Datasheet

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 k ×
(
1.8 V  0.6 V
)
0.6 V
= 2 k
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
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
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


=






+ 


×

1

+ (
(


+ 
)


)


1

+ ( ×  × 

)

×
1


×




×
1






=
25 k

25 k

+ 6.25 k

×

1

+ (2 × 25 k × (
(
1.8/15
)
+ 0.0035) × 0.0011)


1

+ (2 × 25 k × 0.0035 × 0.0011)

×
1.8
0.6
×
1
500 × 10

× 8.3
×
15
1.8
= 60.25 kΩ


=
1
2



=
1
2 × 3.14 × 60.25 × 10

× 6.25 × 10

= 423 pF
Loss Calculations
Duty cycle = 1.8/12 V = 0.15
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,()
=  × 
1()
+
(
1  
)
× 
()
 × 


= (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
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
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
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
COUT
= (I
RMS
)
2
× ESR = (1.5 A)
2
× 1.4 mΩ = 3.15 mW
2
)(
LOAD
LOSSDCR
IDCRP ×=
= 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