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Blend effect

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mr.pepper

New Member
So im not fantastic with electronics and was wondering if you guys could help me out with some schematics. Basically i want/need talking through them, on whats happening where and what everything within the curcuit is doing.



It is a schematic of a looper pedal for guitar, which 'blends' the dry(input) signal with the wet(loop) effected signal.

Thanks for your help,
Much Appreciated
Mr.P
 

johankj

New Member
I'm no ace myself, but I'll say what I know.

Vcc is power, and the three lines that form an up-side-down pyramide is ground. The parallel short and long lines denotes a battery.

Well, the R's are resistors, they limit current and there is a voltage drop over them (Look up Ohms law).

The C's are capacitors, a simplified analogy is that they are like little batteries.

The triangle is an operational amplifier, or an Op-Amp for short. That's an amplifier of sorts, an integrated circuit (IC).

The circuit in the top-left corner looks like the power-supply circuitry, and the other connects input 'in1' and 'in2' to the opamp and possibly amplifies it.
 

Brevor

Member
It's basically just a summing amplifier, The inputs IN-1 and IN-2 are summed at the inverting input of the op amp. The output will be inverted with a gain of 2.2. This is determined by the values of R-1, R-2 and R-10. C11 is to block the DC component and only let the AC component
(the audio signal) pass to the output. It has this dc level because the power is split from the battery by R-12 and R-13 this causes half the battery voltage to appear at the output of the op amp.
 

RadioRon

Well-Known Member
Building on what Johankj has explained, let's go through the circuit operation next.

The core of this thing is the op-amp in the center. It is wired in an inverting amplifier configuration, a very common circuit type. I suggest that you google op-amp and study "inverting amplifier" for more details. In this configuration, there is feedback applied from pin 1 and pin 2 through R10 and C10. This feedback works with R1 and R2 to establish the gain of the amplifier. The feedback also causes the circuit to isolate In1 from In2 by insuring that the voltage at the op-amp's inverting input is always 0. At DC, the gain of this setup is the ratio of R10/R1 for In1 and R10/R2 for In2, so both inputs have a voltage gain of 2.2. C10 is included to decrease the gain at higher frequencies. The "corner frequency" of this low pass filter is 154 KHz. This capacitor is probably included to insure that the amplifier remains stable and that it avoids amplifying high frequency noise that inadvertantly arrives at the input.

R11 is a DC load put on the amplifier. I'm not sure why this is in the circuit, but some very old amplifiers needed something like this to avoid crossover distortion so that may be the reason. It also may assist in helping the quiescent bias point of the amp to be 0 volts, since the bias supply is a bit of a compromise. C11 is a DC blocking capacitor. Rload is, as its name implies, part of the load on the amp. Note that R11 and Rload are both loading the amp at AC frequencies.

R12 and R13 are a voltage divider that attempts to balance the battery voltage around ground so that Vcc is about equal to -Vcc. I believe that C12 is included to help stabilize the difference between -Vcc (and therefore Vcc as well since the battery has a very low internal resistance) and the circuit ground. Since the relationship between +/- Vcc and ground is through high resistances, this capacitor helps insure that AC current draw from the battery doesn't cause AC variation of Vcc and -Vcc with respect to ground.

I'll also mention that the symbols at the left connected to In1 and In2 are ideal voltage sources, apparently AC ones. These are theoretically perfect sources whose output resistance is zero ohms. We don't usually see symbols like this one in schematics unless the schematic is an exercise for school.
 
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Torben

Well-Known Member
Hi Mr. P,

I'm no pro but I'm gonna take a shot at going into some detail about what's going on here. As much for my own education as yours. If I blow something here I hope to be corrected. :)

The subcircuit in the top right corner is creating a virtual ground from a single voltage supply. The idea is that many opamps require a split supply (i.e. instead of just wanting a single supply, say, +15V and ground, they want +7.5V, ground, and -7.5V) but you often just have a single supply available. One solution is what they've done here: take the positive voltage, create a voltage halfway between it and 0V, and call this new middle voltage "ground". R12 and R13 create what's called a "voltage divider", and since they are of equal values, the voltage at the junction between them is halfway between Vcc and -Vcc.

I think it's accurate to say that this works because voltages are relative, not absolute: i.e. when we say a certain point in a circuit is at +5V, what we really mean is that it's 5V more positive than some other point in the circuit (if not specified, usually ground).

If you look at the main circuit, you'll see that the op-amp has connections to Vcc and -Vcc, which means it connects at those points to the points marked Vcc and -Vcc in the top-right subcircuit--which are the battery terminals. The middle supply from the subcircuit is then called "ground" or 0V and is connected to all the points in the main circuit marked ground (inverted triangle thing).

Here's where my knowledge gets a bit shakier but I'll take a stab at it. The main circuit is a simple active mixer (adder) with gain of 2.2 (R10/R1). R1 and R2 present the inputs with an impedance of about 10k. I don't have a lot of experience with these things but that seems low to me. R1 and R2 then join up to the op-amp's inverting input (marked '-'). Gain is set by the ratio of R10 to the input resistor (either of R1 or R2 in this case).

Capacitor C10 sets the high-frequency cutoff for the circuit. This allows lower frequencies to be amplified while the higher frequencies we don't want are attenuated. This is also called a "low-pass filter", since it passes lower frequencies more readily than higher ones. The filter is described by its "cutoff point" or "break-point", which is usually taken as the frequency at which the gain is -3dB. The formula to find the break-point is f = 1/(2piRC). In this case, R is R10 and C is C10, so the break-point is at 1 / (2 * pi * 22,000 * (47*10^-12)), which if I'm doing this right puts the break-point at 154 kHz.

If I'm doing that wrong will someone please beat me with a clue stick?

I'm not sure what R11 is for. C11 blocks DC from going to the output (it "capacitively couples" the output).

I hope this helps! I also hope I got most of it right. . .;)


Torben
 

RadioRon

Well-Known Member
Hey Torben, no mistakes, well done. By the way, 10K ohms at the input isn't particularly low. In the audio world, this would be called a high impedance input, since "low" might be 600 ohms, or 8 ohms, or 1K ohms depending on whose talking.

Ron
 

Torben

Well-Known Member
Thanks Ron! And it's good to know about the input impedance thing.


Cheers,

Torben
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Torben said:
Thanks Ron! And it's good to know about the input impedance thing.
10K isn't really a 'high' impedance input, it's much too low for a passive guitar for example - it's basically a 'line input' - and is probably a little on the low side for that as well?.

But looking at the circuit, it's obviously not intended as a practical design.
 

Torben

Well-Known Member
Nigel Goodwin said:
10K isn't really a 'high' impedance input, it's much too low for a passive guitar for example - it's basically a 'line input' - and is probably a little on the low side for that as well?.

But looking at the circuit, it's obviously not intended as a practical design.
OK, cool. But would it be safe to say that it's a good idea to look at what you're plugging in and adjust the input impedance accordingly? This would at least partly explain line/mic switches/jacks on audio mixer inputs.

I know this is a simple circuit (which is why I dove in) but I haven't really looked into how it could be better. I'll think on that.


Torben
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Torben said:
OK, cool. But would it be safe to say that it's a good idea to look at what you're plugging in and adjust the input impedance accordingly? This would at least partly explain line/mic switches/jacks on audio mixer inputs.
Yes, except the line/mic switch normally bypasses the front end for the mic as well, as you don't need as much gain (or usually a balanced input).

Generally your input impedance wants to be at least five times the output impedance of the source.
 

mr.pepper

New Member
Thanks all, really appreciate it.
A couple specific questions regarding the curcuitary.
1,Where are the 'pots' for the input of the guitar, output of 'blended signal(to amplifier), Loop send and loop return?
2,The Curcuit on the top right, the power curcuit, this goes at Vcc right?
3,The circular things on the left hand side of the curcuit, are these the stomp switches?
4,What do i do with the earths?
5,Which Resistor will i need to change to a variable resistor for the blend effect(from 100% clean(none effected) to 100% effected?

yeah noob questions i know :(
Thanks all!
 
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Nigel Goodwin

Super Moderator
Most Helpful Member
Like I said before, this isn't a full working circuit, merely a representation of a virtual earth mixer for simulation.

If you're wanting to build it, you need a proper circuit, and this isn't one.
 
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