Being a guitar player that is also going to school for an Electrical Engineering degree, I'm tempted to use all the knowledge I'm gaining to build things pertaining to music production and guitar. While in the Circuits series of classes, I learned the basics of op-amp circuits, so I decided to give building a clean boost pedal a try. All a clean boost pedal does is amplify the signal from the guitar without any distortion or other changes in sound. The purpose of a clean boost is to overdrive a tube guitar amplifier's input circuit. Tube distortion is considered pleasing to the ear and is how most guitar players achieve the desired tone from their guitar, and boosting the signal before it enters the amplifier is a nice way to get more tube distortion.
A dual-supply op-amp circuit that simply amplifies a signal is an especially easy circuit, requiring only the power supplies, the op-amp, and two resistors, and it looks something like this:
If you pretend that the source V1 is a guitar pickup, this circuit would do the job, and give a gain dependent on the values of the two resistors: A=1+(Rf/Ri).
The problem is that the most widely used type of power supply used by most pedal manufacturers is a single 9V supply. So while I could design a pedal with a dual supply, it would not be compatible with most supplies on the market. That being the case, I decided to build a circuit using a single 9V power supply.
Circuits never taught us how to design a single-supply op-amp circuit. Thank goodness for Google.
After some searching, I found some great application notes on single-supply op-amp design. The idea is pretty simple. The positive voltage supply is hooked to 9V, the negative to ground. The op-amp can then operate within that voltage range. In order to amplify an alternating signal, it needs to be centered at one half the supply voltage, that way there is the most "room" for the signal to increase to 9V on the positive side of the signal, and decrease to 0 on the lower side of the signal. Luckily for us, a circuit can behave differently for AC and DC signals. The circuit can be designed to have a gain of 1 for the DC signal and a gain any thing we'd like for the AC signal by using coupling capacitors.
The voltage divider network using R1 and R2 create the bias voltage of the circuit, and if they are the same splits the source voltage in half, providing the required 4.5V. The capacitor C1 blocks the DC portion of the negative feedback from going to ground, creating a DC gain of 1, and the DC signal at the output is still simply 4.5V. The AC gain can be set normally using R1 and R2. C3 blocks the bias voltage from entering the source, but allows the AC source voltage to "ride" on top of the bias voltage. Capacitor C2 removes the 4.5V DC portion of the signal from the output, and the output can then be referenced to ground as normal.
In the case of my pedal, I wanted as much boost as possible. The maximum output of a passive guitar pickup (active guitars have built in preamps, negating the need for this pedal, usually) is around 400mV. I want the output voltage to swing from 1V to 8V (the op-amp I am using, OPA134, will swing within a few hundred milivolts of the source, so keeping it at 1V and 8V will keep it from accidentally clipping), so I half that, 7/2=3.5, then divide that by 0.4. 3.5/0.4=8.75. So an Rf value of 87.5k ohms and an Ri value of 10k ohms will give the desired gain.
How to add a volume knob, tone knob, and calculate capacitor values coming up on the next post.