ADP1878ACPZ-0.6-R7(RevA) View Datasheet(PDF) - Analog Devices

Part Name

Description

Manufacturer

ADP1878ACPZ-0.6-R7
(Rev.:RevA)

Synchronous Buck Controller with Constant On Time and Valley Current Mode

Analog Devices 

ADP1878/ADP1879

Feedback Resistor Network Setup

Choosing RB = 1 kΩ as an example. Calculate RT as follows:

1 kΩ

1.8 V 0.6 V

0.6 V

2 kΩ

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

8.33 A/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

25 kΩ

√25 kΩ 6.25 kΩ

1 2π 25 kΩ 1.8/15 0.0035 0.0011

1 2π 25 kΩ 0.0035 0.0011

1.8

1

0.6 500 10

15

8.3 1.8

= 60.25 kΩ

1

2

1

2 3.14 60.25 10 6.25 10

= 423 pF

Data Sheet

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

PDCR(LOSS)



DCR



I

2

LOAD

= 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

Rev. A | Page 30 of 40

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