Fairchild RC5040, RC5042 specifications Converter Efficiency, Design Equations, Example

APPLICATION NOTE AN42

Converter Efficiency

Losses due to parasitic resistance in the switches, inductor, and sense resistor dominate at high load-current levels. The major loss mechanisms under heavy loads, in order of importance, are:

•MOSFET I2R Losses

•Inductor Losses

•Sense Resistor Losses

•Gate-Charge Losses

•Diode-Conduction Losses

•Transition Losses

•Input Capacitor Losses

•Losses Due to the Operating Supply Current of the IC.

Efficiency of the converter under heavy loads can be calculated as follows:

Efficiency =	POUT	=	IOUT	⋅ VOUT	,
	------------pIN-		I-------------------------------------------------------OUT⋅ VOUT + PLOSS-

where PLOSS = PDMOSFET + PDINDUCTOR + PDRSENSE + PDGATE + PDDIODE + PDTRAN + PDCAP + PDIC

Design Equations:

2	⋅ RDS,ON ⋅ DutyCycle , where DutyCycle	VOUT + VD
(1) PDMOSFET = IOUT		= -----------------------------------------	+ VD – VSW
		VIN

(2)PDINDUCTOR = IOUT2 ⋅ RINDUCTOR

(3)PDRSENSE = IOUT2 ⋅ RSENSE

(4)PDGATE = qGATE ⋅ f ⋅ 5V , where qGATE is the gate charge and f is the switching frequency

(5)PDDIODE = Vf ⋅ ID(1 – DutyCycle)

(6) PDTRAN	VIN2 ⋅ CRSS ⋅ ILOAD	⋅ f
	= -------------------------------------------------------------	, where CRSS is the reverse transfer capacitance of the high-side MOSFET.
	IDRIVE

(7)PDCAP = IRMS2 ⋅ ESR

(8)PDIC = VCC ⋅ ICC

Example:

3.3+ 0.5

DutyCycle = -----------------------------= 0.73 5 + 0.5 – 0.3

PDMOSFET = 10 2 ⋅ 0.030 ⋅ 0.73 = 2.19W

PDINDUCTOR = 10 2 ⋅ 0.010 = 1W

PDRSENSE = 102 ⋅ 0.0065 = 0.65W

PDGATE = CV ⋅ f ⋅ 5V = 1.75nf ⋅ (9 – 1)V ⋅ 650Khz ⋅ 5V = 0.045W

PDDIODE = 0.5 ⋅ 10(1 – 0.73) = 1.35W

PDTRAN =	5 2	⋅ 400pf ⋅ 10 ⋅ 650khz	∼ 0.010W
	---------------------------------------------------------------	0.7A

PDCAP = (7.5 – 2.5)2 ⋅ 0.015 = 0.37W

PDIC = 0.2W

9