Datasheet

OPA690
50W
R
F
402W
R
G
200W
R
B
146W
R
M
67W
Source
DIS
+5V
-5V
R
O
50W
0.1 Fm 6.8 Fm
+
0.1 Fm
0.1 Fm
6.8 Fm
+
50 LoadW
OPA690
SBOS223F –DECEMBER 2001–REVISED FEBRUARY 2010 www.ti.com
INVERTING AMPLIFIER OPERATION 100Ω feedback resistor. This has the interesting
advantage that the noise gain becomes equal to 2 for
Since the OPA690 is a general-purpose, wideband
a 50Ω source impedance—the same as the
voltage-feedback op amp, all of the familiar op amp
noninverting circuits considered in the previous
application circuits are available to the designer.
section. The amplifier output, however, will now see
Inverting operation is one of the more common
the 100Ω feedback resistor in parallel with the
requirements and offers several performance
external load. In general, the feedback resistor should
benefits. Figure 43 shows a typical inverting
be limited to the 200Ω to 1.5kΩ range. In this case, it
configuration where the I/O impedances and signal
is preferable to increase both the R
F
and R
G
values,
gain from Figure 36 are retained in an inverting circuit
as shown in Figure 43, and then achieve the input
configuration.
matching impedance with a third resistor (R
M
) to
ground. The total input impedance becomes the
parallel combination of R
G
and R
M
.
The second major consideration, touched on in the
previous paragraph, is that the signal source
impedance becomes part of the noise gain equation
and influences the bandwidth. For the example in
Figure 43, the R
M
value combines in parallel with the
external 50Ω source impedance, yielding an effective
driving impedance of 50Ω || 67Ω = 28.6Ω. This
impedance is added in series with R
G
for calculating
the noise gain (NG). The resultant NG is 2.8 for
Figure 43, as opposed to only 2 if R
M
could be
eliminated as discussed above. The bandwidth will
therefore be slightly lower for the gain of ±2 circuit of
Figure 43 than for the gain of +2 circuit of Figure 36.
The third important consideration in inverting amplifier
design is setting the bias current cancellation resistor
on the noninverting input (R
B
). If this resistor is set
equal to the total dc resistance looking out of the
inverting node, the output dc error, due to the input
bias currents, will be reduced to (Input Offset Current)
× R
F
. If the 50Ω source impedance is dc-coupled in
Figure 43. Gain of –2 Example Circuit
Figure 43, the total resistance to ground on the
inverting input will be 228Ω. Combining this in parallel
with the feedback resistor gives the R
B
= 146Ω used
In the inverting configuration, three key design
in this example. To reduce the additional
considerations must be noted. The first is that the
high-frequency noise introduced by this resistor, it is
gain resistor (R
G
) becomes part of the signal channel
sometimes bypassed with a capacitor. As long as R
B
input impedance. If input impedance matching is
< 350Ω, the capacitor is not required because the
desired (which is beneficial whenever the signal is
total noise contribution of all other terms will be less
coupled through a cable, twisted-pair, long PCB
than that of the op amp input noise voltage. As a
trace, or other transmission line conductor), R
G
may
minimum, the OPA690 requires an R
B
value of 50Ω
be set equal to the required termination value and R
F
to damp out parasitic-induced peaking—a direct short
adjusted to give the desired gain. This is the simplest
to ground on the noninverting input runs the risk of a
approach and results in optimum bandwidth and
very high-frequency instability in the input stage.
noise performance. However, at low inverting gains,
the resultant feedback resistor value can present a
significant load to the amplifier output. For an
inverting gain of 2, setting R
G
to 50Ω for input
matching eliminates the need for R
M
but requires a
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