Editing Buffer amplifier (section)

==Current buffer==
Typically a current buffer amplifier is used to transform a current signal with a low [[output impedance]] from a first circuit into an identical current with high impedance for a second circuit.<ref>{{Cite web|url = http://web.mit.edu/6.012/www/SP07-L20.pdf|title = Lecture 20 - Transistor Amplifiers (II) - Other Amplifier Stages|quote = A current buffer takes the input current which may have a relatively small Norton resistance and replicates the current at the output port, which has a high output resistance ... Input resistance is low ... Output resistance is high ... transform a current source with medium source resistance to an equal current with high source resistance}}</ref> The interposed buffer amplifier prevents the second circuit from loading the first circuit's current unacceptably and interfering with its desired operation. In the ideal current buffer (Figure 1 bottom), the output impedance is infinite (an ideal current source) and the input impedance is zero (a short circuit). Again, other properties of the ideal buffer are: perfect linearity, regardless of signal amplitudes; and instant output response, regardless of the speed of the input signal.

For a current buffer, if the current is transferred unchanged (the current [[gain (electronics)|gain]] ''β<sub>i</sub>'' is 1), the amplifier is again a '''unity gain buffer'''; this time known as a '''current follower''' because the output current ''follows'' or tracks the input current.

As an example, consider a [[Norton's theorem|Norton source]] (current ''I<sub>A</sub>'', parallel resistance ''R<sub>A</sub>'') driving a resistor load ''R<sub>L</sub>''. Because of [[current division]] (also referred to as "loading") the current delivered to the load is only ''{{sfrac|I<sub>A</sub> R<sub>A</sub>|R<sub>L</sub> + R<sub>A</sub>}}''. However, if the Norton source drives a unity gain buffer such as that in Figure 1 (bottom, with unity gain), the current input to the amplifier is ''I<sub>A</sub>'', with ''no current division'' because the amplifier input resistance is zero. At the output the dependent current source delivers current ''β<sub>i</sub> I<sub>A</sub> = I<sub>A</sub>'' to the load, again without current division because the output resistance of the buffer is infinite. A Norton equivalent circuit of the combined original Norton source ''and'' the buffer is an ideal current source ''I<sub>A</sub>'' with infinite Norton resistance.