LV5026M Datasheet, PDF (7/17 Pages) Sanyo Semicon Device – LED Driver IC

LV5026M
LED current and inductance setting
It is available to use both no-isolation and isolation applications.
(For non-isolation application)
The output current value is the average of the current value that flows during one cycle. The current value that flows into
coil is a triangular wave shown in the figure below. Make sure to set IL_PK so that (average of current value at one cycle)
is equal to (LED current value).
Vac
a blockdiagram in outline
R1
Vzd
R2
CLK _Q
REF_IN
+
-
CS
-
RESET
VREF
(0.605V typ)
LED
L
OUT
Rcs
Inductor current
FET
Vac
REF_IN
VREF (0.605V typ)/built-in reference
ILED slope is proportion to
Vac voltage (REF pin voltage)
T
ON
OFF
Ipk
Inductor
current
IL = Vac/L × T
IL = Vf/L × T
Ipk = (Vac-Vf)/L × T_c
= Vf/L × T_d
T_c
FET_on
T_d
FET_off
T (1cycle)
Given that the period when current flows into coil is
T_c+T_d
DutyI = _T
Ipk
×
1
2
×
(Duty
×
T)/T
=
ILED
Ipk
×
2
× ILED
DutyI
(1) since
Ipk
×
VREF_IN
Rcs
Rcs
×
VFEF_IN
Ipk
=
DutyI × VFEF_IN
2ILED
(2)
Ipk: peak inductor current
Vf: LED forward voltage drop
Vac: effective value(R.M.S value)
VREF: Built-in reference voltage (0.605V)
VREF_IN: REF_IN voltage (6 pin)
Rs: External sense resistor
Vzd: Zener diode voltage (REF_IN pin)
Since formula for LED current is different between on period and off period as shown above,
Ipk
×
Vac-Vƒ
L
×
T_c
=
Vƒ
L
×
T_d
(3)
Since T_c + T_d = DutyI × T, T_c = DutyI × T - T_d (4)
Based
on
the
result
of
(3)
and
(4),
T_d
=
DutyI
×
T
×
Vac-Vƒ
Vac
(5)
To obtain L from the equation (1), (3), (5),
L
×
Vƒ × DutyI
2 × ILED
×
DutyI
×
T
=
Vac - Vƒ
Vac
=
2
Vƒ
× ILED
×
1
ƒosc
×
Vac - Vƒ
Vac
×
(DutyI)²
(6)
Since LED and inductor are connected in serial in non-isolation mode, LED current flows only when AC voltage exceed
VF.
No.A1950-7/17