Datasheet
+
R
F
-
C
F
IF R
A
< < R
F
C
F
c =
¨
©
§
1 +
R
B
R
A
¨
©
§
R
A
C
F
c
R
B
R
F
C
IN
A
0
C
F
=
2S
R
F
A
0
1 +
1 + 4S
C
IN
(pF)
0
4
8 12 16
18
22 24
V
CM
(V)
0
5
10
15
20
20
14
106
2
V
S
= 5V
V
S
= 24V
LM6211
SNOSAH2C –FEBRUARY 2006–REVISED MARCH 2013
www.ti.com
Figure 48. Input Common-Mode Capacitance vs. V
CM
An essential component for obtaining a maximally flat response is the feedback capacitor, C
F
. The capacitance
seen at the input of the amplifier, C
IN
, combined with the feedback resistor, R
F
, generates a phase lag which
causes gain-peaking and can destabilize the circuit. C
IN
is usually just the sum of C
D
and C
CM
. The feedback
capacitor C
F
creates a pole, f
P
in the noise gain of the circuit, which neutralizes the zero in the noise gain, f
Z
,
created by the combination of R
F
and C
IN
. If properly positioned, the noise gain pole created by C
F
can ensure
that the slope of the gain remains at 20 dB/decade till the unity gain frequency of the amplifier is reached, thus
ensuring stability. As shown in Figure 50, f
P
is positioned such that it coincides with the point where the noise
gain intersects the op amp’s open loop gain. In this case, f
P
is also the overall 3 dB frequency of the
transimpedance amplifier. The value of C
F
needed to make it so is given by Equation 2. A larger value of C
F
causes excessive reduction of bandwidth, while a smaller value fails to prevent gain peaking and maintain
stability.
(2)
Calculating C
F
from Equation 2 can sometimes return unreasonably small values (<1 pF), especially for high
speed applications. In these cases, it is often more practical to use the circuit shown in Figure 49 in order to
allow more reasonable values. In this circuit, the capacitance C
F
' is (1+ R
B
/R
A
) times the effective feedback
capacitance, C
F
. A larger capacitor can now be used in this circuit to obtain a smaller effective capacitance.
Figure 49. Modifying C
F
For example, if a C
F
of 0.5 pF is needed, while only a 5 pF capacitor is available, R
B
and R
A
can be selected
such that R
B
/R
A
= 9. This would convert a C
F
' of 5 pF into a C
F
of 0.5 pF. This relationship holds as long as R
A
<< R
F
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