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ADP1879ACPZ-0.6-R7 датащи(PDF) 30 Page - Analog Devices

номер детали ADP1879ACPZ-0.6-R7
подробное описание детали  Synchronous Buck Controller
PDF  40 Pages
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производитель  AD [Analog Devices]
домашняя страница  http://www.analog.com
Logo AD - Analog Devices

ADP1879ACPZ-0.6-R7 датащи(HTML) 30 Page - Analog Devices

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ADP1878/ADP1879
Data Sheet
Rev. A | Page 30 of 40
Feedback Resistor Network Setup
Choosing RB = 1 kΩ as an example. Calculate RT as follows:
1kΩ
1.8V
0.6V
0.6V
2kΩ
Compensation Network
To calculate RCOMP, CCOMP, and CPAR, the transconductance
parameter and the current sense gain variable are required. The
transconductance parameter (Gm) is 500 μA/V, and the current
sense loop gain is
1
1
24 0.005
8.33A/V
where ACS and RON 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 1/12th of the switching frequency:
300 kHz/12 = 25 kHz
The zero frequency is 1/4th of the crossover frequency:
25 kHz/4 = 6.25 kHz
1
1
1
1
25kΩ
√25kΩ
6.25kΩ
1
2π 25kΩ
1.8/15
0.0035
0.0011
1
2π 25kΩ 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
RON(N2) = 5.4 mΩ
tBODY(LOSS) = 20 ns (body conduction time)
VF = 0.84 V (MOSFET forward voltage)
CIN = 3.3 nF (MOSFET gate input capacitance)
QN1,N2 = 17 nC (total MOSFET gate charge)
RGATE = 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 × 103 × 15 A × 0.84 × 2
= 151.2 mW
PSW(LOSS) = fSW × RGATE × CTOTAL × ILOAD × VIN × 2
= 300 × 103 × 1.5 Ω × 3.3 × 10−9 × 15 A × 12 × 2
= 534.6 mW
PDR(LOSS) = [VDR × (fSWCupperFETVDR + IBIAS)] + [VREG ×
(fSWClowerFETVREG +IBIAS)]
=(4.62 × (300 ×103 × 3.3 × 10−9 × 4.62 + 0.002)) +
(5.0 × (300 × 103 × 3.3 × 10−9 × 5.0 + 0.002))
= 57.12 mW
PDISS(LDO) = (VIN – VREG) × (fSW × CTOTAL × VREG + IBIAS)
= (13 V – 5 V) × (300 × 103 × 3.3 × 10−9 × 5 + 0.002)
= 55.6 mW
PCOUT = (IRMS)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
PCIN = (IRMS)2 × ESR = (7.5 A)2 × 1 mΩ = 56.25 mW
PLOSS = PN1,N2 + PBODY(LOSS) + PSW + PDCR + PDR + PDISS(LDO) + PCOUT
+ PCIN = 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



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