A8508 View Datasheet(PDF) - Allegro MicroSystems
Part Name
Description
MFG CO.
A8508
Allegro MicroSystems
'A8508' PDF : 28 Pages View PDF
A8508
Wide Input Voltage Range, High Efficiency
8-Channel Fault Tolerant LED Driver
Design Example
This section provides a method for selecting component values
when designing an application using the A8508.
Assumptions: For the purposes of this example, the following are
given as the application requirements:
• VIN: 10 to 16 V
• Quantity of LED channels, #CHANNELS: 8
• Quantity of series LEDs per channel, #SERIESLEDS : 10
• LED current per channel, ILED : 120 mA
• Vf(120) at 120 mA: 3.2 V (max)
• fSW : 600 kHz
• TA(max): 65°C
• PWM dimming frequency: 200 Hz, 1% duty cycle
Step 1: Connect LEDs to pins LED1 through LED8.
Step 2: Determine the LED current by setting resistor RISET . To
do so, apply equation 2:
RISET = (1.000 / ILED ) × 1160
= (1.000 V / 0.120 A ) × 1160
= 9.67 kΩ
Choose a 9.53 kΩ resistor.
STEP 3: Determine the OVP resistor. The OVP resistor is con-
nected between the OVP pin and the output voltage of the con-
verter. The first step is to determine the maximum voltage based
on the LED requirements. Then the regulation voltage of 600 mV
should be added, along with 2 V for noise and regulation. Given
the regulation voltage (VLED) of the A8508 is 850 mV, the mini-
mum required voltage can be determined as follows:
VOUT(OVP) = #SERIESLEDS ×Vf(120) + VLED + 2 V (10)
= 10 × 3.2 V + 0.650 V + 2 V
VOUT(OVP)(min) = 34.65 V
The OVP resistor (ROVP) value can be calculated as:
ROVP =
VOUT(OVP)(min) – VOVP(th)(min)
IOVPH(min)
(11)
= 34.65 V – 1.11 V
45 μA
= 745 kΩ; use the nearest standard value, 750 kΩ
where both IOVP(th)(min) and VOVP(th)(min) are found in the
Electrical Characteristics table. Choose a value of resistor that is
the closest value higher than the calculated ROVP . In this design
example, a value of 750 kΩ is selected.
Below is the actual value of the minimum OVP trip level with the
selected resistor, applying equation 8:
VOVP = ROVP × IOVPH + 1.25 V
= 750 kΩ × 49 μA + 1.25 V
= 38.75 V
STEP 4: Determine the inductor. The inductor must be chosen
such that it can handle the necessary input current. In most appli-
cations, due to stringent EMI requirements, the inductor must
operate in continuous conduction mode throughout the whole
input voltage range.
STEP 4a: Determine the maximum duty cycle of the system:
D(max) = 1– VIN(min) × η
(12)
VOUT(OVP) + Vf(boost)
10 V × 0.9
= 1– 34.65 V+ 0.4 V
= 74.5%
A good approximation of efficiency (η) is 90%. The voltage drop
of the boost diode can be approximated to be about 0.4 V.
STEP 4b: Determine the maximum and minimum input current
to the system. The minimum input current dictates the inductor
value. The maximum current rating dictates the current rating of
the inductor.
To calculate the maximum input current, first determine the
required output current:
IOUT = #CHANNELS × ILED
(13)
= 8 × 120 mA
= 0.960 A
Then substitute into the formula for maximum input current:
IIN(max)
=
VOUT × IOUT
VIN(min) × η
(14)
=
34.65 V × 0.960 A
10 V × 0.90
= 3.7 A
Allegro MicroSystems, Inc.
21
115 Northeast Cutoff
Worcester, Massachusetts 01615-0036 U.S.A.
1.508.853.5000; www.allegromicro.com
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