R2=300 (stock) and R1=100k. Max gain is then 333. And you don't change the freq response.

Of course, I would need to know EXACTLY what value the gain pot is in YOUR pedal, they are sometimes different values. I'm trying to demonstrate why changing R2 is not a good idea, whereas changing the total resistance of the pot is a GOOD idea. Does this make sense?

scaesic

yes i understand fully what you're getting at, however, what i am getting at is if R2 is really low then the gain will be relatively high a lot of the time, which is what i am experiencing as i am getting fuzz (high gain) even with the pot at low values.

for example if X2 tends to a small value, then X/X2 tends to infinite no matter how low X is (the pot), i could fiddle around with the corresponding cap values to keep the frequency response similar. anyways, i know what you're getting at, i'll give it a try and let you know how it goes.

Seeing how my pot measures 100k i'm willing to bet that the network to ground going into pin2 also has different values, i'll measure them and see what the gain is.

again, thanks for the advice.