ADP1879-1.0-EVALZ Analog Devices, ADP1879-1.0-EVALZ Datasheet - Page 29

• Manufacturer Part Number: ADP1879-1.0-EVALZ
• Description: Power Management IC Development Tools 1Mhz Switching Frequency Converter
• Manufacturer: Analog Devices
• Series: ADP1879r

Specifications of ADP1879-1.0-EVALZ

• Rohs: yes
• Product: Evaluation Boards
• Factory Pack Quantity: 1

Data
The rise in package temperature (for a 14-lead LFCSP_WD) is
Assuming a maximum ambient temperature environment of 85°C,
which is below the maximum junction temperature of 125°C.
DESIGN EXAMPLE
The
design criteria. For example, the example outlined in this section
uses only four design criteria: V
V
Input Capacitor
The maximum input voltage ripple is usually 1% of the
minimum input voltage (11.8 V × 0.01 = 120 mV).
V
V
= 120 mV − (15 A × 0.001) = 45 mV
��
= 120 µF
Choose five 22 µF ceramic capacitors. The overall ESR of five
22 µF ceramic capacitors is less than 1 mΩ.
Inductor
Determining inductor ripple current amplitude:
Then, calculating for the inductor value
The inductor peak current is approximately
Therefore, an appropriate inductor selection is 1.0 µH with
DCR = 3.3 mΩ (Würth Elektronik 7443552100) with a peak
current handling of 20 A.
����,������
RIPP
MAX,RIPPLE
IN
= 12 V (typical), and f
ADP1878/ADP1879
T
= 30°C × 132.05 mW
= 4.0°C
T
I
P
∆��
�� =
=
= 1.03 µH
15 A + (5 A × 0.5) = 17.5 A
��
= 0.003 × (15 A)
= 120 mV
RMS
������(��������)
CIN
R
J
��
= T
5 V × 300 × 10
= θ
=
13.2 V – 1.8 V
Sheet
≈
= I
= (I
���
= V
4��
R
JA
��
����,������
LOAD
��������
× T
����
× P
RMS
3
��
RIPP
��������,������
��
/2 = 7.5 A
)
= ������ × ��
A
∆��
DR(LOSS)
������,������������
2
= 5 A
= 4.0°C + 85°C = 89.0°C,
− (I
× ESR = (7.5 A)
− ��
��
2
LOAD,MAX
������
= 675 mW
3
×
are easy to use, requiring only a few
�
SW
��
2
13.2 V
×
1.8 V
=
= 300 kHz.
× ESR)
��
OUT
4 × 300 × 10
����,������
��
������
= 1.8 V, I
2
× 1 mΩ = 56.25 mW
15 A
LOAD
3
= 15 A (pulsing),
× 105 mV
Rev. B | Page 29 of 40
Current-Limit Programming
The valley current is approximately
Assuming a low-side MOSFET R
valley current limit from Table 7 and Figure 71 indicate, a pro-
gramming resistor (RES) of 100 kΩ corresponds to an A
of 24 V/V.
Choose a programmable resistor of R
sense gain of 24 V/V.
Output Capacitor
Assume that a load step of 15 A occurs at the output and no more
than 5% output deviation is allowed from the steady state operating
point. In this case, the advantage of the
the frequency is pseudo fixed, the converter is able to respond
quickly because of the immediate, though temporary, increase
in switching frequency.
Assuming the overall ESR of the output capacitor ranges from
5 mΩ to 10 mΩ,
Therefore, an appropriate inductor selection is five 270 µF
polymer capacitors with a combined ESR of 3.5 mΩ.
Assuming an overshoot of 45 mV, determine if the output
capacitor that was calculated previously is adequate
Choose five 270 µF polymer capacitors.
The rms current through the output capacitor is
The power loss dissipated through the ESR of the output
capacitor is
15 A − (5 A × 0.5) = 12.5 A
ΔV
��
= 2 ×
= 1.11 mF
��
=
= 1.4 mF
��
=
P
������
������
������
COUT
(1.8 − 45 mV)
1
2
DROOP
×
1 × 10
=
= 2 ×
=
= (I
300 × 10
√ 3
1
1
2
( (��
= 0.05 × 1.8 V = 90 mV
×
RMS
( 13.2 V – 1.8 V )
1 µF×300×10
������
√ 3
−6
��
1
)
�� ��
2
× ESR = (1.5 A)
× (15 A)
���
15 A
− ∆��
3
× ( ∆��
∆��
2
����,������
× ( 90 mV )
( �� × ��
− (1.8)
��������
�� × ��
����������
����������
2
2
− ��
��������
3
�� ��
×
)
2
2
������
13.2 V
)
ON
− (��
1.8 V
)
ADP1878/ADP1879
2
of 4.5 mΩ and 13 A, as the
�
× 1.4 mΩ = 3.15 mW
RES
×
������
ADP1878
=1.49 A
��
= 100 kΩ for a current
����,������
��
)
������
2
)
is that because
CS