OPA2889 datasheet(23/38 Pages) NSC

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INVERTING AMPLIFIER OPERATION

50W

750W

375W

261W

57.6W

Source

+5V

-5V

50W

0.1 F

6.8 F

0.1 F

6.8 F

Load

= 2

1/2

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

paragraph,

that

the

signal

source

gain resistor (RG) 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 RM 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), RG may

driving impedance of 50

Ω || 57.6Ω = 26.7Ω. This

be set equal to the required termination value and RF

impedance is added in series with RG 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 RM could be

bandwidth and noise performance. However, at low

eliminated

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 RG to 50Ω for

of Figure 50.

input matching eliminates the need for RM 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

2V/V

for

Ω

source

on the noninverting input (RB). If this resistor is set

impedance—the same as the noninverting circuits

equal to the total dc resistance looking out of the

considered

Figure

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) × RF. If the 50Ω 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 RF and RG values (see Figure 60), and then

Combining

this

resistance

parallel

with

the

achieve the input matching impedance with a third

feedback resistor gives the RB = 261Ω used in this

resistor (RM) to ground. The total input impedance

example. To reduce the additional high-frequency

becomes the parallel combination of RG and RM.

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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