参数资料
型号: LMZ14201EVAL/NOPB
厂商: National Semiconductor
文件页数: 15/27页
文件大小: 0K
描述: EVAL BOARD FOR LMZ14201
产品培训模块: SIMPLE SWITCHER Power Modules
标准包装: 1
系列: SIMPLE SWITCHER®
主要目的: DC/DC,步降
输出及类型: 1,非隔离
输出电压: 3.3V
电流 - 输出: 1A
输入电压: 6 ~ 24 V
稳压器拓扑结构: 降压
频率 - 开关: 最高 1MHz
板类型: 完全填充
已供物品:
已用 IC / 零件: LMZ14201
LMZ14201
www.ti.com
SNVS649F – JANUARY 2010 – REVISED OCTOBER 2013
Following is a comparison pair of waveforms of the showing both CCM (upper) and DCM operating modes.
Figure 33. CCM and DCM Operating Modes
V IN = 12V, V O = 3.3V, I O = 1 A / 0.25 A
500 mA/Div
2.00 2 s/Div
The approximate formula for determining the DCM/CCM boundary is as follows:
I DCB ? V O *(V IN –V O )/(2*10 μ H*f SW(CCM) *V IN )
Following is a typical waveform showing the boundary condition.
Figure 34. Transition Mode Operation
V IN = 24V, V O = 3.3V, I O = 0.29 A
(16)
500 mA/Div
2.00 2 s/Div
The inductor internal to the module is 10 μ H. This value was chosen as a good balance between low and high
input voltage applications. The main parameter affected by the inductor is the amplitude of the inductor ripple
current (I LR ). I LR can be calculated with:
I LR P-P =V O *(V IN - V O )/(10μH*f SW *V IN )
(17)
Where V IN is the maximum input voltage and f SW is determined from Equation 10 .
If the output current I O is determined by assuming that I O = I L , the higher and lower peak of I LR can be
determined. Be aware that the lower peak of I LR must be positive if CCM operation is required.
POWER DISSIPATION AND BOARD THERMAL REQUIREMENTS
For the design case of V IN = 24V, V O = 3.3V, I O = 1A, T AMB(MAX) = 85°C , and T JUNCTION = 125°C, the device must
see a thermal resistance from case to ambient of less than:
θ CA < (T J-MAX — T AMB(MAX) ) / P IC-LOSS - θ JC
(18)
Given the typical thermal resistance from junction to case to be 1.9 °C/W. Use the 85°C power dissipation curves
in the Typical Performance Characteristics section to estimate the P IC-LOSS for the application being designed. In
this application it is 0.52W.
θ CA = (125 — 85) / 0.52W — 1.9 = 75
To reach θ CA = 75, the PCB is required to dissipate heat effectively. With no airflow and no external heat, a good
estimate of the required board area covered by 1 oz. copper on both the top and bottom metal layers is:
Board Area_cm 2 = 500°C x cm 2 /W / θ JC
(19)
Copyright ? 2010–2013, Texas Instruments Incorporated
Product Folder Links: LMZ14201
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