Datasheet
MCP660/1/2/3/4/5/9
DS22194D-page 24 2009-2012 Microchip Technology Inc.
4.4.2 GAIN PEAKING
Figure 4-8 shows an op amp circuit that represents
non-inverting amplifiers (V
M
is a DC voltage and V
P
is
the input) or inverting amplifiers (V
P
is a DC voltage
and V
M
is the input). The capacitances C
N
and C
G
rep-
resent the total capacitance at the input pins; they
include the op amp’s Common mode input capacitance
(C
CM
), board parasitic capacitance and any capacitor
placed in parallel.
FIGURE 4-8: Amplifier with Parasitic
Capacitance.
C
G
acts in parallel with R
G
(except for a gain of +1 V/V),
which causes an increase in gain at high frequencies.
C
G
also reduces the phase margin of the feedback
loop, which becomes less stable. This effect can be
reduced by either reducing C
G
or R
F
.
C
N
and R
N
form a low-pass filter that affects the signal
at V
P
. This filter has a single real pole at 1/(2R
N
C
N
).
The largest value of R
F
that should be used, depends
on noise gain (see G
N
in Section 4.4.1 “Capacitive
Loads”), C
G
and the open-loop gain’s phase shift.
Figure 4-9 shows the maximum recommended R
F
for
several C
G
values. Some applications may modify
these values to reduce either output loading or gain
peaking (step response overshoot).
FIGURE 4-9: Maximum Recommended
R
F
vs. Gain.
Figure 2-35 and Figure 2-36 show the small signal and
large signal step responses at G = +1 V/V. The unity
gain buffer usually has R
F
= 0 and R
G
open.
Figure 2-37 and Figure 2-38 show the small signal and
large signal step responses at G = -1 V/V. Since the
noise gain is 2 V/V and C
G
10 pF, the resistors were
chosen to be R
F
= R
G
= 401 and R
N
= 200.
It is also possible to add a capacitor (C
F
) in parallel with
R
F
to compensate for the destabilizing effect of C
G
.
This makes it possible to use larger values of R
F
. The
conditions for stability are summarized in Equation 4-6.
EQUATION 4-6:
V
P
R
F
V
OUT
R
N
C
N
V
M
R
G
C
G
MCP66X
1.E+02
1.E+03
1.E+04
1.E+05
110100
Noise Gain; G
N
(V/V)
Maximum Recommended R
F
(Ω)
G
N
> +1 V/V
100
10k
100k
1k
C
G
= 10 pF
C
G
= 32 pF
C
G
= 100 pF
C
G
= 320 pF
C
G
= 1 nF
f
F
f
GBWP
2G
N2
, G
N1
G
N2
We need:
G
N1
1 R
F
R
G
+=
G
N2
1 C
G
C
F
+=
f
F
12
R
F
C
F
=
f
Z
f
F
G
N1
G
N2
=
Given:
f
F
f
GBWP
4G
N1
, G
N1
G
N2