Rh5rh – Ricoh RH5RH13B User Manual
Page 15
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RH5RH
When the output current (I
OUT
) 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 I
OUT
is gradually increased, topen eventually becomes
equal to toff (topen=toff), and when I
OUT
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–V
IN
/V
OUT
) ................................................................................................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 P
ON
charged in the inductor is provided by Equation 3
as follows :
P
ON
=
∫
0
ton
V
IN
· IL (t) dt =
∫
0
ton
(V
IN
2
· t/L) dt
=V
IN
2
· ton
2
/(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, P
OFF
=
∫
0
topen
V
IN
· IL (t) dt =
∫
0
topen
((V
OUT
–V
IN
) · t/L)dt
=V
IN
· (V
OUT
–V
IN
) · topen
2
/(2 · L)
Here, topen=V
IN
· ton/(V
OUT
–V
IN
) from Equation 1, and when this is substituted into the above equation.
=V
IN
3
· ton
2
/(2 · L · (V
OUT
–V
IN
)..........................................................................Equation 4
Input power is (P
ON
+P
OFF
)/T. When this is converted in its entirely to the output.
P
IN
=(P
ON
+P
OFF
)/T=V
OUT
· I
OUT
=P
OUT
.....................................................................Equation 5
Equation 6 can be obtained as follows by solving Equation 5 for I
OUT
by substituting Equations 3 and 4 into
Equation 5 :
I
OUT
=V
IN
2
· ton
2
/(2 · L · T · (V
OUT
–V
IN
)).....................................................................Equation 6
The peak current which flows through L · LxTr · SD is
ILmax=V
IN
· ton/L ...................................................................................................... Equation 7