欢迎访问ic37.com |
会员登录 免费注册
发布采购

RH5RH602B-T1 参数 Datasheet PDF下载

RH5RH602B-T1图片预览
型号: RH5RH602B-T1
PDF下载: 下载PDF文件 查看货源
内容描述: PWM升压型DC / DC转换器 [PWM STEP-UP DC/DC CONVERTER]
分类和应用: 转换器
文件页数/大小: 29 页 / 198 K
品牌: RICOH [ RICOH ELECTRONICS DEVICES DIVISION ]
 浏览型号RH5RH602B-T1的Datasheet PDF文件第8页浏览型号RH5RH602B-T1的Datasheet PDF文件第9页浏览型号RH5RH602B-T1的Datasheet PDF文件第10页浏览型号RH5RH602B-T1的Datasheet PDF文件第11页浏览型号RH5RH602B-T1的Datasheet PDF文件第13页浏览型号RH5RH602B-T1的Datasheet PDF文件第14页浏览型号RH5RH602B-T1的Datasheet PDF文件第15页浏览型号RH5RH602B-T1的Datasheet PDF文件第16页  
RH5RH
When the output current (I
OUT
) is relatively small, topen<toff as illustrated in the above diagram. In this case,
the energy charged in the inductor during the time period of ton is discharged in its entirely during the time peri-
od of toff, so that ILmin becomes zero (ILmin=0). When I
OUT
is gradually increased, topen eventually becomes
equal to toff (topen=toff), and when I
OUT
is further increased. ILmin becomes larger than zero (ILmin>0). The
former mode is referred to as the discontinuous mode and the latter mode is referred to as the continuous mode.
In the continuous mode, when Equation 1 is solved for ton and the solution is tonc,
tonc =T · (1–V
IN
/V
OUT
) ................................................................................................Equation 2
When ton<tonc, the mode is the discontinuous mode, and when ton=tonc, the mode is the continuous mode.
Output Current in Discontinuous Mode
In the discontinuous mode, when LxTr is on, the energy P
ON
charged in the inductor is provided by Equation 3
as follows :
ton
P
ON
=∫
0
V
IN
· IL (t) dt =∫
0
(V
IN
2
· t/L) dt
=V
IN
2
· ton
2
/(2 · L) ................................................................................................. Equation 3
ton
In the case of the step-up DC/DC converter, the energy is also supplied from the input power source at the time
of OFF.
topen
topen
V
IN
· IL (t) dt =∫
0
((V
OUT
–V
IN
) · t/L)dt
Thus, P
OFF
=∫
0
=V
IN
· (V
OUT
–V
IN
) · topen
2
/(2 · L)
Here, topen=V
IN
· ton/(V
OUT
–V
IN
) from Equation 1, and when this is substituted into the above equation.
=V
IN
3
· ton
2
/(2 · L · (V
OUT
–V
IN
) ..........................................................................Equation 4
Input power is (P
ON
+P
OFF
)/T. When this is converted in its entirely to the output.
P
IN
=(P
ON
+P
OFF
)/T=V
OUT
· I
OUT
=P
OUT
.....................................................................Equation 5
Equation 6 can be obtained as follows by solving Equation 5 for I
OUT
by substituting Equations 3 and 4 into
Equation 5 :
I
OUT
=V
IN
2
· ton
2
/(2 · L · T · (V
OUT
–V
IN
)) .....................................................................Equation 6
The peak current which flows through L · LxTr · SD is
ILmax=V
IN
· ton/L ...................................................................................................... Equation 7
11