In Equation 1, VOUT´topen/L and (VIN-VOUT)´ton/L are respectively show 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´VIN/VOUT××× 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 LxTr is ON:
(Wherein, Ripple Current P-P value is described as IRP, ON resistance of LXTr 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 LxTr 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 ¼Equation 6
wherein, peak current that flows through L, LxTr, and SD is as follows;
ILmax=IOUT+IRP/2
¼Equation 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.
n External Components
1. Inductor
Select an inductor that peak current does not exceed ILmax. If larger current than allowable current flows,
magnetic saturation occurs and make transform efficiency worse.
When the load current is same, the smaller value of L, the larger the ripple current.
Provided that the allowable current is large in that case and DC current is small, therefore, for large output current,
efficiency is better than using an inductor with a large value of L and vice versa,
2. Diode
Use a diode with low VF (Schottky type is recommended.) and high switching speed.
Reverse voltage rating should be more than VIN and current rating should be equal or more than ILmax.
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Rev. 1.11
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RICOH [ RICOH ELECTRONICS DEVICES DIVISION ]
