Datasheet
+
¨
©
§
¨
©
§
-1
2C
IN
P
1,2
=
1
R
1
1
R
2
r
1
R
1
1
R
2
+
2
-
4 A
0
C
IN
R
2
-R
2
/R
1
1 +
s
¨
©
§
¨
©
§
+
s
2
A
0
C
IN
R
2
¨
©
§
¨
©
§
V
OUT
V
IN
(s) =
A
0
R
1
R
1
+
R
2
C
IN
R
1
R
2
V
OUT
+
-
+
-
V
IN
+
-
V
OUT
V
IN
R
2
R
1
A
V
=
-
=
-
C
F
LPV531
www.ti.com
SNOSAK5B –MARCH 2006–REVISED MARCH 2013
INPUT CAPACITANCE AND FEEDBACK CIRCUIT ELEMENTS
The LPV531 has a very low input bias current (50 fA). To obtain this performance a large CMOS input stage is
used, which adds to the input capacitance of the op amp, C
IN
. Though this does not affect the DC and low
frequency performance, at higher frequencies the input capacitance interacts with the input and the feedback
impedances to create a pole, which results in lower phase margin and gain peaking. The gain peaking can be
reduced by carefully choosing the appropriate feedback resistor, as well as, by using a feedback capacitance,
C
F
. For example, in the inverting amplifier shown in Figure 43, if C
IN
and C
F
are ignored and the open loop gain
of the op amp is considered infinite then the gain of the circuit is −R
2
/R
1
. An op amp, however, usually has a
dominant pole, which causes its gain to drop with frequency. Hence, this gain is only valid for DC and low
frequency. To understand the effect of the input capacitance coupled with the non-ideal gain of the op amp, the
circuit needs to be analyzed in the frequency domain using a Laplace transform.
Figure 43. Inverting Amplifier
For simplicity, the op amp is modeled as an ideal integrator with a unity gain frequency of A
0
. Hence, its transfer
function (or gain) in the frequency domain is A
0
/s. Solving the circuit equations in the frequency domain, ignoring
C
F
for the moment, results in the following equation for the gain:
(6)
It can be inferred from the denominator of the transfer function that it has two poles, whose expressions can be
obtained by solving for the roots of the denominator:
(7)
Equation 7 shows that as the values of R
1
and R
2
are increased, the magnitude of the poles is reduced, and
hence the bandwidth of the amplifier is decreased. Furthermore, R
1
and R
2
are related by the gain of the
amplifier.
A
V
= −R
2
/R
1
, or alternatively
R
2
= −A
V
R
1
It is the presence of pairs of poles in Equation 7 that causes gain peaking. In order to eliminate this effect, the
poles should be placed in Butterworth position, since poles in Butterworth position do not cause gain peaking. To
achieve a Butterworth pair, the quantity under the square root in Equation 7 should be set to equal −1. Using this
fact and the relation between R
1
and R
2
, the optimum value for R
1
can be found. This is shown in Equation 8. If
R
1
is chosen to be larger than this optimum value, gain peaking will occur.
Copyright © 2006–2013, Texas Instruments Incorporated Submit Documentation Feedback 19
Product Folder Links: LPV531