AD8033AKSZ Datasheet, PDF (22/25 Pages) Analog Devices – Low Cost, 80 MHz FastFET Op Amps

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 fC = 4 MHz can be constructed that has a relatively
low Q of 0.707 while still maintaining 15 dB of attenuation an
octave above fC and 35 dB of stop-band attenuation. The filter
and response are shown in Figure 60 and Figure 61, respectively.
C3
22pF
+V_S
R1
R2
AD8033
V_OUT
V_IN
2.49kΩ 2.49kΩ
R5
C1
–V_S
49.9Ω
10pF
Figure 60. 2-Pole Butterworth Active Filter
5
0
–5
–10
–15
–20
–25
–30
–35
–40
–45
100k
1M
10M
FREQUENCY (Hz)
100M
Figure 61. 2-Pole Butterworth Active Filter Response
AD8033/AD8034
WIDEBAND PHOTODIODE PREAMP
Figure 62 shows an I/V converter with an electrical model of a
photodiode.
The basic transfer function is
V_OUT
=
I _PHOTO × R_F
1 + sC _F R_F
where IPHOTO is the output current of the photodiode, and the
parallel combination of RF and CF sets the signal bandwidth.
C_F
R_F
I_PHOTO
R_SH = 10¹¹â„¦
C_D ^CM
C_S
C_M
V_OUT
V_B
C_F + C_S
R_F
Figure 62. Wideband Photodiode Preamp
The stable bandwidth attainable with this preamp is a function
of RF, the gain bandwidth product of the amplifier, and the total
capacitance at the summing junction of the amplifier, including
CS and the amplifier input capacitance. RF and the total capacitance
produce a pole in the loop transmission of the amplifier that
can result in peaking and instability. Adding CF 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
f ₍₄₅₎ =
f _CR
2π × R_F × C_S
where:
fCR is the amplifier crossover frequency.
RF is the feedback resistor.
CS is the total capacitance at the amplifier summing junction
(amplifier + photodiode + board parasitics).
The value of CF that produces f(45) is
C_F =
C_S
2π × R_F × f _CR
The frequency response in this case shows about 2 dB of
peaking and 15% overshoot. Doubling CF and cutting the
bandwidth in half results in a flat frequency response, with
about 5% transient overshoot.
Rev. D | Page 21 of 24