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CS51413EDR8 参数 Datasheet PDF下载

CS51413EDR8图片预览
型号: CS51413EDR8
PDF下载: 下载PDF文件 查看货源
内容描述: [Switching Regulator, Current/voltage-mode, 4A, 594kHz Switching Freq-Max, PDSO8, 0.150 INCH, SO-8]
分类和应用: 开关光电二极管
文件页数/大小: 16 页 / 137 K
品牌: CHERRY [ CHERRY SEMICONDUCTOR CORPORATION ]
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Application Information  
VO  
VIN  
WSAT  
=
× IS × VSAT  
where VSAT = saturation voltage of the power switch which  
is shown in Figure 3.  
The switching loss occurs when the switch experiences  
both high current and voltage during each switch transi-  
tion. This regulator has a 30ns turn-off time and associated  
power loss is equal to  
IS × VIN  
WS =  
× 20ns × fS  
2
The turn-on time is much shorter and thus turn-on loss is  
not considered here.  
The total power dissipated by the IC is sum of all the above  
W
IC = WQ + WDRV + WBASE + WSAT + WS  
Figure 7: In short circuit, the foldback current and foldback frequency  
limit the switching current to protect the IC, inductor and catch diode.  
The IC junction temperature can be calculated from the  
ambient temperature, IC power dissipation and thermal  
resistance of the package. The equation is shown as follows,  
Thermal Considerations  
A calculation of the power dissipation of the IC is always  
necessary prior to the adoption of the regulator. The cur-  
rent drawn by the IC includes quiescent current, pre-driver  
current, and power switch base current. The quiescent cur-  
rent drives the low power circuits in the IC, which include  
comparators, error amplifier and other logic blocks.  
Therefore, this current is independent of the switching cur-  
rent and generates power equal to  
TJ = WIC × RθJA + TA  
The maximum IC junction temperature shall not exceed  
125°C to guarantee proper operation and avoid any dam-  
ages to the IC.  
Minimum Load Requirement  
As pointed out in the previous section, a minimum load is  
required for this regulator due to the pre-driver current  
feeding the output. Placing a resistor equal to VO divided  
by 12mA should prevent any voltage overshoot at light  
load conditions. Alternatively, the feedback resistors can be  
valued properly to consume 12mA current.  
W
Q = VIN × IQ  
where IQ= quiescent current.  
The pre-driver current is used to turn on/ off the power  
switch and is approximately equal to 12mA in worst case.  
During steady state operation, the IC draws this current  
from the Boost pin when the power switch is on and then  
receives it from the VIN pin when the switch is off. The pre-  
driver current always returns to the VSW pin. Since the pre-  
driver current goes out to the regulators output even when  
the power switch is turned off, a minimum load is required  
to prevent overvoltage in light load conditions. If the Boost  
pin voltage is equal to VIN + VO when the switch is on, the  
power dissipation due to pre-driver current can be calculat-  
ed by  
Component Selection  
Input Capacitor  
In a buck converter, the input capacitor witnesses pulsed  
current with an amplitude equal to the load current. This  
pulsed current and the ESR of the input capacitors deter-  
mine the VIN ripple voltage, which is shown in Figure 8.  
For VIN ripple, low ESR is a critical requirement for the  
input capacitor selection. The pulsed input current possess-  
es a significant AC component, which is absorbed by the  
input capacitors. The RMS current of the input capacitor  
can be calculated using:  
2
VO  
W
DRV = 12mA × ( VIN – VO +  
)
VIN  
The base current of a bipolar transistor is equal to collector  
current divided by beta of the device. Beta of 60 is used  
here to estimate the base current. The Boost pin provides  
the base current when the transistor needs to be on. The  
power dissipated by the IC due to this current is  
I
RMS = IO D(1 – D)  
where D = switching duty cycle which is equal to VO/ VIN.  
IO = load current.  
2
VO  
IS  
60  
WBASE  
=
×
VIN  
where IS = DC switching current.  
When the power switch turns on, the saturation voltage  
and conduction current contribute to the power loss of a  
non-ideal switch. The power loss can be quantified as  
8
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