Help With Reading Schematic - Very Beginner

[jacobcarlgrant]

I'm tackling my first build, which is the Moodring reverb. I just received all of the components, as well as the PCB. Before I solder everything into place, I'd like to test it on a breadboard, which is where I'm stuck. On the schematic for the Moodring, there are certain "tags" that say VB and VC, and they usually come from the positive side of an op-amp or from a transistor. Where do they go? Also, there are a few circles with an X through them, which usually say "BYP.1_IN" or something similar. Where do these go? I know, these questions are beyond noob level. Any help is appreciated!

[diablochris6]

I would say that this circuit would be pretty intense to breadboard, especially if you are just testing it out. I would just go for it and put it together, using sockets for ICs and the 5V power regulator.

As for the VC and VB question, look at the power section of the schematic on the top right corner. The +9V power jack connection goes through a protection diode and VC is on the other side; VC is your +9V. Keep following the VC until it hits R35 and R36. Both of the resistors form a voltage divider, creating a +4.5V source. This is your VB. If any part of the schematic has a VC tag on it (the + power pins of the TL072), that means +9V is going to it. The parts of the circuit with the VB tag is receiving the +4.5V reference voltage.

I don't see the circles with "BYP.1_IN" on the schematic I am looking at, but it sounds like those are the solder pads to activate the effect and cut out the dry signal. Both of the bypass pins should be connected for the effect to be heard, especially if you are just testing out the effect. Deciding if you want a buffered bypass or true bypass will determine how you wire the pads up to the footswitch later. I could be wrong on this part though...

[Boba7]

A moodring is not an easy build, I would not recommend it as a first build!

Why breadboard it? Populate the board, solder the pots with wires, and test it directly, it'll be easier

VB, VC etc are usually connected to power, whether it is 4.5v, 9v, 18v, 5v with a regulator)
Dont have the build docs on hand, Ill look them up tomorrow

[jacobcarlgrant]

Thanks for the info! I'm just gonna go for it. I've soldered before, wired up a few guitars, but nothing quite this involved. Sockets are on the way, so I don't fry the ICs. Thanks again for the advice!

[midwayfair]

Part 1.
Start in the upper right corner.

This is the power section. All electronics need a power source, and a ground, or common, which is 0V. The power source is measured in relation to ground, and is the supply voltage. It might be positive or negative; in this case, you're asked to supply 9V of DC (direct current), which is the standard for most pedal circuits because historically many ran on 9V batteries.

The circuit is all about controlling how the electricity flows from the supply voltage to ground. Many things in the circuit will be voltage dividers, which are made up of resistors, which prevent the flow of current. If you put two resistors in order and connect them to each other +...-R1-o-R2-...- the point in the middle, o, will be the ratio of voltage V = v_in * R2/(R1+R2). One of the most important things anyone can possibly realize about electronics is that Everything useful in a circuit is a voltage divider.

The other component you will see often in guitar pedals is a capacitor. Capacitors block the flow of DC but allow AC to pass. They can also store a charge, and some of them have a + and - end.

If you combine a capacitor and a resistor, you get what's called an RC network; it will pass "high" frequencies when arranged like this (high pass filter):
-C-|---out
    R
    |
    Ground

And will pass "low" frquencies when arranged like this (a low-pass filter):

-R-|---out
    C
    |
    Ground

Nearly all tone shaping is done with these in a guitar pedal. (I won't go into inductors here).

Then there are semicondctors. Resistors and capacitors allow DC to pass both ways; semiconductors allow it to pass only one way. A diode, like D1 immediately after the 9V input, points toward the lower voltage. So the positive (+ve) voltage comes in, is allowed to pass from the anode (+) to the cathode (-) of the diode, and only a tiny bit of voltage is lost. If you instead plugged in -9V, then the DC would not be allowed to pass from the anode, because it's the wrong polarity. Ok?

The LED is also a semiconductor. When the right amount of direct current passes through it, it lights up. Too much and it blows up. How do we prevent that? We restrict the flow of direct current while still allowing it to pass. What part does that? Right, a resistor. R34.

C34 and C35 are connected to the power source and ground. There is a hidden resistor here -- it's in the power supply before it even gets to the circuit. This is an RC filter. What frequencies does it cut? (Look above.) Why? It turns out that power supplies sometimes make noise. This cleans it up.

I'll come back to the 8, 4 stuff.

R35 and R36 form a voltage divider. They're the same value. What's that? It's a voltage divider. Calculate the voltage that will appear on the little Vb flag based on the formula given above.

C36 is the last thing in the circuit. It's connected to a "power source" Vb and ground. Does its job look familiar to you? What kind of filter is it forming?

[midwayfair]

Part 2.

We're going to move onto the next section, but a good thing to keep in mind about schematics is that you don't have to physically draw a line from one section to another. You can just label things. So anything that connects to "Vb" refers back to that little flag from the power section. Anything that connects to "Vc" refers to that flag from the power section.

Let's go to the top left. This is the audio input, starting with "In."

Your guitar pickups produce an AC voltage. A capacitor, C1, will allow that to pass, but it will block DC. Why block DC? Well, we're going to carefully manage the DC that our transistors (short for "transference resistor" -- yes, they will work like a resistor) are allowed to see on their various pins. Remember how I said capacitors hold a little charge? Well ... without getting too deep into it, if there's any on the input side of C1, even unintentionally (including static!), C1 would hold a little bit of it. We want to bleed that off to ground but not create a heavy voltage divider with the pickups (which have a hidden resistance not shown in the circuit). R1 forms a voltage divider with the pickup, but it won't lose much signal. It will, however, allow DC to pass (slowly) and go away to ground. That gets rid of built-up charge on C1. The reason is that otherwise a change in voltage when the pedal's switch is thrown would manifest as a pop -- because such a change in voltage wouldn't look much different from the AC voltage your guitar pickups are producing. It's just a different frequency and it's pretty loud and sudden.

The next little block we encounter is this little triangle thing. It's an IC, integrated circuit, and it's made up of a bunch of transistors. This type, TL072, is called an operational amplifier. Its job is to use the pedal's supply voltage to amplify either, or both of, the input signal's volume (which is smaller than the 9V supply) and current (which is far weaker than the 9V supply). I don't feel like fully explaining op amps, but in this case, we're going into the - pin, which is known as an "inverting" op amp. It will flip the polarity of the signal. There are two of them in the circuit, so we'll only really look at one.

Op amps use something called "negative feedback." Look at R2 and R3. R2 is about half the size of R3. If we look up the datasheet or some literature on this type of chip, we would discover that the gain (signal voltage amplification) of this arrangement will be R3/R2, or about 2x. The guitar signal will be twice as loud on the output pin (marked 1). Think about the term "negative feedback." You're probably familiar with positive feedback -- put something in a system and it stays. Like pointing a microphone at a speaker. More positive feedback means more signal. The more negative feedback you provide, the opposite happens -- less signal. R3 is preventing negative feedback. If you make it bigger, you will get less negative feedback -- and thus more signal. If you make it smaller, you will get more negative feedback. What do you think might happen if it was a wire (0 Ohms)?

Op amps are happy to operate when you give them three things: A supply voltage, in this case on pin 8, a ground, in this case on pin 4, and a reference voltage, in this case on pins 3 and 5, which will dictate the resting voltage for the other pins. If the reference voltage is halfway between the supply voltage and ground, then the input signal will be allowed to swing from the middle all the way up to 9V and all the way down to 0V equally. If it's closer to one or the other, one side of the wave might misbehave. What's Vb in this circuit?

C2 filters some high frequencies. The AC voltage it allows to pass is never amplified. Or more to the point, it looks like a wire for high frequencies. Again: What do you think would happen if R3 was a wire? What do you think would happen if C2 was bigger? Smaller?

The amplifier circuitry in IC1_B is similar, except that the "gain" control is a variable resistor. What do you think happens when you turn this control so that the resitance is bigger? Smaller? What are the maximum and minimum gain factors you can get out of this (look at the formula above).

IC4_A is a different kind of amplifier. If we look at the datasheet for reference, we'll find that this is called a non-inverting amplifier when we go into the + pin (pins 3 or 5). Here there's no resistance at all between pins 2 and 3. Specifically, this is a type of circuit usually just called a buffer. Or more formally, a current buffer. Its point is to provide a strong signal with a low source impedance at the output pin1, so that any circuit that follows won't make a big voltage divider and lose signal.

Next we'll look at the power supply for the brick and the PT2399 and then the brick and PT2399 themselves.

[midwayfair]

There are two regulators in this circuit, REG1 and REG2. Looking up the part number, these are 5V regulators. Their job is to take an input voltage, Vc in this case, and knock it down to a steady output voltage, 5V in this case.

5V is the typical operating voltage for a lot of digital stuff (3.3v is also popular). At their o pin, you can just think of that as another voltage source, like Vb or Vc. What are C29 and C30 doing? Do you recognize their function from elsewhere in the circuit? Knowing that, can you guess what C17 and C18 are probably doing. Go one step further: What do you think must Pin 2 of the BTDR-2 integrated circuit module be acting like? (If you were to remove all the epoxy and trace the insides of the brick, you'd discover that there is a very, very tiny resistor connecting it to pin 4).

Some stuff with the digital chips will be mysterious, because they aren't completely open. But we can go along the bottom of the PT2399 first. The datasheet (always look up the datasheet if you want to know how something works) tells us that if we increase the resistance between Pin6 and ground, the chip will provide a longer delay. There's a variable resistor, "SPACE," that will do this for us.

Most of the other components along the bottom of the chip are suggested by the datasheet. They get rid of digital noise or help smooth out supply voltages or something. D2 -- not something found in the datasheet -- limits the input signal size that the little digital circuit can play with. Since the digital circuit only gets 5V to play with, not the 9V that the audio circuit got, it can distort -- that's what happens when the signal size exceeds the supply voltage. If you're curious about the LED, I discuss it somewhere a long time ago in my Hamlet delay, which is the source of this modification to the PT2399's wrapper circuitry. I discovered it by playing around on a breadboard. I encourage you to play around on a breadboard even if a circuit seems complicated, but break the circuit down into manageable parts so you can stay organized.

The other side of the PT2399 actually has a couple op amps; pins 13-16 are inverting op amp stages. They look a little funny compared to IC1_A, but maybe you can spot the similarities.

So the guitar input goes through C1, into IC1_A, out pin1, through C5, R11, and R12, through IC4_A, out pin1, through C9 and R13, and then it's split: through R14 into the inverting op amp stage (pin16 input, pin15 output), and then mixes with the pin15 output and the output from R31/C27 coming out of a part type we haven't seen yet: A JFET, a type of transistor.

This is just a JFET buffer. It has a supply voltage, Vc, and a limiting resistor, R8, to prevent all the supply voltage from rushing off down to ground, which would put a lot of current through the JFET and dry it pretty quickly. We always need some resistance somewhere. The input pin, called the gate, can be "wiggled" up and down by the audio signal, which will change the resistance between the other two pins (top and bottom in the schematic, called the drain and gate respectively) of the JFET. The result is that the DC supply voltage becomes an AC voltage, just like all he other audio signals in the circuit, except that it has lots of strong current like the supply voltage, and a low output impedance (to prevent losing signal to voltage dividers at the beginning of the next piece of a circuit it encounters).

Anyway, if you look closely, you can see that the inverting op amp stages of the PT2399, Pins 13-16, are actually labeled as LPF, for low pass filter. Think about what C10 and C11 are doing.

The signal out from Pin 14 goes through R19. What are R19 and C15 doing? (Look up a couple posts if you've forgotten.)

C16 blocks the DC from pin14. Kind of like how C1 blocked any DC before the audio reacted the pedal. We need to do this so that the LEDs D3 and D4 will do their job.

D3 and D4 are providing a similar function to D2, except where D2 was limiting the signal size for the digital part, these are limiting the signal size for the audio part. When diodes (LEDs are a type of diode) are arranged like this, they will conduct -- and clip -- both sides of the AC audio waveform when it exceeds their forward voltage, which is about 1.7V. This means that if your audio signal ever gets to be about 3.4V peak to peak in the delay circuitry, it'll distort. But the idea is that LEDs sound better than the PT2399's op amps distorting.

Finally, after R20, the circuit goes through the belton brick. This is actually several PT2399 chips in a row inside there, but it's hidden -- you can't play with it!

Next we'll look at the final parts of the circuit: the mood control and the dwell control.

[midwayfair]

Part 4.

The Belton brick has two outputs, a dry signal and a delayed signal, which are mixed together through R23 and R24. C19 is yet another coupling capacitor to block the DC that was on the brick's pins.

Now look at C20 and R26 all by themselves. What type of filter is this?

Now look at R25 and C21 by themselves. What type of filter is this?

Now look at the mood control. This is a potentiometer wired as a voltage divider -- the other ones we've looked at so far have just been variable resistors. Think of this as two resistors sitting next to each other like in the first post, and the wiper is the little o between them moving back and forth. What happens when the wiper (lug 2) is all the way at the top? What filter is the signal going to see primarily? What happens when the wiper is all the way at the bottom? What filter will the signal see primarily?

C22 leads to another op amp stage, this time with a Gain control.

After R30, the signal is split again: The Ring control is a voltage divider -- in this case just a volume control. As the wiper moves toward the bottom, the signal gets closer and closer to ground, which is silence. Turn it up and more of the delayed signal will get sent to pin 6, which is the input of the IC1_B amplifier stage. The dwell control is also a voltage divider. As you turn it up (less signal goes to ground), more signal is allowed to feedback into the PT2399. When the guitar signal gets too quiet, the PT2399 won't make a delay signal from it, so allowing more signal to feedback means the delay "works" longer -- that is, you'll get more repeats. The output of this control goes to the JFET we already talked about.

At long last, we have our delayed signal from the Ring control through R32 mix with our original dry signal, which came out of pin 1, through R4 and R5. (The "dry" switch prevents the dry signal from getting through by severing this connection.) The mixed signal is then amplified by IC1_B. (See if you can figure out the minimum and maximum gain similar to what we did with IC1_A.) R6 provides a little buffering and safety for the op amp (it's complicated, actually), and then C4 blocks the DC from the op amp so that the next circuit in line is safe. (Bonus questions: What kind of filter is C4 and R7? Go find a calculator or the formula -- what is the cutoff frequency of this filter?)

N.B. the symbol labeled Byp1_out is just an extra pad on the PCB.

[benny_profane]

Jon, thank you for the incredibly thorough and educational analysis. It's very appreciated.

[Boba7]

That is great Jon, thanks a lot for your work, patience and time

[Caedarn]

Jon - Great walk-thru, helpful to folks like me who are new and learning.  Thanks very much!

[OhmBrewmeister]

I joined the forum just to agree with this - a really excellent beginner tutorial that will no doubt be invaluable to many! Thanks!