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ILD8150 80 V high side buck LED driver IC with hybrid dimming
Circuit description
2.5.2
Select switching frequency
80 kHz to 100 kHz may be a reasonable compromise between switching loss and inductor size. This needs to be
verified by measurement of the finished design.
2.5.1
Inductor value and its current calculation
0.36 Ω (51 V − 70 �� ( 120 + 270 ) · 10
�� =
R
C
fltr
fltr
The inductor average current is 1 A, while the peak current is I
Saturation current I
of the inductor must be higher than this value.
SAT
2.5.2
Diode selection
To calculate diode currents the duty cycle D is needed:
RMS diode current: ��
��,������
Average diode current: ��
For a design with variable number of LEDs, the highest diode currents will occur at the lowest output voltage.
"1-D" is close to 1 under such conditions, and diode currents are close to the output current. In general a 1 A
diode is recommended as long as the output current is below 1 A.
The reverse blocking voltage V
is sufficient in this example.
2.5.3
C
capacitor selection
IN
��
≥
����
��
������,������
A 4.7 µF MLCC capacitor with a rated voltage of 100 V with RMS current rate of more than 0.73 A would be
recommended in this case.
2.5.4
C
capacitor selection
OUT
The value of C
can be estimated as:
OUT
Application Note
100000 ���� (0.39 �� − 0.33 ��)
assumed as 270 ns (1.5 kΩ and 180 pF accordingly F
�� =
= 1 ∙ √ 0.2714 ∙ (1 +
= 1 ∙ 0.2714 = 0.27 A
��,������
of the diode must be higher than the maximum input voltage V
BR
1
∙ 0.7285 ∙ 0.2714 = 3.60 ����, decided to choose 4.7 ����
3
80∙10
∙0.01∙70
= 1 ∙ √ 0.7285 ∙ (1 − 0.2714 +
��
≫
������
2�� ∙ 80 ������ ∙ 6.8 Ω
−9
�� · 100000 ����)
−
70 �� · 100000 ���� (0.39 �� − 0.33 ��)
L≈860 µH
= I
Pk
LED,AVG
��
51
������
=
= 0.7285
��
70
����
1 − �� = 0.2714
2
1
0.167
∙ (
)
) = 0.52 A
12
1
2
1
0.167
∙ (
)
12
1
5
= 1.46 µ��
13 of 28
0.36 Ω · 51 �� · 51 V
≈ 600 kHz)
fltr
+ I
∙1.083 = 1.083 A.
/2 = I
OUT
LED,AVG
), ��
= 0.73 ��
������,������
. A 100 V diode
IN
V 1.0
2019-01-23
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