Datasheet

–13–
REV. F
AD8009
Driving a Capacitive Load
A capacitive load, like that presented by some A/D converters,
can sometimes be a challenge for an op amp to drive depending
on the architecture of the op amp. Most of the problem is caused
by the pole created by the output impedance of the op amp and
the capacitor that is driven. This creates extra phase shift that
can eventually cause the op amp to become unstable.
One way to prevent instability and improve settling time when
driving a capacitor is to insert a resistor in series between the
op amp output and the capacitor. The feedback resistor is still
connected directly to the output of the op amp, while the series
resistor provides some isolation of the capacitive load from the
op amp output.
10F
+
0.1F0.001F
10F
+
0.1F
0.001F
AD8009
49.9
+5V
–5V
3
2
4
R
T
R
S
C
L
50pF
2V
STEP
7
6
R
F
R
G
G = +2: R
F
= 301 = R
G
G = +10: R
F
= 200, R
G
= 22.1
Figure 5. Capacitive Load Drive Circuit
Figure 5 shows such a circuit with an AD8009 driving a 50 pF
load. With R
S
= 0, the AD8009 circuit will be unstable. For a
gain of +2 and +10, it was found experimentally that setting R
S
to 42.2 Ω will minimize the 0.1% settling time with a 2 V step at
the output. The 0.1% settling time was measured to be 40 ns with
this circuit.
For smaller capacitive loads, a smaller R
S
will yield optimal
settling time, while a larger R
S
will be required for larger capacitive
loads. Of course, a larger capacitance will always require more
time for settling to a given accuracy than a smaller one, and this
will be lengthened by the increase in R
S
required. At best, a
given RC combination will require about seven time constants
by itself to settle to 0.1%, so a limit will be reached where too
large a capacitance cannot be driven by a given op amp and still
meet the system’s required settling time specification.