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參數(shù)資料
| 型號(hào)： | R5210D28A-TR |
| 廠商： | RICOH COMPANY LTD |
| 元件分類(lèi)： | 穩(wěn)壓器 |
| 英文描述： | 0.7 A SWITCHING REGULATOR, 960 kHz SWITCHING FREQ-MAX, PDSO6 |
| 封裝： | 0.90 MM HEIGHT, HSON-6 |
| 文件頁(yè)數(shù)： | 11/13頁(yè) |
| 文件大小： | 136K |
| 代理商： | R5210D28A-TR |
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Rev. 1.11
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same as those when Lx Tr. turns on and when it turns off.
The difference between ILmax and ILmin, which is represented by
I;
I = ILmax – ILmin = VOUT × topen / L = (VIN-VOUT)×ton/LEquation 1
Where, T=1/fosc=ton+toff
duty (%)=ton/T
×100=ton× fosc × 100
topen
≤ toff
In Equation 1, VOUT
×topen/L and (VIN-VOUT)×ton/L are respectively shown the change of the current at
ON, and the change of the current at OFF.
When the output current (IOUT) is relatively small, topen < toff as illustrated in the above diagram. In this case,
the energy is charged in the inductor during the time period of ton and is discharged in its entirely during the
time period of toff, therefore ILmin becomes to zero (ILmin=0). When Iout is gradually increased, eventually,
topen becomes 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
continuous mode.
In the continuous mode, when Equation 1 is solved for ton and assumed that the solution is tonc,
tonc=T
×VOUT/VIN Equation 2
When ton<tonc, the mode is the discontinuous mode, and when ton=tonc, the mode is the continuous mode.
n OUTPUT CURRENT AND SELECTION OF EXTERNAL COMPONENTS
When Lx Tr. is “ON”:
(Wherein, Ripple Current P-P value is described as IRP, ON resistance of LX Tr. is described as Rp the direct
current of the inductor is described as RL.)
VIN=VOUT+(Rp+RL)
×IOUT+L×IRP/ton Equation 3
When Lx Tr. is “OFF”:
L
×IRP/toff = VF+VOUT+RL×IOUT Equation 4
Put Equation 4 to Equation 3 and solve for ON duty, ton/(toff+ton)=DON,
DON=(VOUT+VF+RL
×IOUT)/(VIN+VF-Rp×IOUT)Equation 5
Ripple Current is as follows;
IRP=(VIN-VOUT-Rp
×IOUT-RL×IOUT)×DON/f/L …Equation 6
Wherein, peak current that flows through L, Lx Tr., and SD is as follows;
ILmax=IOUT+IRP/2
…Equation 7
Consider ILmax, condition of input and output and select external components.
HThe above explanation is directed to the calculation in an ideal case in continuous mode.
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