ADP1879-0.6-EVALZ Analog Devices, ADP1879-0.6-EVALZ Datasheet - Page 30

• Manufacturer Part Number: ADP1879-0.6-EVALZ
• Description: Power Management IC Development Tools 600khz Switching Frequency Converter
• Manufacturer: Analog Devices
• Series: ADP1879r

Specifications of ADP1879-0.6-EVALZ

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

Feedback Resistor Network Setup
Choosing R
Compensation Network
To calculate R
parameter and the current sense gain variable are required. The
transconductance parameter (G
sense loop gain is
where A
(see the Programming Resistor (RES) Detect Circuit section
and the Valley Current-Limit Setting section).
The crossover frequency is 1/12
The zero frequency is 1/4
25 kHz/4 = 6.25 kHz
ADP1878/ADP1879
��
��
300 kHz/12 = 25 kHz
��
��
��
�1
1.8
0.6
= 60.25 kΩ
��
=
= 423 pF
1
��������
����
��������
��
��������
��
2 × 3.14 × 60.25 × 10
×
2
= 1 kΩ ×
CS
×
=
+ (2π × 25 kΩ × ( ( 1.8/15 ) + 0.0035) × 0.0011)
��
��
and R
500 × 10
������
��
������
�1
=
B
=
=
����
���
= 1 kΩ as an example. Calculate R
COMP
2����
√25 kΩ
1
2
��
×
����������
+ (2π × 25 kΩ × 0.0035 × 0.0011)
ON
����
��
( 1.8 V − 0.6 V )
, C
��������
1
are taken from setting up the current limit
��
��
=
����������
−6
1
��
1
COMP
2
25 kΩ
2
24 × 0.005
����
+ ��
× 8.3
0.6 V
+ 6.25 kΩ
��
��������
, and C
1
th
��������
1
of the crossover frequency:
×
3
2
1.8
15
× 6.25 × 10
th
m
×
PAR
) is 500 µA/V, and the current
of the switching frequency:
= 2 kΩ
2
= 8.33 A/V
�1
, the transconductance
�1
×
2
2
+ (�� ( ��
+ (�� × ������ × ��
3
��
T
+ ������ ) ��
as follows:
2
������
������
2
)
)
2
×
Rev. B | Page 30 of 40
2
×
Loss Calculations
Duty cycle = 1.8/12 V = 0.15
R
t
V
C
Q
R
BODY(LOSS)
ON(N2)
GATE
F
IN
N1,N2
= 0.84 V (MOSFET forward voltage)
= 3.3 nF (MOSFET gate input capacitance)
��
= (0.15 × 0.0054 + 0.85 × 0.0054) × (15 A)
= 1.215 W
��
= 20 ns × 300 × 10
= 151.2 mW
P
= 300 × 10
= 534.6 mW
P
(f
=(4.62 × (300 ×10
(5.0 × (300 × 10
= 57.12 mW
P
= (13 V – 5 V) × (300 × 10
= 55.6 mW
P
P
P
+ P
55.6 + 3.15 mW + 675 mW + 56.25 mW = 2.655 W
P
= 1.5 Ω (MOSFET gate input resistance)
��1,��2(����)
��������(��������)
SW(LOSS)
DR(LOSS)
DISS(LDO)
COUT
CIN
LOSS
= 17 nC (total MOSFET gate charge)
SW
DCR
= 5.4 mΩ
CIN
C
= (I
(
= 20 ns (body conduction time)
= P
lowerFET
LOSS
= (I
= 1.215 W + 151.2 mW + 534.6 mW + 57.12 mW +
= [V
RMS
N1,N2
= f
)
= (V
RMS
=
= ��� × ��
3
V
SW
)
DCR
× 1.5 Ω × 3.3 × 10
2
=
+ P
REG
)
DR
× ESR = (7.5 A)
IN
2
× R
× ESR = (1.5 A)
��
× (f
– V
BODY(LOSS)
��������(��������)
+I
3
×
GATE
× 3.3 × 10
3
BIAS
I
��
3
SW
REG
����
× 3.3 × 10
LOAD
��1(����)
2
× 15 A × 0.84 × 2
C
)]
× C
) × (f
upperFET
+ P
= 0.003 × (15 A)
TOTAL
+ ( 1 − �� ) × ��
× ��
SW
3
SW
−9
× 3.3 × 10
V
2
+ P
−9
× C
��������
× 5.0 + 0.002))
DR
× I
−9
× 1 mΩ = 56.25 mW
2
× 4.62 + 0.002)) +
× 1.4 mΩ = 3.15 mW
× 15 A × 12 × 2
+ I
DCR
LOAD
TOTAL
× ��
BIAS
+ P
× V
��
−9
× V
)] + [V
DR
× 2
× 5 + 0.002)
��2(����)
2
IN
+ P
REG
Data
= 675 mW
2
× 2
DISS(LDO)
REG
+ I
� × ��
BIAS
×
Sheet
)
+ P
��������
2
COUT