ADP1879 Analog Devices, ADP1879 Datasheet - Page 30

• Manufacturer Part Number: ADP1879
• Description: Synchronous Buck Controller with Constant On-Time and Valley Current Mode with Power Saving Mode
• Manufacturer: Analog Devices

ADP1878/ADP1879
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
300 kHz/12 = 25 kHz
1.8
0.6
= 60.25 kΩ
= 423 pF
1
1
2
CS
and R
1 kΩ
500
3.14
B
2π
= 1 kΩ as an example. Calculate R
1
COMP
2
√25 kΩ
1
ON
10
, C
25 kΩ
60.25
1
are taken from setting up the current limit
1.8 V
2π
1
COMP
1
25 kΩ
24
0.6 V
8.3
, and C
25 kΩ
1
th
6.25 kΩ
0.6 V
10
1
of the crossover frequency:
0.005
1.8/15
1.8
15
th
m
PAR
) is 500 μA/V, and the current
of the switching frequency:
6.25
0.0035
, the transconductance
1
2 kΩ
8.33 A/V
1
0.0035
10
0.0011
T
as follows:
0.0011
Rev. 0 | Page 30 of 40
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)
SW(LOSS)
DR(LOSS)
DISS(LDO)
COUT
CIN
LOSS
= 17 nC (total MOSFET gate charge)
SW
DCR
= 5.4 mΩ
1,
CIN
C
= (I
(
= 20 ns (body conduction time)
= P
LOSS
lowerFET
= (I
= 1.215 W + 151.2 mW + 534.6 mW + 57.12 mW +
= [V
= f
RMS
N1,N2
)
= (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
BIAS
3
I
3
SW
REG
2
LOAD
× 3.3 × 10
× 15 A × 0.84 × 2
1
C
)]
× C
) × (f
upperFET
+ P
= 0.003 × (15 A)
TOTAL
SW
3
SW
−9
× 3.3 × 10
V
2
+ P
× C
1
−9
× 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
× V
−9
)] + [V
DR
× 5 + 0.002)
2
IN
+ P
REG
Data Sheet
2
= 675 mW
2
× 2
DISS(LDO)
+ I
REG
BIAS
×
)
+ P
COUT