参数资料
型号: RH5RH303B-T2
厂商: RICOH COMPANY LTD
元件分类: 稳压器
英文描述: PWM STEP-UP DC/DC CONVERTER
中文描述: 0.25 A SWITCHING REGULATOR, 120 kHz SWITCHING FREQ-MAX, PDSO5
封装: SOT-89, 5 PIN
文件页数: 4/29页
文件大小: 199K
代理商: RH5RH303B-T2
11
RH5RH
When the output current (IOUT) 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 IOUT is gradually increased, topen eventually becomes
equal to toff (topen=toff), and when IOUT 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–VIN/VOUT) ................................................................................................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 PON charged in the inductor is provided by Equation 3
as follows :
PON=
0
ton VIN IL (t) dt =
0
ton (VIN2 t/L) dt
=VIN
2 ton2/(2 L).................................................................................................Equation 3
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.
Thus, POFF=
0
topen VIN IL (t) dt =
0
topen ((VOUT–VIN) t/L)dt
=VIN (VOUT–VIN) topen
2/(2 L)
Here, topen=VIN ton/(VOUT–VIN) from Equation 1, and when this is substituted into the above equation.
=VIN
3 ton2/(2 L (VOUT–VIN)..........................................................................Equation 4
Input power is (PON+POFF)/T. When this is converted in its entirely to the output.
PIN=(PON+POFF)/T=VOUT IOUT=POUT .....................................................................Equation 5
Equation 6 can be obtained as follows by solving Equation 5 for IOUT by substituting Equations 3 and 4 into
Equation 5 :
IOUT=VIN
2 ton2/(2 L T (VOUT–VIN)).....................................................................Equation 6
The peak current which flows through L LxTr SD is
ILmax=VIN ton/L ......................................................................................................Equation 7
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