Datasheet

INVERTING AMPLIFIER OPERATION
50W
R
F
750W
R
G
375W
R
B
261W
R
M
57.6W
Source
+5V
-5V
R
O
50W
0.1 Fm 6.8 Fm
+
0.1 Fm
0.1 Fm
6.8 Fm
+
50 LoadW
V
O
V
I
= 2-
V
O
V
I
1/2
OPA2889
OPA2889
www.ti.com ....................................................................................................................................................... SBOS373B – JUNE 2007 – REVISED AUGUST 2008
with a 750 Ω resistor across the two op amp inputs,
the voltage follower response is similar to the gain of
+2V/V response of Figure 51 . Reducing the value of
the resistor across the op amp inputs further limits the
frequency response due to increased noise gain.
The OPA2889 exhibits minimal bandwidth reduction
going to single-supply (+5V) operation as compared
with ± 5V. This behavior occurs because the internal
bias control circuitry retains nearly constant quiescent
current as the total supply voltage between the
supply pins is changed.
The OPA2889 is a general-purpose, wideband,
voltage-feedback op amp; therefore, all of the familiar
op amp application circuits are available to the
designer. Inverting operation is one of the more
common requirements and offers several
performance benefits. See Figure 60 for a typical
inverting configuration where the I/O impedances and
signal gain from Figure 50 are retained in an inverting
Figure 60. Gain of – 2V/V Example Circuit
circuit configuration.
In the inverting configuration, three key design
The second major consideration, touched on in the
considerations must be noted. The first is that the
previous paragraph, is that the signal source
gain resistor (R
G
) becomes part of the signal channel
impedance becomes part of the noise gain equation
input impedance. If input impedance matching is
and influences the bandwidth. For the example in
desired (which is beneficial whenever the signal is
Figure 60 , the R
M
value combined in parallel with the
coupled through a cable, twisted-pair, long PCB
external 50 Ω source impedance yields an effective
trace, or other transmission line conductor), R
G
may
driving impedance of 50 Ω || 57.6 Ω = 26.7 Ω . This
be set equal to the required termination value and R
F
impedance is added in series with R
G
for calculating
adjusted to give the desired gain. This consideration
the noise gain (NG). The resulting NG is 2.86V/V for
is the simplest approach and results in optimum
Figure 60 , as opposed to only 2V/V if R
M
could be
bandwidth and noise performance. However, at low
eliminated as discussed above. Therefore, the
inverting gains, the resultant feedback resistor value
bandwidth is slightly lower for the gain of – 2V/V
can present a significant load to the amplifier output.
circuit of Figure 60 than for the gain of +2V/V circuit
For an inverting gain of – 2V/V, setting R
G
to 50 Ω for
of Figure 50 .
input matching eliminates the need for R
M
but
requires a 100 Ω feedback resistor. This approach has
The third important consideration in inverting amplifier
the interesting advantage that the noise gain
design is setting the bias current cancellation resistor
becomes equal to 2V/V for a 50 Ω source
on the noninverting input (R
B
). If this resistor is set
impedance — the same as the noninverting circuits
equal to the total dc resistance looking out of the
considered in Figure 60 The amplifier output,
inverting node, the output dc error, as a result of the
however, now sees the 100 Ω feedback resistor in
input bias currents, is reduced to (Input Offset
parallel with the external load. In general, the
Current) × R
F
. If the 50 Ω source impedance is
feedback resistor should be limited to the 200 Ω to
dc-coupled in Figure 60 , the total resistance to
1.5k Ω range. In this case, it is preferable to increase
ground on the inverting input is 402 Ω .
both the R
F
and R
G
values (see Figure 60 ), and then
Combining this resistance in parallel with the
achieve the input matching impedance with a third
feedback resistor gives the R
B
= 261 Ω used in this
resistor (R
M
) to ground. The total input impedance
example. To reduce the additional high-frequency
becomes the parallel combination of R
G
and R
M
.
noise introduced by this resistor, it is sometimes
bypassed with a capacitor. As long as R
B
< 350 Ω , the
capacitor is not required because the total noise
contribution of all other terms will be less than that of
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