RH5RH552B-T2 datasheet(12/29 Pages) RICOH

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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 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=

∫

ton VIN · IL (t) dt =

∫

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=

∫

topen VIN · IL (t) dt =

∫

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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