參數(shù)資料
型號(hào): OPA2674IDR
英文描述: Dual Wideband, High Output Current Operational Amplifier with Current Limit
中文描述: 雙寬帶,高輸出電流運(yùn)算放大器,具有電流限制
文件頁(yè)數(shù): 20/30頁(yè)
文件大?。?/td> 472K
代理商: OPA2674IDR
"#$%
SBOS270 AUGUST 2003
www.ti.com
20
inductance can have a major effect on circuit performance.
A SPICE model for the OPA2674 is available through the
TI web site (www.ti.com). This model does a good job of
predicting small-signal AC and transient performance
under a wide variety of operating conditions, but does not
do as well in predicting the harmonic distortion or dG/dP
characteristics. This model does not attempt to distinguish
between the package types in small-signal AC
performance, nor does it attempt to simulate channel-to-
channel coupling.
OPERATING SUGGESTIONS
SETTING RESISTOR VALUES TO OPTIMIZE
BANDWIDTH
A current-feedback op amp such as the OPA2674 can hold
an almost constant bandwidth over signal gain settings
with the proper adjustment of the external resistor values,
which are shown in the Typical Characteristics; the
small-signal bandwidth decreases only slightly with
increasing gain. These characteristic curves also show
that the feedback resistor is changed for each gain setting.
The resistor values on the inverting side of the circuit for a
current-feedback op amp can be treated as frequency
response compensation elements, whereas the ratios set
the signal gain. Figure 10 shows the small-signal
frequency response analysis circuit for the OPA2674.
V
O
R
G
V
I
R
I
Z
(S)
I
ERR
α
R
F
I
ERR
Figure 10. Current-Feedback Transfer Function
Analysis Circuit
The key elements of this current-feedback op amp model
are:
α
= buffer gain from the noninverting input to the
inverting input
R
I
= buffer output impedance
I
ERR
= feedback error current signal
Z(s) = frequency dependent open
-
loop transimpe-
dance gain from I
ERR
to V
O
NG
NoiseGain
1
R
F
R
G
The buffer gain is typically very close to 1.00 and is normal-
ly neglected from signal gain considerations. This gain,
however, sets the CMRR for a single op amp differential
amplifier configuration. For a buffer gain of
α
< 1.0, the
CMRR = 20
log(1
α
)dB.
R
I
, the buffer output impedance, is a critical portion of the
bandwidth control equation. The OPA2674 inverting
output impedance is typically 22
.
A current-feedback op amp senses an error current in the
inverting node (as opposed to a differential input error
voltage for a voltage-feedback op amp) and passes this on
to the output through an internal frequency dependent
transimpedance gain. The Typical Characteristics show
this open-loop transimpedance response, which is
analogous to the open-loop voltage gain curve for a
voltage-feedback op amp. Developing the transfer
function for the circuit of Figure 10 gives Equation 14:
V
O
V
I
1
RF
RG
1
RF
RI
1
RF
RG
Z(s)
NG
RI
Z(s)
1
RF
NG
This is written in a loop-gain analysis format, where the
errors arising from a non-infinite open-loop gain are shown
in the denominator. If Z(s) were infinite over all frequen-
cies, the denominator of Equation 14 reduces to 1 and the
ideal desired signal gain shown in the numerator is
achieved. The fraction in the denominator of Equation 14
determines the frequency response. Equation 15 shows
this as the loop-gain equation:
Z(s)
R
I
R
F
NG
LoopGain
If 20 log(R
F
+ NG
×
R
I
) is drawn on top of the open-loop
transimpedance plot, the difference between the two
would be the loop gain at a given frequency. Eventually,
Z(s) rolls off to equal the denominator of Equation 15, at
which point the loop gain has reduced to 1 (and the curves
have intersected). This point of equality is where the
amplifier closed-loop frequency response given by
Equation 14 starts to roll off, and is exactly analogous to
the frequency at which the noise gain equals the open-loop
voltage gain for a voltage-feedback op amp. The
difference here is that the total impedance in the
denominator of Equation 15 may be controlled somewhat
separately from the desired signal gain (or NG). The
OPA2674 is internally compensated to give a maximally
flat frequency response for R
F
= 402
at NG = 4 on
±
6V
(14)
(15)
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