OPA690IDBVT datasheet(20/37 Pages) TI1

20 / 37 page

OPA690

50W

402W

200W

146W

67W

Source

DIS

+5V

-5V

50W

0.1 F

6.8 F

0.1 F

6.8 F

Load

OPA690

SBOS223F – DECEMBER 2001 – REVISED FEBRUARY 2010

www.ti.com

INVERTING AMPLIFIER OPERATION

100

Ω feedback resistor. This has the interesting

advantage that the noise gain becomes equal to 2 for

Since the OPA690 is a general-purpose, wideband

source

impedance—the

same

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 100

Ω feedback resistor in parallel with the

requirements

and

offers

several

performance

external load. In general, the feedback resistor should

benefits.

Figure

shows

typical

inverting

be limited to the 200

Ω to 1.5kΩ range. In this case, it

configuration where the I/O impedances and signal

is preferable to increase both the RF and RG 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 (RM) to

ground. The total input impedance becomes the

parallel combination of RG and RM.

The second major consideration, touched on in the

paragraph,

that

the

signal

source

impedance becomes part of the noise gain equation

and influences the bandwidth. For the example in

Figure 43, the RM value combines in parallel with the

external 50

Ω source impedance, yielding an effective

driving impedance of 50

Ω || 67Ω = 28.6Ω. This

impedance is added in series with RG for calculating

the noise gain (NG). The resultant NG is 2.8 for

Figure 43, as opposed to only 2 if RM 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 (RB). 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)

× RF. 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 228

Ω. Combining this in parallel

with the feedback resistor gives the RB = 146Ω used

In the inverting configuration, three key design

this

example.

reduce

the

additional

considerations must be noted. The first is that the

high-frequency noise introduced by this resistor, it is

gain resistor (RG) becomes part of the signal channel

sometimes bypassed with a capacitor. As long as RB

input impedance. If input impedance matching is

< 350

Ω, 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), RG may

minimum, the OPA690 requires an RB value of 50Ω

be set equal to the required termination value and RF

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 RG to 50Ω for input

matching eliminates the need for RM but requires a

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