Microchip Technology MCP1631RD-MCC2 Data Sheet
MCP1631HV MULTI-CHEMISTRY
BATTERY CHARGER
REFERENCE DESIGN
© 2009 Microchip Technology Inc.
DS51791A-page 47
Appendix D. Design Example
D.1
DESIGN EXAMPLE
D.1.1
Design Parameters:
• P
OUT
= 10W (typical)
• V
IN
= 6V to 16V for MCP1631HV. Choose V
IN
= 6V for worst case (WC).
• V
OUT
= 8.4V for two Li-Ion batteries or 6.6V for four NiMH batteries
• I
OUT
= 1.5A for 4 NiMH batteries. Choose DI
L
= 20%, I
OUT
= 300 mA.
• η = Switcher Efficiency @ 80% (typical)
• V
• V
F
= Schottky Diode Drop @ 0.35V (typical)
• I
IN_WC
= I
OUT
* V
OUT_WC
/(V
IN_WC
* ç) = 1.5A * 6.6V/(6V * 0.8) = 2.06A
• F
OSC
= 500 kHz, T
SWITCH
= 1/F
SWITCH
= 2 µs
• D
OSC
= 25%, t
OSC_ON
= 0.5 µs, t
OSC_OFF
= 1.5 µs
• Choose C20 (Artificial Ramp) = 1500 pF
D.1.2
Design:
• Duty Cycle (D) = ((V
OUT
+ V
F
)/η)/(((V
OUT
+ V
F
)/η) +
VIN
)
= ((8.4V + 0.35)/0.8)/(((8.4V + 0.35)/0.8) + 6V)
= 65% (worst case is two Li-Ion batteries)
= 65% (worst case is two Li-Ion batteries)
• t
ON
= D * 1/F
SWITCH
= 0.65/500 kHz = 1.3 µs
• t
OFF
= T
SWITCH
- t
ON
= 2 µs – 1.3 µs = 0.7 µs
• L = V
IN
* D * T
SWITCH
/
ΔI
L
= 6V * 0.65 * 2 µs/300 mA = 26 µH.
For SEPIC coupled inductor, use ½ L because V
IN
“sees” 2 inductors.
L
COUPLED
= 26 µH/2 = 13 µH. Choose 10 µH as standard value.
•
ΔI
L
= 2 * V
IN
* D * T
SWITCH
/ (4 * L
COUPLED
) = 2*6V*0.65*2 µs/(4*10 µH)
ΔI
L
≅ 390 mA
•
ΔI
IN
=
ΔI
L
= 390 mA
• I
IN
= I
OUT
* V
OUT
/(V
IN
* η)= 1.5A * 6.6V/( 6V * 0.8) = 2.06A
I
IN PEAK
= I
IN
+ ½
ΔI
L
= 2.06A + 0.195A = 2.26A
• P
OUT
= V
OUT
* I
OUT
= 6.6V * 1.5A = 9.9W
D.1.3
Slope Compensation:
• M
Artificial Ramp
=
ΔI
L
* R
SENSE
/t
OSC_FF
= 390 mA * 0.11
Ω/1.5 us = 0.0429V/1.5 µs
But the ramp divider is (R10 + R11)/(R10 + R11 + R35)
= (1k + 1k)/(1k + 1k + 20k) = 2k/22k = 1/11
= (0.0429V/1.5 µs)/(1/11) = 471.9 mV/1.5 µs
= (1k + 1k)/(1k + 1k + 20k) = 2k/22k = 1/11
= (0.0429V/1.5 µs)/(1/11) = 471.9 mV/1.5 µs
•
τ
Artificial Ramp
(R33 & C20) - Using V
OUT
= V
IN
(1-e
-(T/
τ)
)
= -t
OFF
/(Ln(1-(V
OUT
/V
IN)
))
= -1.5 µs/(Ln(1 – (0.4719V/5.0V)))
= -1.5 µs/Ln (0.90562) = 15.13 µs
= -1.5 µs/Ln (0.90562) = 15.13 µs