Help understanding a Schaller TR-68 clone schematic

[bushidov]

I have built a Schaller TR-68 clone pedal based on this schematic:


It works great, so no issues there.

However, I am still new to these analog circuits and am trying to understand all that is going on here. Sorry if these questions are simple ones.

1. I get that R1 is an anti-pop resistor, and R2 is to control the current about to go into the C2 coupler and go into the Q1 Booster circuit. However, what is R3 for and why is it going to ground? It makes it look like a voltage divider for the input signal.

2. What is the reason for the voltage divider R5 and R6 (along with the C3 capacitor). I am just used to seeing a single 10K-ish resistor pulling up to 9V in booster circuits like this, so I was curious.

3. The big question. How does this LFO circuit (I am guessing it is an LFO) work? I see how one pot changes the frequency and one changes the amplitude. However, I am more familiar with the dual op-amp type LFO circuits, like the ones in chorus pedal clones based around the Boss CE-2. I guess there isn't any signal filtering here, as all this is doing is using Q2 as a variable pot to pull in the input signal back and forth to ground via the oscillation, but I am still curious of the maths used to determine the cap and resistor values in that left hand side of the circuit.

Any help would be gladly appreciated. I just would like a good understanding of this circuit, as I think it probably contains a lot of good analog fundamentals (such as a discrete oscillator).

Thanks in advance!

[anotherjim]

I think R3 could be considered to be optional. It was probably intended to cut the level when a hot input is used -  either an active instrument/keyboard/effect send or preceding pedal. For Q2, I don't suppose you would want the base-collector ever getting forward biased. The way R2/R3 lower the input impedance, a Telecaster may not mind, but an overwound humbucker would get decidedly dull - but when the original pedal was designed, what pickups did it need to sound good with?

The LFO is a phase shift oscillator although I'm not all that familiar with the way the frequency is controlled -  it's drawn funny -  the RC phase shift network is usually shown all on the L/H side of the transistor input.

[antonis]

IMHO, it's a weird design (according to my taste..) 'cause C3 partially decoupling collector resistor should be omited for such a low Q1 gain..
(which gain lies between R5/R7 & R4/(R2//R3) - with tendency to former one..)

C3 has an impedance of less than 100R even at the lowest frequnecy of interest so it permanently AC grounds R6..
DC Collector bias should be obtained with a single Collector resistor without practically altering Q1 gain..)

The whole circuitry around Q1 seems to me more in retrospect fix than initiall design task..
(or, maybe, C3 was intended for R4 partially decoupling for raising total input impedance..)

[amz-fx]

I believe this old design was the basis for the Diaz Tremodillo (and the Swamp Thang).

http://www.dirk-hendrik.com/swamp_thang.pdf

R1 is not needed, you can remove it.

The R2/R3 voltage divider is used since the transistor provides gain. The idea is cut the input and then amplify it back to unity (or so).

The Tremodillo always has a pulsing/throbbing sound in the background. I reviewed it here: http://www.muzique.com/trem.htm

regards, Jack

[PRR]

> How does this LFO circuit..work?

Phase Shift Oscillator.

[Rob Strand]

1. R2 is to control the current about to go into the C2 coupler and go into the Q1 Booster circuit. However, what is R3 for and why is it going to ground? It makes it look like a voltage divider for the input signal.

It pretty much is a divider to set the overall gain *BUT* it's addition changes the way the gain changes.   Think of the transistor as a variable resistor.  If the transistor resistance is 4.7k then the presence of the R3 won't affect the gain.  Then if the transistor doubles it's resistance to say 10k  again the effect of R3 is small.   However if the transistor resistance is at 47k then changes to 100k  the presence of the 47k has a much larger effect.

There's a few subtle things about that design.   The form of the amplifier is like an inverting amplifier,

Ri = R2  and Rf = R4.

If you tried to build this circuit with a real opamp R3 and the transistor would have little effect and the circuit wouldn't work.   The difference is the opamp has a very high gain (ideally infinite) whereas the transistor has a very finite gain.  That means the feedback cannot hide the effect of R3 and the transistor's resistance.

2. What is the reason for the voltage divider R5 and R6 (along with the C3 capacitor). I am just used to seeing a single 10K-ish resistor pulling up to 9V in booster circuits like this, so I was curious.

That decreases the gain of the transistor stage and also stops ripple from the supply getting into the audio. 

3. The big question. How does this LFO circuit (I am guessing it is an LFO) work? I see how one pot changes the frequency and one changes the amplitude. However, I am more familiar with the dual op-amp type LFO circuits, like the ones in chorus pedal clones based around the Boss CE-2. I guess there isn't any signal filtering here, as all this is doing is using Q2 as a variable pot to pull in the input signal back and forth to ground via the oscillation, but I am still curious of the maths used to determine the cap and resistor values in that left hand side of the circuit.

The basic idea is a phase shift oscillator.   These are linear oscillators.  The feedback network is C6, through C10 and R16+VR2.   The network is essentially a high-pass filter.   There is a frequency where the feedback provides 180deg phase shift.  When that is fed into the inverting amplifier (Q3 + Q4) is provide the condition for positive feedback.   At that frequency the feedback network has a certain gain, or loss, the transistor amplifier must provide enough gain to make the overall gain around the loop more than one.   For a basic Phase-Shift oscillator  the gain needs more than about 29.   The maths is basically what I've said here but it will take a bit of work to chug through the equations.    A real circuit like this doesn't quite fit the ideal amplifier so there non-ideal parts affect the behaviour of the real circuit a bit.  For example Q4 doesn't have zero output impedance.   You could try to analyse the circuit with the output impedance in there but it will complicate the equations a lot -so it's best to ignore that for a simple analysis.

There's a couple subtle things about the amplifier as well.  Q3 is a buffer.  The reason that is there is it lets you adjust VR2 without affecting the biasing too much.    The phase-shift oscillator normally has VR2+R16 going to ground.   When you wrap VR2 + R16 around an amplifier the effective resistance is less due to the Miller effect.  So when you analyse the RC feedback network you would need to use the Miller multiplied resistance.

Another point about real circuits, especially ones where the frequency is varied by a single pot, is the (minimum) required amplifier gain varies depending on the frequency.    So you need an amplifier which is going to have enough gain for *all* settings of the frequency pot.   Doing that with maths is a *lot* of hard work so these days you would probably simulate (or build) the circuit and make sure it will always oscillate.   Transistor gain for example is one variable which can stuff-up the circuit.

[PRR]

> feedback cannot hide the effect of R3 and the transistor's resistance.

Q1 gain is <6.8k/2.2k or maybe 3.

So the R2 R4 "10:1 NFB" is not doing anything.

R4 is just biasing Q1.

R2 (and R3) set the source impedance for Q2 to work against. It's probably divided because the peak voltage across Q2 without considerable distortion is dozens of mV.

Ignoring Q2, the audio gain is like 0.5*3 or 1.5.

I'm startled that there is no apparent limit to Q2 base drive current. R18 C5 set some upper bound so it won't smoke, but how low can Q2 go? It can certainly go to 500 ohms, which makes audio gain like 0.01*3 or 0.03, 50:1 or 34dB drop from raw gain, which is excessive; I guess you diddle VR1 to taste.