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OPA2889 Datasheet, PDF (23/38 Pages) Burr-Brown (TI) – Dual, Low-Power, Wideband, Voltage Feedback OPERATIONAL AMPLIFIER with Disable
OPA2889
www.ti.com ....................................................................................................................................................... SBOS373B – JUNE 2007 – REVISED AUGUST 2008
with a 750Ω 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.
0.1mF
RB
261W
+5V
+
0.1mF
6.8mF
1/2
OPA2889
RO
VO 50W
50W Load
INVERTING AMPLIFIER OPERATION
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
circuit configuration.
In the inverting configuration, three key design
considerations must be noted. The first is that the
gain resistor (RG) becomes part of the signal channel
input impedance. If input impedance matching is
desired (which is beneficial whenever the signal is
coupled through a cable, twisted-pair, long PCB
trace, or other transmission line conductor), RG may
be set equal to the required termination value and RF
adjusted to give the desired gain. This consideration
is the simplest approach and results in optimum
bandwidth and 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 –2V/V, setting RG to 50Ω for
input matching eliminates the need for RM but
requires a 100Ω feedback resistor. This approach has
the interesting advantage that the noise gain
becomes equal to 2V/V for a 50Ω source
impedance—the same as the noninverting circuits
considered in Figure 60 The amplifier output,
however, now sees the 100Ω feedback resistor in
parallel with the external load. In general, the
feedback resistor should be limited to the 200Ω to
1.5kΩ range. In this case, it is preferable to increase
both the RF and RG values (see Figure 60), and then
achieve the input matching impedance with a third
resistor (RM) to ground. The total input impedance
becomes the parallel combination of RG and RM.
50W
Source
VI
RG
375W
RM
57.6W
RF
750W
VO
VI
= -2
0.1mF
6.8mF
+
-5V
Figure 60. Gain of –2V/V Example Circuit
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 60, the RM value combined in parallel with the
external 50Ω source impedance yields an effective
driving impedance of 50Ω || 57.6Ω = 26.7Ω. This
impedance is added in series with RG for calculating
the noise gain (NG). The resulting NG is 2.86V/V for
Figure 60, as opposed to only 2V/V if RM could be
eliminated as discussed above. Therefore, the
bandwidth is slightly lower for the gain of –2V/V
circuit of Figure 60 than for the gain of +2V/V circuit
of Figure 50.
The third important consideration in inverting amplifier
design is setting the bias current cancellation resistor
on the noninverting input (RB). If this resistor is set
equal to the total dc resistance looking out of the
inverting node, the output dc error, as a result of the
input bias currents, is reduced to (Input Offset
Current) × RF. If the 50Ω source impedance is
dc-coupled in Figure 60, the total resistance to
ground on the inverting input is 402Ω.
Combining this resistance in parallel with the
feedback resistor gives the RB = 261Ω used in this
example. To reduce the additional high-frequency
noise introduced by this resistor, it is sometimes
bypassed with a capacitor. As long as RB < 350Ω, the
capacitor is not required because the total noise
contribution of all other terms will be less than that of
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