# AC Coupling using an Emitter Follower

Problem

Input signals often have a DC component that is not of interest and inconvenient to work with.  The DC component might be unknown or of a value that doesn’t match the circuit the input is driving.  Under standard linear circuit  assumptions, the response to the time-varying component of the input is independent of the DC component, so we can strip away the DC component and replace it with a more convenient value.  A blocking capacitor generally provides the “AC Coupling” of the input signal.

A problem with blocking capacitors arises when I try to select one for my audio circuits.  To the signal source, my circuit appears to have some input impedance, R, and the equivalent circuit becomes a single-pole high-pass filter:

The commonly-defined “cut-off” frequency for such a circuit is $f_{c}=\frac{1}{2\pi RC}$, and I often try to have a cut-off frequency one decade below my lowest frequency of interest.  To accomplish this I need a very large RC product, but the input impedance of my circuit is generally too low.

Here’s a recent example:

The input impedance of this circuit is $R\approx 8k\Omega$.  For ideal low-end response (down to 20 Hz or so), I would need a coupling capacitor of at least 10µF.  With a more practical coupling capacitor of 0.1µF as shown, I end up with $f_{c} \approx 200 Hz$ , which may be too high for some applications.

Idea

Because my biasing resistors, R1 and R2, largely determine the input impedance, I want  to replace them.  Simply using bigger values in the same proportion will not work (Q1 base current will begin to load the divider significantly).  What if I use the bigger resistors in conjunction with a transistor in emitter follower configuration?

Here’s the result:

Design discussion coming next…