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LM67660 Datasheet(PDF) 39 Page - Texas Instruments

Part # LM67660
Description  LM65680, LM65660 and LM65640 65V, 8A/6A/4A, Synchronous Buck DC/DC Converter Family With Low-EMI ZEN 1 Switcher Technology Optimized for High Power Density
PDF  66 Pages
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Manufacturer  TI2 [Texas Instruments]
Direct Link  https://www.ti.com
Logo TI2 - Texas Instruments

LM67660 Datasheet(HTML) 39 Page - Texas Instruments

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RRTkΩ = 16.4
FSWMHz −0.633=16.40.4−0.633=40.36kΩ
(27)
8.2.1.2.3 Buck Inductor Selection
1. Use Equation 28 to calculate the buck inductance based on an inductor peak-to-peak ripple current of 30%
to 40% of the converter maximum-rated current. Select a standard value for LO of 3.3µH.
LO= VOUT
FSW×∆ILO× 1− VOUT
VINnom = 5V
400kHz×3.2A× 1− 5V48V =3.5µH
(28)
2. Use Equation 29 to calculate the peak inductor current at maximum steady-state input voltage. Subharmonic
oscillation occurs with peak current-mode control at a duty cycle greater than 50%. For design simplification,
the LM65680 has an internal slope compensation ramp proportional to the switching frequency that adds to
the current-sense signal to damp any tendency toward subharmonic oscillation.
ILOpk =IOUT+ VOUT
2×FSW×LO× 1− VOUT
VINmax =8A+ 5V
2×400kHz×3.3µH× 1− 5V65V =9.75A
(29)
3. To avoid subharmonic oscillation, choose a buck inductance greater than that given by Equation 30. An
effective maximum inductance value sets the minimum ripple current amplitude required for current-mode
control to perform correctly. As a general rule, the minimum inductor ripple current must be no less than
approximately 10% of the converter maximum-rated current under nominal conditions. This limit applies
to applications where the switch duty cycle becomes greater than or equal to 50% under any operating
condition.
LO,min≥M×VOUTFSW
(30)
where
• M = 0.16, 0.21 and 0.29 for the LM65680, LM65660 and LM65640, respectively.
8.2.1.2.4 Input Capacitor Selection
A power supply input typically has a relatively high source impedance at the switching frequency. Good-quality
input capacitors are necessary to limit the input ripple voltage. In general, the ripple current splits between the
input capacitors based on the relative impedance of the capacitors at the switching frequency.
1. Select the input capacitors with sufficient voltage and RMS current ratings. Use Equation 31 to calculate
the RMS current in the input capacitors, where the worst-case operating point is at an input voltage of 10V,
corresponding to 50% duty cycle.
ICINrms = IOUT× D× 1−D =8A ×  0.5 × 1−0.5 =4A 
(31)
2. Use Equation 32 to find the required input capacitance, assuming an approximate 10% duty cycle for
48V-to-5V conversion:
CIN ≥  D× 1−D ×IOUT
FSW× ΔVIN−RESR,Cin×IOUT = 0.1× 1−0.1 ×8A
400kHz × 480mV−2mΩ × 8A = 4.8µF
(32)
where
• ΔVIN is the specification for the peak-to-peak input ripple voltage.
• RESR,Cin is the effective ESR of the input capacitors.
3. Recognizing the voltage coefficient of ceramic capacitors, select four 4.7µF, 100V, X7R, 1210 ceramic input
capacitors. Each capacitor has an effective capacitance value of approximately 1.3µF at 48VDC. Place
these capacitors adjacent to the input pin pairs, [VIN1, PGND1] and [VIN2, PGND2].
4. Use Equation 33 to calculate the peak-to-peak ripple voltage amplitude.
∆VIN=IOUT×D× 1−D
CIN×FSW +RESR,Cin×IOUT=8A×0.1× 1−0.1
4.2µF×400kHz +2mΩ×8A=0.44V
(33)
www.ti.com
LM65680
SNVSCF2 – NOVEMBER 2025
Copyright © 2025 Texas Instruments Incorporated
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Product Folder Links: LM65680



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