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LTC3642IMS8E-3.3E-PBF 参数 Datasheet PDF下载

LTC3642IMS8E-3.3E-PBF图片预览
型号: LTC3642IMS8E-3.3E-PBF
PDF下载: 下载PDF文件 查看货源
内容描述: 高艾菲效率,高电压50毫安同步降压型转换器 [High Effi ciency, High Voltage 50mA Synchronous Step-Down Converter]
分类和应用: 转换器
文件页数/大小: 20 页 / 275 K
品牌: Linear [ Linear ]
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LTC3642  
APPLICATIONS INFORMATION  
Efficiency Considerations  
Toincreasethedurationofthereferencevoltagesoft-start,  
place a capacitor from the SS pin to ground. An internal  
5ꢀA pull-up current will charge this capacitor, resulting in  
a soft-start ramp time given by:  
Theefficiencyofaswitchingregulatorisequaltotheoutput  
power divided by the input power times 100%. It is often  
useful to analyze individual losses to determine what is  
limiting the efficiency and which change would produce  
the most improvement. Efficiency can be expressed as:  
0.8V  
5ꢀA  
tSS = CSS  
Efficiency = 100% – (L1 + L2 + L3 + ...)  
When the LTC3642 detects a fault condition (input supply  
undervoltage or overvoltage) or when the RUN pin falls  
below 1.1V, the SS pin is quickly pulled to ground and the  
internal soft-start timer is reset. This ensures an orderly  
restart when using an external soft-start capacitor.  
where L1, L2, etc. are the individual losses as a percent-  
age of input power.  
Although all dissipative elements in the circuit produce  
losses, two main sources usually account for most of  
2
the losses: V operating current and I R losses. The V  
IN  
IN  
The duration of the 1ms internal peak current soft-start  
operating current dominates the efficiency loss at very  
may be increased by placing a capacitor from the I pin  
SET  
2
low load currents whereas the I R loss dominates the  
toground.Thepeakcurrentsoft-startwillrampfrom25mA  
efficiency loss at medium to high load currents.  
to the final peak current value determined by a resistor  
from I  
SET  
to ground. A 1ꢀA current is sourced out of the  
1. The V operating current comprises two components:  
SET  
IN  
I
pin. With only a capacitor connected between I  
The DC supply current as given in the electrical charac-  
teristics and the internal MOSFET gate charge currents.  
The gate charge current results from switching the gate  
capacitance of the internal power MOSFET switches.  
Each time the gate is switched from high to low to  
SET  
and ground, the peak current ramps linearly from 25mA  
to 115mA, and the peak current soft-start time can be  
expressed as:  
0.8V  
1A  
tSS(ISET) = CISET  
high again, a packet of charge, dQ, moves from V to  
IN  
ground. The resulting dQ/dt is the current out of V  
IN  
that is typically larger than the DC bias current.  
A linear ramp of peak current appears as a quadratic  
waveform on the output voltage. For the case where the  
2
2. I R losses are calculated from the resistances of the  
internal switches, R , and external inductor R . When  
peak current is reduced by placing a resistor from I  
SET  
SW  
L
to ground, the peak current offset ramps as a decaying  
switching, the average output current flowing through  
the inductor is “chopped” between the high side PMOS  
switch and the low side NMOS switch. Thus, the series  
resistance looking back into the switch pin is a function  
exponential with a time constant of R  
• C . For this  
ISET  
ISET  
case, the peak current soft-start time is approximately  
3 • R • C  
.
ISET  
ISET  
of the top and bottom switch R  
values and the  
DS(ON)  
Unlike the SS pin, the I  
pin does not get pulled to  
SET  
duty cycle (DC = V /V ) as follows:  
OUT IN  
ground during an abnormal event; however, if the I  
SET  
pin is floating (programmed to 115mA peak current),  
R
= (R )DC + (R  
DS(ON)TOP  
)(1 – DC)  
DS(ON)BOT  
SW  
the SS and I pins may be tied together and connected  
SET  
The R  
for both the top and bottom MOSFETs can  
DS(ON)  
to a capacitor to ground. For this special case, both the  
peak current and the reference voltage will soft-start on  
power-up and after fault conditions. The ramp time for  
be obtained from the Typical Performance Characteris-  
2
tics curves. Thus, to obtain the I R losses, simply add  
R
to R and multiply the result by the square of the  
SW  
L
this combination is C  
• (0.8V/6ꢀA).  
SS(ISET)  
average output current:  
2
2
I R Loss = I (R + R )  
O
SW  
L
3642f  
14  
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