Physics 173 / BGGN 266 Primer on OpAmp Basics
David Kleinfeld, Spring 2008
An Operation Amplifier may be considered as a block that takes the difference of two-signals and
produces an output that is a high-gain version of this difference.
In the idealized case:
•
The output is Vout = A (V+ !!!V! ) with the gain, A, tending to infinity (maybe 106 in practice).
•
The inputs have infinite impedance (as much as 1012 Ω in practice).
•
The output has zero impedance (as low as 10 W in practice).
Ca se 1. Non -in v erti n g bu ff er
We have Vout = A (V+ !!!Vout ) or Vout =
A
V+ or Vout #A!"
##
! V+ .
A +1
USE AS BUFFER FILTER EXERCISE
Ca se 2. No n-i nv erti n g bu f fer w ith gain (u sed i n ex trac ellu la r/in tra c ellula r a mps )
Here
only
a
fraction
$
R '
Vout #A!"
##
! & 1 + 1 ) V+ .
R2 (
%
of
the
output
voltage,
R2
Vout
R1 + R2
,
is
sensed.
We
have
Ca se
3.
Cu rren t- to -vo lta ge
conv erter
(us ed
as
pho to diod e/p hotomu ltiplier
tub e
amplif iers)
We
have
(V! !!!Vout ) = I
R
Vout = A (V+ !!!V! )
and
so
that
Vout =
A
(V+ !!!IR )
A +1
or
Vout #A!"
##
! V+ !$IR . The input V+ is either grounded or set to an oddest voltage Voffset.
It is instructive to calculate the input impedance seen by the current source. We take V+ = 0 for
simplicity and find the impedance as Rin =
V!
R
. This as the gain does to infinity, the input
=
I 1+ A
to the current measurement goes to zero.
Case 4. Summing Amplifier with gain.
Here we convert input voltages into currents by passing them through a resistor, and sum these
currents with the above circuit.
The summing junction makes use of the effective low input
impedance Rin ≈ Rf/A. The current from input voltage V1 is V1/R1, that from V2 is V2/R2, etc. These
%V
V
V (
current sum so that Vout ###
! V+ !$R f ' 1 + 2 + 3 * .
& R1 R2 R3 )
A!"