# Common-collector gain for guitar pedal

I am currently working on a distortion pedal. I have just made the initial input buffer to prevent loading the source signal. It is simply an NPN transistor in an emitter follower (common collector) mode. Here is the circuit:

And here is the simulation of the circuit:

The green wave is node 1 (just an oscillating sine source), and the blue wave is node 2. Why does the blue wave get boosted to oscillate about 3.7 V? I am not sure why the sine source is not adding with the 4.5 V to produce an oscillation about 4.5 V?

• Marko, do you need the output to be offset by some particular voltage? Because the circuit you have will have a DC bias that is highly variable depending upon the specific BJT (in practice, anyway.)
– jonk
Commented Aug 22, 2022 at 5:22
• Marko, for example, I probably would expect about $\pm \,10\%$ change in the output voltage over the typical range of a given BJT's $\beta$ values within the same family (2N2222, for example) given your values. About 80% of the VBE over temperature variations will appear at the output, as well. (Close to -2 mV per degree Kelvin.) Both VBE and $\beta$ will play in the variations. All together you may be looking at around $\pm 400\:\text{mV}$ variation at the output. It's not a precision circuit.
– jonk
Commented Aug 22, 2022 at 5:34
• Use AC coupling on your output to eliminate the DC component. I.e. add a capacitor in series with your output. The capacitor will have a high pass effect, so should be large enough so that frequencies of interest to you are not attenuated. The break frequency depends on both the capacitor size and the effective impedance at that point. $f_0=\frac{1}{2\pi RC}$ Commented Aug 22, 2022 at 5:57

DC Current ($$\I_{B}\$$) flows from 4.5V through $$\R_{2}\$$, through the base emitter junction, the though $$\R_{3}\$$ to ground. The current ($$\I_{E}\$$) through $$\R_{3}\$$ is actually $$\I_{B}\$$ + $$\I_{C}\$$ the collector current. The voltage from point 2 to ground is: $$V_{2} =V_{R3}+V_{R3} = 3.7V$$ The correct way to say: The signal is level shifted from 0V dc at node 1 to 3.7V dc at node 2. Boost usually implies amplification. The resistor $$\R_{2}\$$ drops the voltage from 4.5V to 3.7V.

Why does the blue wave get boosted to oscillate about 3.7 V?

The base-emitter junction of the BJT needs to be properly forward-biased and. this produces a volt-drop of anything between 0.5 volts to 1 volts between the bias voltage on the base and the emitter voltage. Inevitably, the average emitter voltage is about 0.7 or 0.8 volts lower than the average base voltage.