What does this resistor do? - GGG - Guv'nor

[Mpesta]

All,

I recently built the GGG - Guv'nor pedal and am loving it!  I had an idea to modify the 680k R6 resistor and when I used 120k and 440k, the pedal didn't seem to work.  I thought I blew the op amp!  So, I whistled past the graveyard and reinstalled the 680k resistor and the pedal came back to life!  Whew.

But in looking at that resistor, I'm thinking it sets the gain on teh second stage.  Am I correct?  If so, shouldn't I be able to use smaller values or maybe even a 500k linear pot in line with w 180k resistor?

This is my first build and my first mod idea.


Thanks for any help,
Mike

[armdnrdy]

Hi Mike,

You might need this!

http://www.pic101.com/op_amp.htm

[Kesh]

negative feed back

essential in every op amp audio circuit

(apart from the insane ones)

but, because there is no other resistor at -ve in, it probably does some weird sh*t too.

[Mpesta]

Thanks guys!

Two follow up questions.

1) Should I be able to adjust that resistor and modify the amount of gain?  If it's negative feedback it's sending amplified signal back into the Op Amp to reduce the amount of gain correct?  So less resistance ought to feed more signal which would reduce the gain?

2) what's the capacitor in parallel doing?  Is it some sort of filter?

Thanks again,
Mike

[Kesh]

1. yes, normally, but the lack of a dc in resistor confuses me, so i'm not sure. the 0.1uf gets in the way of a normal inverting op amp gain stage. the 680k seems also to be biasing the input. someone with more knowledge will probably show up soon.

2. yes, some frequencies (high ones) will go through the cap, bypassing the resistor, so give more feedback, so less gain for those frequencies

[Tony Forestiere]

Hi Mike,

You might need this!

http://www.pic101.com/op_amp.htm

Thanks for that. Well written, and after a few more reads, it might even sink through my thick skull. A good explanation. 
Bookmarked

[nick d]

    Just been looking at a couple of Guvnor schemos - this looks a bit odd . The  inverting op-amp input should connect to a 10k resistor , the cap C4 should not be there . The 10k and the 680k would then give a gain of 68 , which you could change by varying the 680k . Try replacing C4 with a link , see what happens .

[Gus]

People are missing how the gain is set
NOTE VR1 clever use of a gain that adjust two gain stages.  IIRC Myself and others have posted about this type of gain control over the years.

As you adjust VR1 both stages gain changes the wiper to right side adds to R2

So at min gain the 2nd stage gain is 680K/110K and max gain 2nd stage gain it is 680K/10K(and what ever output resistance the IC1a opamp has)  Also keep in mind the RCs that form the highpass filter between stage

You can work out the input noninverting opamp gain
Google Bing etc inverting opamp circuit and noninverting opamp circuit



http://www.mif.pg.gda.pl/homepages/tom/files/guvnor.gif

Note the original switching for why C3 and C4 are in the circuit

[Mike Burgundy]

@ Nick - what Gus said, plus it might help if you think of the caps as resistors variable with frequency. For frequencies high enough, they are essentially a short, so R2 (and a part of VR1) sets gain together with R6. For lower frequencies, C4 and 3 start to look like increasingly large resistors.
DC conditions are satisfied since the + input is referenced to 4.5V, which sets the output DC level at 4.5V (and the - input as well through R6).
@ OP: I don't see any reason why lower values shouldn't work for R6. I think an error snuck in there somewhere.

[nick d]

 Thanks Mike - looking and thinking more closely , it all makes sense ( sort of ! ) . Also , thanks to Gus I now see how  the gain stages in the Marshall pedals work , compared to other op-amp dirt-boxes - subtly different , clever stuff ! God bless Jim , RIP .

[PRR]

Yes, you should be able to replace the 680K with smaller values for less gain.

That it "didn't seem to work" is odd. How "not work"; low gain? nothing not even hisss? Steam came out?

Bad solder joints are #2, wrong connections are #1 most popular cause of trouble. (However it working with the 680K suggests you are too clever to go for the popular mistakes.)

DC Voltmeter is a friend. Most audio-only changes won't change the DC voltages. You expect most opamp pins near 4.5V; anything way-off is a Clue.

[Mpesta]

Wow!  Thank you everyone!  There's a lot to digest here.
@PRR - the sound went low and sputtered and then nothing.  I will give you bad connections for sure as I wasn't soldering the test resistors in, I was just inserting them into the PCB.  I thought I would get something and then solder that or maybe insert a 500k linear pot in line with a 180k resistor but I didn't have either in stock. 

Thanks again everyone, I may still have questions after I digest.
Mike

[samhay]

The 2-gain-stages-with-1-pot feature of this circuit is a nice touch. I might have to start doing that.
If you reduce R6, the gain of the second op-amp will drop. However, also bear in mind that R6 and c8 form a 1064 Hz low-pass filter and do most of the high frequency roll-off. If you reduce R6, you will increase the frequency of this filter (if R6 = 180K, f = 4020 Hz) and it might get a little fizzy.
If I understand correctly, the order of R2, C3 and C4 do not matter - at least at radio frequencies - and C3 and C4 could be replaced with a single 68n cap if you can find one. Is there any reason why they are ordered the way they are, or is this more likely to be legacy from a circuit design-by-modules approach?

[Mpesta]

Samhay,

Thanks for the response, I'm not sure why C3 and C4 are ordered that way but I suspect your idea about the design by modules approach to be the case.

Let me ask you this, my original post was about R6 and C8.  I see R6 controlling the negative feedback from the output (pin 7) back to the inverting input (pin 6).  But are you also saying that the audio signal is traveling in the other direction from pin 6 to pin 7 and through the R6 and C8 filter?

Thanks,
Mike

[ashcat_lt]

Remember that thing about how the capacitors are like resistors which vary by frequency?  At high frequencies C9 looks like a small resistor in parallel with R6, thus more negative feedback and less gain.

Also...

http://www.mif.pg.gda.pl/homepages/tom/files/guvnor.gif

Note the original switching for why C3 and C4 are in the circuit

[jymaze]

Mike,

From an intellectual point of view it makes the most sense to consider the signal going from 7 to 6 since it is a feedback.

But for real it is more a wave going in both directions and the signal in the feedback loop is just reversed in phase compared to the input (so the sum of both generates no signal at pin 6).

[samhay]

Mike - as ashcat_it said, but seeing as I opened this can of worms, I will explain too.
We typically use capacitors for 2 reasons - either to block or filter DC or as frequency-dependent resistors for AC signals (look up capacitive reactance if you want more info). In this case, we are concerned with what C8 is doing to an audio (AC) signal, so the cap can be thought of as behaving like a resistor. C8 is parallel to R6, so the op-amp gain is set collectively by R6 and C8 - whenever you see 2+ components connected in parallel, it is probably best to consider them as a single unit.
Anyway, the effective resistance of C8 drops as the frequency increases (resistance is inversely proportional to both C and freq). At low frequency, the feedback resistance is dominated by R6 (gain = 680/(10 + whatever VR1 is doing) ), but as the frequency increases, the effective resistance of C8 drops until at about 1000 Hz it is equal to that of R6 and the gain is now half ( (680||680)/(10 + VR1)). Above 1000 Hz, the effective resistance of C8 drops further and the gain approaches 0.

[Ice-9]

Samhay,

Thanks for the response, I'm not sure why C3 and C4 are ordered that way but I suspect your idea about the design by modules approach to be the case.

C3and C4 are like this as they were part of the original bypass method, there would of also have been a second resistor in there.

There is no need for a DC rsistor on the input as it is a inverting opamp that ihas its + pin biased directly to Vbias.