EVAL-AD5398AEBZ AD [Analog Devices], EVAL-AD5398AEBZ Datasheet - Page 12

• Manufacturer Part Number: EVAL-AD5398AEBZ
• Description: 120 mA, Current Sinking, 10-Bit, I2C DAC
• Manufacturer: AD [Analog Devices]

AD5398A
POWER SUPPLY BYPASSING AND GROUNDING
When accuracy is important in an application, it is beneficial
to consider power supply and ground return layout on the PCB.
The PCB for the AD5398A should have separate analog and
digital power supply sections. Where shared AGND and DGND
is necessary, the connection of grounds should be made at only
one point, as close as possible to the AD5398A.
Pay special attention to the layout of the AGND return path
and track it between the voice coil motor and I
any series resistance. Figure 20 shows the output current sink
of the AD5398A and illustrates the importance of reducing the
effective series impedance of AGND, and the track resistance
between the motor and I
Inductor L
coil is effectively a dc current that results in a voltage drop, V
when the AD5398A is sinking current; the effect of any series
inductance is minimal. The maximum voltage drop allowed
across R
voltage of Q1 is 200 mV. This means that the AD5398A output
has a compliance voltage of 600 mV. If V
600 mV, the output transistor, Q1, can no longer operate
properly and I
As the current increases through the voice coil, V
and V
specified compliance voltage of 600 mV. The ground return
path is modelled by the R
tance between the voice coil and the AD5398A is modelled as
R
and because the current is maintained as a constant, it is not as
critical as the purely resistive component of the ground return path.
T
. The inductive effects of L
DROP
SENSE
Figure 20. Effect of PCB Trace Resistance and Inductance
decreases and eventually approaches the minimum
C
INDUCTANCE
RESISTANCE
and Resistor R
is 400 mV, and the minimum drain to source
SINK
GROUND
GROUND
might not be maintained as a constant.
R
AGND
SENSE
3.3Ω
R
L
G
ACTUATOR
G
SINK
D1
G
Q1
C
V
and L
VOICE
. The current through the voice
. The voice coil is modelled as
DD
V
COIL
G
G
influence R
G
I
SINK
components. The track resis-
R
R
L
V
C
T
C
BAT
V
V
DROP
T
DROP
V
SENSE
C
TRACE
RESISTANCE
falls below
SINK
and R
C
to minimize
increases
C
equally,
Rev. 0 | Page 12 of 16
C
,
When the maximum sink current is flowing through the motor,
the resistive elements, R
voltage headroom of Q1 and may, in turn, limit the maximum
value of R
For example, if
then the largest value of resistance of the voice coil, R
For this reason, it is important to minimize any series impedance
on both the ground return path and interconnect between the
AD5398A and the motor.
The power supply of the AD5398A should be decoupled with
0.1 μF and 10 μF capacitors. These capacitors should be kept as
physically close as possible, with the 0.1 μF capacitor serving as
a local bypass capacitor, and therefore should be located as close
as possible to the V
tantalum bead-type; the 0.1 μF capacitor should be a ceramic
type with a low effective series resistance and effective series
inductance. The 0.1 μF capacitor provides a low impedance path
to ground for high transient currents.
The power supply line should have as large a trace as possible to
provide a low impedance path and reduce glitch effects on the
supply line. Clocks and other fast switching digital signals
should be shielded from other parts of the board by digital
ground. Avoid crossover of digital and analog signals if possible.
When traces cross on opposite sides of the board, ensure that
they run at right angles to each other to reduce feedthrough
effects through the board. The best board layout technique is to
use a multilayer board with ground and power planes, where
the component side of the board is dedicated to the ground
plane only and the signal traces are placed on the solder side.
However, this is not always possible with a 2-layer board.
V
R
R
I
V
R
3.6
SINK
G
T
BAT
DROP
C
= 0.5 Ω
= 0.5 Ω
=
V
= 120 mA
= 3.6 V
C
= 600 mV (the compliance voltage)
V
−
because of voltage compliance.
BAT
[600
−
[
mV
V
DD
DROP
120
pin. The 10 μF capacitor should be a
+
T
2
and R
mA
+
×
(
(120
I
SINK
I
SINK
G
, may have an impact on the
mA
×
R
×
T
)
0.5
+
(
Ω)]
I
SINK
=
×
24
R
G
Ω
)]
C
, is
=