# Purpose of 100 ohm resistor on guitar pedal output buffer

On the schematic for an Ibanez "Tube Screamer" there is a 100 ohm resistor in series with the output emitter follower:

The output impedance of this pedal is 1.2K:

It looks like RB has a negligable effect as it increases Zout to 1.3K

An explanation I've seen for this is:

To start with, the series resistance at the emitter limits the amount of drive available to drive an amplifier input (although not by much), and in concert with the series capacitor forms a voltage divider with the output shunt resistor and the input impedance of the amplifier plugged into it. This drops the available signal only a trivial amount, probably inaudibly so. (http://www.geofex.com/article_folders/tstech/tsxtech.htm)

Considering the next stage after this buffer is likely to be either a guitar amplifier or pedal with input impedance of approximately 1M ohm I don't understand why it is needed.

It is there to protect the transistor.

The collector of the transistor is connected directly to the 9V battery.

The base is biased such that there is probably around 4.5V at the emitter at all times.

A short circuit from the emitter to ground would destroy the transistor.

A phone plug like you use on guitar amplifiers and pedals can short its pins while plugging or unplugging a cable.

A large capacitor will allow a large current to flow for a short moment when you change the voltage across it.

Shorting the phone plug would short the emitter to ground through the output capacitor and destroy the transistor if that 100 ohm resistor weren't there.

It protects the transistor from damage if the output is accidentally shorted to ground or to another output.

It also isolates the emitter from capacitive loads on the output that the transistor may not be able to handle. Wires that connect equipment together have some amount of capacitance.