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5962R0724901QXC

更新时间： 2024-01-13 00:14:50
品牌 Logo 应用领域
英特矽尔 - INTERSIL 控制器
页数 文件大小 规格书
11页 648K
描述
Radiation Hardened, High Performance IndustryStandard Single-Ended Current Mode PWM Controller

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5962R0724901QXC 技术参数

生命周期：Active零件包装代码：DFP
包装说明：DFP,针数：8
Reach Compliance Code：unknownECCN代码：EAR99
HTS代码：8542.39.00.01风险等级：5.73
模拟集成电路 - 其他类型：SWITCHING CONTROLLER控制模式：CURRENT-MODE
控制技术：PULSE WIDTH MODULATION最大输入电压：13.2 V
最小输入电压：9 VJESD-30 代码：R-CDFP-F8
JESD-609代码：e4长度：6.485 mm
功能数量：1端子数量：8
最高工作温度：125 °C最低工作温度：-55 °C
最大输出电流：1 A封装主体材料：CERAMIC, METAL-SEALED COFIRED
封装代码：DFP封装形状：RECTANGULAR
封装形式：FLATPACK认证状态：Qualified
筛选级别：MIL-PRF-38535 Class V座面最大高度：2.92 mm
表面贴装：YES切换器配置：SINGLE
最大切换频率：1000 kHz温度等级：MILITARY
端子面层：GOLD端子形式：FLAT
端子节距：1.27 mm端子位置：DUAL
宽度：6.475 mmBase Number Matches：1

5962R0724901QXC 数据手册

 浏览型号5962R0724901QXC的Datasheet PDF文件第5页浏览型号5962R0724901QXC的Datasheet PDF文件第6页浏览型号5962R0724901QXC的Datasheet PDF文件第7页浏览型号5962R0724901QXC的Datasheet PDF文件第9页浏览型号5962R0724901QXC的Datasheet PDF文件第10页浏览型号5962R0724901QXC的Datasheet PDF文件第11页 
ISL78840ASRH, ISL78841ASRH, ISL78843ASRH, ISL78845ASRH  
where D is the percent of on time during a switching  
cycle. Setting Q = 1 and solving for Se yields  
Equation 10:  
RTCT signal with the current sense feedback and applies  
the result to the CS pin as shown in Figure 6.  
1
π
1
⎛⎛  
⎝⎝  
⎞
⎞
– 1  
(EQ. 10)  
--  
-------------  
S
= S  
+ 0.5  
e
n
⎠
⎠
1 – D  
V
REF  
Since Sn and Se are the on time slopes of the current  
ramp and the external ramp, respectively, they can be  
R9  
R6  
multiplied by t  
to obtain the voltage change that  
.
ON  
CS  
occurs during t  
ON  
1
π
1
RTCT  
⎛⎛  
⎝⎝  
⎞
⎞
– 1  
--  
-------------  
V
= V  
+ 0.5  
(EQ. 11)  
e
n
⎠
⎠
1 – D  
C4  
where V is the change in the current feedback signal  
n
(ΔI) during the on time and Ve is the voltage that must  
be added by the external ramp.  
FIGURE 6. SLOPE COMPENSATION  
For a flyback converter, Vn can be solved for in terms of  
input voltage, current transducer components, and  
primary inductance, yielding Equation 12:  
Assuming the designer has selected values for the RC  
filter (R and C ) placed on the CS pin, the value of R  
6
4
9
required to add the appropriate external ramp can be  
found by superposition.  
D ⋅ T  
⋅ V ⋅ R  
IN CS  
1
π
1
SW  
⎛⎛  
⎞
⎞
– 1  
---------------------------------------------------- --  
-------------  
V
=
+ 0.5  
V
e
⎝⎝  
⎠
⎠
(EQ. 12)  
L
1 – D  
p
2.05D ⋅ R  
6
---------------------------  
(EQ. 16)  
V
=
V
e
R
+ R  
9
6
where R  
CS  
is the current sense resistor, f is the  
sw  
switching frequency, L is the primary inductance, V is  
p
IN  
The factor of 2.05 in Equation 16 arises from the peak  
amplitude of the sawtooth waveform on RTCT minus a  
base-emitter junction drop. That voltage multiplied by  
the maximum duty cycle is the voltage source for the  
the minimum input voltage, and D is the maximum duty  
cycle.  
The current sense signal at the end of the ON time for  
CCM operation is Equation 13:  
slope compensation. Rearranging to solve for R yields  
9
Equation 17:  
(1 – D) ⋅ V ⋅ f  
sw  
N
⋅ R  
CS  
N
P
⎛
⎜
⎝
⎞
⎟
⎠
O
(2.05D – V ) ⋅ R  
S
e
6
------------------------  
-------------------------------------------  
V
=
I
+
O
V
(EQ. 13)  
---------------------------------------------  
R
=
Ω
(EQ. 17)  
CS  
2L  
s
9
V
e
The value of R  
CS  
determined in Equation 15 must be  
rescaled so that the current sense signal presented at the  
where V  
is the voltage across the current sense  
resistor, L is the secondary winding inductance, and I is  
CS  
s
O
CS pin is that predicted by Equation 13. The divider  
the output current at current limit. Equation 13 assumes  
the voltage drop across the output rectifier is negligible.  
created by R and R makes this necessary.  
6
9
R
+ R  
9
R
9
6
--------------------  
R′  
=
⋅ R  
(EQ. 18)  
Since the peak current limit threshold is 1.00V, the total  
current feedback signal plus the external ramp voltage  
must sum to this value when the output load is at the  
current limit threshold as shown in Equation 14.  
CS  
CS  
Example:  
= 12V  
V
IN  
V
+ V  
= 1  
CS  
(EQ. 14)  
e
V = 48V  
O
L = 800µH  
s
Substituting Equations 12 and 13 into Equation 14 and  
solving for R  
yields Equation 15:  
Ns/Np = 10  
Lp = 8.0µH  
CS  
1
----------------------------------------------------------------------------------------------------------------------------------------------------  
R
=
CS  
1
π
--  
⎛
+ 0.5  
⎞
⎟
D ⋅ f ⋅ V  
N
(1 – D) ⋅ V ⋅ f  
O sw  
⎛
⎜
⎝
⎞
⎟
⎠
I = 200mA  
O
sw  
IN  
s
⎜
------------------------------- -----------------  
------  
-------------------------------------------  
⋅
–1  
+
⋅
I
+
O
⎜
⎝
⎟
⎠
L
1 – D  
N
p
2L  
s
p
Switching Frequency, f  
sw  
= 200kHz  
(EQ. 15)  
Duty Cycle, D = 28.6%  
R = 499Ω  
6
Adding slope compensation is accomplished in the  
ISL7884xASRH using an external buffer transistor and  
the RTCT signal. A typical application sums the buffered  
Solve for the current sense resistor, R , using  
CS  
Equation 15.  
FN6991.0  
December 21, 2009  
8
1...567891011

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