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參數(shù)資料
| 型號(hào)： | R1230D281E-TR |
| 廠商： | RICOH COMPANY LTD |
| 元件分類： | 穩(wěn)壓器 |
| 英文描述： | SWITCHING REGULATOR, 575 kHz SWITCHING FREQ-MAX, PDSO8 |
| 封裝： | 0.9 MM HEIGHT, SON-8 |
| 文件頁數(shù)： | 4/26頁 |
| 文件大?。?/td> | 284K |
| 代理商： | R1230D281E-TR |
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R1230D
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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 × 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 AND SELECTION OF EXTERNAL COMPONENTS
When P-channel Tr. of LX is ON:
(Wherein, Ripple Current P-P value is described as IRP, ON resistance of P-channel Tr. and N-channel Tr. of LX are re-
spectively described as Ronp and Ronn, and the DC resistor of the inductor is described as RL.)
VIN = VOUT + (Ronp + RL) × IOUT + L × IRP / ton.................................................................Equation 3
When P-channel Tr. of LX is "OFF"(N-channel Tr. is "ON"):
L × IRP / toff = RL × IOUT + VOUT + Ronn × IOUT ...................................................................Equation 4
Put Equation 4 to Equation 3 and solve for ON duty of P-channel transistor, ton / (toff + ton) = DON,
DON = (VOUT – Ronn × IOUT + RL × IOUT) / (VIN + Ronn × IOUT – Ronp × IOUT) .................. Equation 5
Ripple Current is as follows;
IRP = (VIN – VOUT – Ronp × IOUT – RL × IOUT) × DON / fosc / L................................................Equation 6
wherein, peak current that flows through L, and LX Tr. is as follows;
ILmax = IOUT + IRP/2.............................................................................................................Equation 7
Consider ILmax, condition of input and output and select external components.
#The above explanation is directed to the calculation in an ideal case in continuous mode.
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