Datasheet

AD8033/AD8034
Rev. D | Page 21 of 24
When selecting components, the common-mode input capacitance
must be taken into consideration.
Filter cutoff frequencies can be increased beyond 1 MHz using the
AD8033/AD8034 but limited open-loop gain and input impedance
begin to interfere with the higher Q stages. This can cause early
roll-off of the overall response.
Additionally, the stop-band attenuation decreases with decreasing
open-loop gain.
Keeping these limitations in mind, a 2-pole Sallen-Key Butterworth
filter with f
C
= 4 MHz can be constructed that has a relatively
low Q of 0.707 while still maintaining 15 dB of attenuation an
octave above f
C
and 35 dB of stop-band attenuation. The filter
and response are shown in Figure 60 and Figure 61, respectively.
–V
S
+V
S
V
IN
R1
2.49kΩ
C3
22pF
V
OUT
AD8033
R2
2.49kΩ
R5
49.9Ω
C1
10pF
02924-060
Figure 60. 2-Pole Butterworth Active Filter
100M100k 1M
FREQUENCY (Hz)
–45
GAIN (dB)
–40
–35
–30
–25
–20
–15
–10
–5
0
5
0
2924-061
10M
Figure 61. 2-Pole Butterworth Active Filter Response
WIDEBAND PHOTODIODE PREAMP
Figure 62 shows an I/V converter with an electrical model of a
photodiode.
The basic transfer function is
FF
F
PHOTO
OUT
RsC
R
I
V
+
×
=
1
where I
PHOTO
is the output current of the photodiode, and the
parallel combination of R
F
and C
F
sets the signal bandwidth.
C
S
R
SH
= 10
11
Ω
V
B
I
PHOTO
02924-062
R
F
C
F
V
OUT
C
M
R
F
C
M
C
D
C
F
+ C
S
Figure 62. Wideband Photodiode Preamp
The stable bandwidth attainable with this preamp is a function
of R
F
, the gain bandwidth product of the amplifier, and the total
capacitance at the summing junction of the amplifier, including
C
S
and the amplifier input capacitance. R
F
and the total capacitance
produce a pole in the loop transmission of the amplifier that
can result in peaking and instability. Adding C
F
creates a zero
in the loop transmission that compensates for the effect of the
pole and reduces the signal bandwidth. It can be shown that the
signal bandwidth resulting in a 45°phase margin (f
(45)
) is defined
by the expression
S
F
CR
CR
f
f
××π
=
2
)45(
where:
f
CR
is the amplifier crossover frequency.
R
F
is the feedback resistor.
C
S
is the total capacitance at the amplifier summing junction
(amplifier + photodiode + board parasitics).
The value of C
F
that produces f
(45)
is
CR
F
S
F
fR
C
C
××π
=
2
The frequency response in this case shows about 2 dB of
peaking and 15% overshoot. Doubling C
F
and cutting the
bandwidth in half results in a flat frequency response, with
about 5% transient overshoot.