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ADP1874ARQZ-0.3-R7 датащи(PDF) 33 Page - Analog Devices |
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ADP1874ARQZ-0.3-R7 датащи(HTML) 33 Page - Analog Devices |
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33 / 44 page ![]() ADP1874/ADP1875 Rev. 0 | Page 33 of 44 Assuming an overshoot of 45 mV, determine if the output capacitor that was calculated previously is adequate. ZERO COMP COMP f R C π = 2 1 () () 2 2 2 6 2 2 2 ) 8 . 1 ( ) mV 45 8 . 1 ( ) A 15 ( 10 1 ) ( ) ( − − × × = − Δ − × = − OUT OVSHT OUT LOAD OUT V V V I L C = 1.4 mF = 3 3 10 25 . 6 10 25 . 60 14 . 3 2 1 × × × × × = 423 pF Loss Calculations Duty cycle = 1.8/12 V = 0.15 Choose five 270 μF polymer capacitors. RON (N2) = 5.4 mΩ The rms current through the output capacitor is tBODY(LOSS) = 20 ns (body conduction time) A 49 . 1 V 2 . 13 V 8 . 1 10 300 μF 1 ) V 8 . 1 V 2 . 13 ( 3 1 2 1 ) ( 3 1 2 1 3 , , = × × × − × = × × − × = MAX IN OUT SW OUT MAX IN RMS V V f L V V I 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) () [ ] 2 1 LOAD N2(ON) N1(ON) N1,N2(CL) I R D R D P × × − + × = = (0.15 × 0.0054 + 0.85 × 0.0054) × (15 A)2 = 1.215 W The power loss dissipated through the ESR of the output capacitor is PCOUT = (IRMS)2 × ESR = (1.5 A)2 × 1.4 mΩ = 3.15 mW Feedback Resistor Network Setup 2 ) ( ) ( × × × = F LOAD SW LOSS BODY LOSS BODY V I t t P = 20 ns × 300 × 103 × 15 A × 0.84 × 2 = 151.2 mW Choosing RB = 1 kΩ as an example, calculate RT as follows: kΩ 2 V 6 . 0 V) 6 . 0 V 8 . 1 ( kΩ 1 = − × = T R PSW(LOSS) = fSW × RGATE × CTOTAL × ILOAD × VIN × 2 = 300 × 103 × 1.5 Ω × 3.3 × 10−9 × 15 A × 12 × 2 = 534.6 mW 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 ( ) [ ] () [] )) 002 . 0 0 . 5 10 3 . 3 10 300 ( 0 . 5 ( )) 002 . 0 62 . 4 10 3 . 3 10 300 ( 62 . 4 ( 9 3 9 3 ) ( + × × × × × + + × × × × × = + × + + × = − − BIAS lowerFET SW BIAS DR upperFET SW DR LOSS DR I VREG C f VREG I V C f V P A/V 33 . 8 005 . 0 24 1 1 = × = × = ON CS CS R A G 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). = 57.12 mW mW 6 . 55 ) 002 . 0 5 10 3 . 3 10 300 ( ) V 5 V 13 ( ) ( ) ( 9 3 ) ( = + × × × × × − = + × × × − = − BIAS total SW IN LDO DISS I VREG C f VREG V P The crossover frequency is 1/12 the switching frequency. 300 kHz/12 = 25 kHz PCOUT = (IRMS)2 × ESR = (1.5 A)2 × 1.4 mΩ = 3.15 mW The zero frequency is 1/4 the crossover frequency. 2 ) ( LOAD LOSS DCR I DCR P × = = 0.003 × (15 A)2 = 675 mW 25 kHz/4 = 6.25 kHz () () CS M REF OUT L OUT OUT L ZERO CROSS CROSS COMP G G V V R C ESR s C ESR R s f f f R 1 1 1 ) ( 1 2 2 2 2 2 2 × × × × × + + + × + = PCIN = (IRMS)2 × ESR = (7.5 A)2 × 1 mΩ = 56.25 mW () () 8 . 1 15 3 . 8 10 500 1 6 . 0 8 . 1 0011 . 0 0035 . 0 25 2 1 0011 . 0 ) 0035 . 0 ) 15 8 . 1 (( 25 2 1 25 . 6 25 25 6 2 2 2 2 2 2 × × × × × × × × π + × + × × π + × + = − k k k k K R COMP = 60.25 kΩ 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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