ADP2102 Analog Devices, ADP2102 Datasheet - Page 20

• Manufacturer Part Number: ADP2102
• Description: Low Duty Cycle, 600 mA, 3 MHz Synchronous Step-Down DC-to-DC Converter
• Manufacturer: Analog Devices

ADP2102
Inductor Losses
Inductor conduction losses are caused by the flow of current
through the inductor, which has an internal resistance (DCR)
associated with it. Larger sized inductors have smaller DCR,
which may decrease inductor conduction losses.
Inductor core losses are related to the magnetic permeability
of the core material. Because the ADP2102 is a high switching
frequency dc-to-dc converter, shielded ferrite core material is
recommended for its low core losses and low EMI.
The total amount of inductor power loss can be calculated by
Switching Losses
Switching losses are associated with the current drawn by the
driver to turn on and turn off the power devices at the switching
frequency. Each time a power device gate is turned on and
turned off, the driver transfers a charge ΔQ from the input
supply to the gate and then from the gate to ground.
The amount of power loss can be calculated by
Transition Losses
Transition losses occur because the P-channel switch cannot
turn on or turn off instantaneously. In the middle of an LX node
transition, the power switch provides all the inductor current.
The source to drain voltage of the power switch is half the input
voltage, resulting in power loss. Transition losses increase with
load current and input voltage and occur twice for each
switching cycle.
The amount of power loss can be calculated by
where:
t
t
THERMAL CONSIDERATIONS
In most applications, the ADP2102 does not dissipate a lot of
heat, due to its high efficiency. However, in applications with
maximum loads at high ambient temperature, low supply voltage,
and high duty cycle, the heat dissipated in the package is great
enough that it may cause the junction temperature of the die to
exceed the maximum junction temperature of 125°C. Once the
junction temperature exceeds 150°C, the converter goes into
thermal shutdown. It recovers only after the junction temperature
has decreased to below 135°C to prevent any permanent damage.
Therefore, thermal analysis for the chosen application solution is
very important to guarantee reliable performance over all
conditions.
R
F
where:
C
C
f
SW
is the rise time of the LX node.
is the fall time of the LX node.
GATE_P
GATE_N
is the switching frequency.
P
P
P
L
SW
TRAN
= DCR × I
is the gate capacitance of the internal high-side switch.
is the gate capacitance of the internal low-side switch.
= (C
= V
GATE_P
IN
/2 × I
OUT
+ C
2
OUT
+ Core Losses
GATE_N
× (t
) × V
R
+ t
F
IN
) × f
2
× f
SW
SW
(16)
(17)
(18)
Rev. B | Page 20 of 24
The junction temperature of the die is the sum of the ambient
temperature of the environment and the temperature rise of the
package due to power dissipation, shown in the following equation:
where:
T
T
T
dissipation in it.
The rise in temperature of the package is directly proportional
to the power dissipation in the package. The proportionality
constant for this relationship is defined as the thermal resistance
from the junction of the die to the ambient temperature, as shown
in the following equation:
where:
T
θ
ambient temperature of the package.
P
DESIGN EXAMPLE
The calculations in this section provide only a rough estimate
and are no substitute for bench evaluation.
Consider an application where the ADP2102 is used to step
down from 3.6 V to 1.8 V with an input voltage range of 2.7 V
to 4.2 V.
Inductor
Choose a 2.2 μH inductor for this application.
JA
D
J
A
R
R
is the junction temperature.
is the ambient temperature.
is the rise in temperature of the package due to power
is the rise in temperature of the package.
is the power dissipation in the package.
is the thermal resistance from the junction of the die to the
T
T
V
Pulsed Load = 300 mA
V
f
T
ΔI
L =
I
P
SW
PK
L
J
R
A
OUT
IN
L
= T
= I
= θ
= 85°C
= I
=
= 2.7 V to 4.2 V (3.6 V typical)
= 3 MHz (typical)
1.90 μH
V
(0.6 A)
= 1.8 V @ 600 mA
OUTMAX
A
V
OUT
LOAD(MAX)
JA
f
+ T
OUT
SW
× P
V
×
×
IN
2
R
D
×
1 (
2
× DCR =
0
× 0.08 Ω (FDK MIPF2520D) = 29 mW
(
×
−
3 .
V
+ ΔI
V
IN
f
×
SW
OUT
I
−
LOAD
L
×
/2 = 0.6 + 0.2/2 = 0.7 A
V
/
L
OUT
V
(
INMAX
MAX
)
≈
)
)
I
=
LOAD
3 (
1
3
8 .
(MAX
×
×
10
1 (
6
)
−
×
= 0.6/3 = 200 mA
. 1
0
/ 8
3 .
×
. 4
0
) 2
) 6 .
=
(19)
(20)