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W ide Input Voltage Range, High Ef ficiency
Fault Tolerant LED Driver
A8502
30
Allegro MicroSystems, Inc.
115 Northeast Cutoff
Worcester, Massachusetts 01615-0036 U.S.A.
1.508.853.5000; www.allegromicro.com
Design Example for SEPIC Configuration
This section provides a method for selecting component values
when designing an application using the A8502 in SEPIC (Sin-
gle-Ended Primary-Inductor Converter) circuit. SEPIC topology
has the advantage that it can generate a positive output voltage
either higher or lower than the input voltage. The resulting design
is diagrammed in figure 34.
Assumptions: For the purposes of this example, the following are
given as the application requirements:
• VBAT: 6 to 14 V ( VIN(min): 5 V and VIN(max): 16 V )
• Quantity of LED channels, #CHANNELS : 2
• Quantity of series LEDs per channel, #SERIESLEDS : 4
• LED current per channel, ILED
: 120 mA
• LED Vf at 120 mA: 3.3 V
• fSW
: 2 MHz
• TA(max): 65°C
• PWM dimming frequency: 200 Hz, 1% duty cycle
Procedure: The procedure consists of selecting the appropriate
configuration and then the individual component values, in an
ordered sequence.
Step 1 Connect LEDs to pins LED1 and LED2. Note: if only one
LED channel is needed, the unused LEDx pin should be pulled to
ground using a 1.5 k resistor. Alternatively, short the LED1 and
LED2 pins together, and half the LED current, to 60 mA
per channel.
Step 2 Determining the LED current setting resistor RISET:
RISET = (VISET × AISET) / ILED (30)
= (1.003 (V) × 980) / 120 mA = 8.19 k
Choose a 8.25 k resistor 1% resistor (or 16.2 k if ILED is
60mA/channel).
Step 3 Determining the OVP resistor. The OVP resistor is
connected between the OVP pin and the output voltage of the
converter.
Step 3a The first step is determining the maximum voltage based
on the LED requirements. The regulation voltage, VLED , of the
A8502 is 720 mV. A constant term, 2 V, is added to give margin
to the design due to noise and output voltage ripple.
VOUT(OVP) = #SERIESLEDS × Vf + VLED + 2 (V) (31)
= 4 × 3.2 (V) + 0.720 (V) + 2 (V) = 15.9 V
Then the OVP resistor is:
ROVP = (VOUT(OVP) – VOVP(th) ) / IOVPH (32)
= (15.9 (V) – 8.1 (V)) / 0.199 (mA) = 39.196 k
where both I
OVPH and VOVP(th) are taken from the Electrical
Characteristics table.
In this case a value of 39.2 k was selected. Below is the actual
value of the minimum OVP trip level with the selected resistor:
VOUT(OVP) = 39.2 (k) × 0.199 (mA) + 8.1 (V) = 15.9 V
Step 3b At this point a quick check must be done to determine if
the conversion ratio is acceptable for the selected frequency.
Dmax = 1 – tSWOFFTIME × fSW (33)
= 1 – 68 (ns) × 2 (MHz) = 86.4%
where the minimum off-time (tSWOFFTIME) is found in the Electri-
cal Characteristics table.
The Theoretical Maximum VOUT is then calculated as:
VOUT(max) =Vd
–
1 – Dmax
Dmax
VIN(min)
0.4 (V) 30.3 V
==–
1 – 0.86
0.86
5 (V)
(34)
where Vd is the diode forward voltage.
The Theoretical Maximum VOUT value must be greater than the
value VOUT(OVP) . If this is not the case, the switching frequency
of the boost converter must be reduced to meet the maximum
duty cycle requirements.