Analog Devices ADP1879 Evaluation Board ADP1879-1.0-EVALZ ADP1879-1.0-EVALZ Hoja De Datos
Los códigos de productos
ADP1879-1.0-EVALZ
ADP1878/ADP1879
Data
Sheet
Rev. B | Page 30 of 40
Feedback Resistor Network Setup
Choosing R
B
= 1 kΩ as an example. Calculate R
T
as follows:
��
��
= 1 kΩ ×
(1.8 V − 0.6 V)
0.6 V
= 2 kΩ
Compensation Network
To calculate R
COMP
, C
COMP
, and C
PAR
, the transconductance
parameter and the current sense gain variable are required. The
transconductance parameter (G
transconductance parameter (G
m
) is 500 µA/V, and the current
sense loop gain is
��
����
=
1
��
����
��
����
=
1
24 × 0.005 = 8.33
A/V
where A
CS
and R
ON
are taken from setting up the current limit
(see the Programming Resistor (RES) Detect Circuit section
and the Valley Current-Limit Setting section).
The crossover frequency is 1/12
and the Valley Current-Limit Setting section).
The crossover frequency is 1/12
th
of the switching frequency:
300 kHz/12 = 25 kHz
The zero frequency is 1/4
th
of the crossover frequency:
25 kHz/4 = 6.25 kHz
��
��������
=
��
����������
���
����������
2
+ ��
��������
2
×
�1
2
+ (��(��
��
+ ������)��
������
)
2
�1
2
+ (�� × ������ × ��
������
)
2
×
1
��
��
×
��
������
��
������
×
1
��
��
��
����
��
��������
=
25 kΩ
√25 kΩ
2
+ 6.25 kΩ
2
×
�1
2
+ (2π × 25 kΩ × ((1.8/15) + 0.0035) × 0.0011)
2
�1
2
+ (2π × 25 kΩ × 0.0035 × 0.0011)
2
×
1.8
0.6 ×
0.6 ×
1
500 × 10
−6
× 8.3 ×
15
1.8
= 60.25 kΩ
��
��������
=
1
2����
��������
��
��������
=
1
2 × 3.14 × 60.25 × 10
3
× 6.25 × 10
3
= 423 pF
Loss Calculations
Duty cycle = 1.8/12 V = 0.15
R
R
ON(N2)
= 5.4 mΩ
t
BODY(LOSS)
= 20 ns (body conduction time)
V
F
= 0.84 V (MOSFET forward voltage)
C
IN
= 3.3 nF (MOSFET gate input capacitance)
Q
N1,N2
= 17 nC (total MOSFET gate charge)
R
GATE
= 1.5 Ω (MOSFET gate input resistance)
��
��1,��2(����)
= ��� × ��
��1(����)
+ (1 − ��) × ��
��2(����)
� × ��
��������
2
= (0.15 × 0.0054 + 0.85 × 0.0054) × (15 A)
2
= 1.215 W
��
��������(��������)
=
��
��������(��������)
��
����
× ��
��������
× ��
��
× 2
= 20 ns × 300 × 10
3
× 15 A × 0.84 × 2
= 151.2 mW
P
P
SW(LOSS)
= f
SW
× R
GATE
× C
TOTAL
× I
LOAD
× V
IN
× 2
= 300 × 10
3
× 1.5 Ω × 3.3 × 10
−9
× 15 A × 12 × 2
= 534.6 mW
P
P
DR(LOSS)
= [V
DR
× (f
SW
C
upperFET
V
DR
+ I
BIAS
)] + [V
REG
×
(f
SW
C
lowerFET
V
REG
+I
BIAS
)]
=(4.62 × (300 ×10
3
× 3.3 × 10
−9
× 4.62 + 0.002)) +
(5.0 × (300 × 10
3
× 3.3 × 10
−9
× 5.0 + 0.002))
= 57.12 mW
P
P
DISS(LDO)
= (V
IN
– V
REG
) × (f
SW
× C
TOTAL
× V
REG
+ I
BIAS
)
= (13 V – 5 V) × (300 × 10
3
× 3.3 × 10
−9
× 5 + 0.002)
= 55.6 mW
P
P
COUT
= (I
RMS
)
2
× ESR = (1.5 A)
2
× 1.4 mΩ = 3.15 mW
2
)
(
LOAD
LOSS
DCR
I
DCR
P
×
=
= 0.003 × (15 A)
2
= 675 mW
P
CIN
= (I
RMS
)
2
× ESR = (7.5 A)
2
× 1 mΩ = 56.25 mW
P
LOSS
= P
N1,N2
+ P
BODY(LOSS)
+ P
SW
+ P
DCR
+ P
DR
+ P
DISS(LDO)
+ P
COUT
+ P
CIN
= 1.215 W + 151.2 mW + 534.6 mW + 57.12 mW +
55.6 + 3.15 mW + 675 mW + 56.25 mW = 2.655 W