I have seen lots of posts where people state that they cannot read schematics, and lots of people asking for sources of basic electronics knowledge. I wrote this post to fill some of that need. I hope it helps someone. It is far from complete and contains many approximations, but I think it may fill in some of the gaps that cause beginners to get stuck. Feel free to ask questions, correct mistakes, or dispute anything I've written.
How to read schematics:
The standard pedal schematic can be read from left to right and top to bottom:
----------------------------------------------------------------------------------------------
High voltages at the top
Inputs on the left Outputs on the right
Low voltages at the bottom.
----------------------------------------------------------------------------------------------
Not coincidentally, this is a list of the four connections that are necessary for any guitar pedal: V+, V-, input, output.
Look at
this beautiful schematic. In many ways, this schematic is drawn perfectly. The supply voltage is given by VCC, but a value isn't given, its just some positive number. In principle it could be a negative number, but it is a convention (not universal, but more common than not) to connect ground to the negative part of the circuit and therefore to the bottom of the picture. If VCC
was a negative number, you would not be considered odd for drawing the circuit vertically flipped. Likewise, with a bipolar power supply, you usually draw V+ on the top, ground in the middle and V- on the bottom.
It is considered good design practice not allow DC voltages on inputs and outputs, so if an instrument is plugged in but not playing anything, we can take it as a given that the input of the circuit is held at zero volts. But if the circuit is bypassed in a way that doesn't ground the input (or otherwise tie it to some reference voltage) then we can't automatically assume those nodes are held at zero unless they are directly connected to ground in the schematic. Otherwise they are free to float to any value. With a pulldown resistor or other ground reference in place, any excess voltage will bleed to ground and the input or output node will come to rest at zero.
So what's a node? For the sake of simplicity, we'll pretend that jumpers have no resistance, so you can consider all connected jumpers to have exactly the same voltage at all times. That means that in the beautiful schematic, all directly connected straight lines will have the same voltage. That set of connected straight lines is called a 'node'.
If the input voltage is held at zero, then every node in the schematic will sit at a rest voltage. This is called the 'idle' or 'quiescent' state. In the idle state, electrons flow steadily upward from ground and into the VCC node. Since electrons have negative charge, traditional current points the opposite direction, flowing downward. If you pick any vertical line in the schematic and follow it upwards, voltages will increase as you cross components. Notice that the source and drain of the jfets follow a single vertical line, but the gate is offset to the left. Thus we can assume that voltage increases across the body of the jfet from source to drain, but we can't assume that the voltage increases from source to gate.
If we calculate the voltages of the nodes at idle, we are said to be doing a DC analysis. The non-black straight lines in the beautiful schematic indicate nodes of reasonably well known DC voltage that don't depend very much on the specifics of the jfets used to build the circuit. Note that each color is a single node. The black lines indicate nodes where the voltage will depend on the jfet particulars. The black nodes will not all necessarily have the same voltage as each other. Only the points that are directly connected by straight lines will. Notice that all the points marked 'GND' can be considered to be connected as well.
So how do you calculate the DC voltages? For specifics, you'll need to learn the math. But we can make some broad generalizations:
Resistors drop voltage. They slow down electric current, essentially by placing barriers in the paths of the electrons that travel through them. Voltage drops are proportional to resistance and you can assume a higher voltage on any leg pointing up in the beautiful schematic. So though you might have to do some calculations to find the exact voltage of given node, you can often tell which of any two nodes will be have a higher voltage at idle. Resistors behave the same for AC and DC, so they're an easy component to learn.
Capacitors: At the simplest level of approximation, capacitors can be considered closed circuits (jumpers) for AC, and open circuits for DC. So when we do a DC analysis, we can mentally remove all of the caps, leaving a hole in their absence. In other words, the cap values have absolutely no bearing on the DC voltages in the circuit. If you think about it, that also means that you can place a cap across absolutely any two random points in the circuit without changing the idle voltages. This is because capacitors don't allow electrons to flow through them. Physically, a cap is two parallel conductive plates, separated by a dielectric (a polarizable insulator). The electrons can't get past the insulator, but they can sense changing voltages on the opposite side of the gap through the polarization of the material inside the gap.
If you want to determine the correct orientation of a polarized cap in a schematic that follows the voltages-increase-upward convention, it's pretty easy. The positive leg points upward. If the schematic doesn't obey the convention, you can just redraw it so that it does. With a little practice you can do it in your head.
But what if the input isn't held at zero? Mentally remove the caps from the schematic, and you will see that the input is no longer connected to any other known voltage, and is free to float to whatever voltage the prevailing conditions see fit. If you add a DC offset to the input or output, the idle voltages of the rest of the circuit won't change.
But if you add a time-varying voltage to the input, then we switch from DC to AC analysis. Caps act like jumpers for AC, so as a first approximation, mentally replace all of the caps that touch the signal path with jumpers, But because caps are open circuit for DC, we have to ignore any pre-existing voltage differences across the caps. When the voltage on one side of the cap (let's call it the 'input side') goes up, the voltage on the opposite side of the cap goes up at the same time. Physically what happens, is that by raising the voltage on one side (we'll call this side the 'input side', and the other side the 'output side'), we are removing electrons from the cap plate on the input side and simultaneously adding an equivalent number of electrons on the output plate of the cap. Those electrons were extracted from whatever the output leg of that cap is connected to, which means that the voltage on the output node increases do to the removal of negatively charged electrons. There is a slight time lag for the transmission of this signal across the cap. The lag for larger capacitors because capacitance (C) is defined by how much charge (Q) will collect on each of the capacitor plates at a given voltage (V), or C = Q/V. Large caps require more electrons to accumulate on one plate of the cap and simultaneously vacate the opposite plate to become fully charged, so they take a longer time to charge up than smaller caps.
But caps aren't really jumpers. A better approximation is that caps are frequency-dependent resistors, but remember, electrons don't actually flow through caps, even though it looks like they do. What happens is that charge flows into one side of the cap and builds up on the plate, while simultaneously an equal number of electrons leave the opposite plate. That makes it look like the electrons are passing through the cap, but the electrons added to one plate of the cap are not usually the same electrons that are removed from the other plate. Caps provide more resistance (technically called capacitive reactance) at low frequencies than at high ones. The resistance comes from the electric field created by the electric charges on the plates. Like charges repel, so it's harder to push electrons onto a cap plate that is already full of electrons and harder to remove electrons from a cap plate that is already depleted. Low-frequency signals give the caps more time to charge up before they reverse polarity, while higher frequencies don't allow the charge to accumulate very much, so they don't push back against the current as much. The larger the cap, the longer it takes to fill up the plates with charge, so the larger a cap is, the less it will resist low frequencies. This is why caps are useful for frequency-filters. By intelligently applying resistance to high or low frequencies, we can attenuate them separately.
Notice on the beautiful schematic, that VCC is drawn at the top left. Obviously it should be drawn at the top because it is the highest voltage in the circuit, by why is it drawn on the left? Because some power supplies are not perfectly stable. Crappy wall-warts often carry high-frequency noise, so it makes sense to draw VCC on the left, because it is a source of signal. Notice that VCC is attached to a low-pass filter. (A resistor followed by a cap to ground with output taken at the junction is a frequency-dependent voltage divider. Homework: calculate the output voltage of this divider for 20 Hz, 1000 Hz, and 10 kHz. Here's a hint: Vout=Vin*Xc/(R+Xc). R=100R, Xc=1/(2*pi*f*C), C=100u). Three of the jfet stages use this filtered voltage as supply, then the supply is filtered again and fed to the remaining three stages. This filters the signal voltage out of the supply line so that the output doesn't get fed back into the input, which could cause oscillation.
As mentioned above, if the circuit is drawn according to the convention shown at the top of this post, you can often read it like a chart of voltages going from high to low as you follow the lines downward. Often, however, it is inconvenient to follow the convention, so you will find schematics like
this one that mostly follow the convention but break it here and there to make it easier to print. Note that the circuit is broken in half, with the left half on top and the right half on the bottom, connected by the two dots labelled A. This is done so it can be printed on a standard piece of printer paper. Notice that the top op-amp is drawn with ground pointing upward. This part would have been more awkward to draw with ground pointing downward. Sometimes components are rotated to make the schematic more dense and thus more compact, which can have the side-effect of moving lower voltages to higher placement on the schematic.
Any schematic with an IC drawn like a physical chip will have a hard time following the convention, but the internal functions of the IC can often be replaced by other symbols that fit into schematics more symbolically, and thus make it easier to trace voltages by sight.
If a schematic is difficult for you to read, or to deduce DC voltages. Try redrawing it so that the higher voltages are at the top, lower voltages are at ground, input is at the left, and output is at the right.
Locked
). As long as the signal voltage is below the forward voltage of the diode (let's say +0.7V), no current flows and it's like the diode is an open circuit (i.e. not there). Once the signal voltage tries to go above +0.7V, current starts to flow to ground and the voltage is prevented from going any higher (ideally) and hence stays near 0.7V. Therefore any part of your signal above 0.7V is clipped. Now, place another diode downstream of that and reverse the orientation (cathode connected to the signal path). In this case, all voltages below -0.7V are clipped. Hey, symmetric clipping! Of course, you can go hog wild at this point and use different diodes, diodes in series, etc to get different clipped waveforms. Note that the more the signal is clipped, the lower the amplitude of the signal will be, so you may need to add a gain stage to pump up the signal level again.
