Datasheet
21
Atmel MSL3080 Datasheet
8 String 60mA LED Driver with Integrated Boost Controller
Inductors have two types of maximum current ratings, RMS current and saturation current. Make sure that the peak inductor current is
less than the saturation current rating. Note that during load current transients, which occur whenever the LEDs are turned on or off (due
to PWM dimming), the inductor current may overshoot its steady state value. How much it overshoots depends on the boost regulator
loop dynamics. If unsure of the loop dynamics, a typical value to use for the overshoot is 50% of the steady-state current. Add half of the
inductor ripple current to this value to determine the peak inductor current. With inductor ripple current in the 25% to 50% range, estimate
the inductor RMS current as 115% of the DC steady state inductor current.
9.10 Setting the External MOSFET Current Limit
The current sense resistor, connected from the switching MOSFET source to GND, sets the boost regulator current limit. The cycle-by-
cycle current limit turns-off the boost regulator switching MOSFET when the current sense input detects instantaneous current above the
current limit threshold. This causes the current to drop until the end of the switching cycle. The current limit threshold is 100mV typical,
and 75mV minimum. Choose the current sense resistor value to set the current limit using:
where I
L(MAX)
is the maximum inductor current. When R
CS
is not equal to a standard 1% resistor value use the next lower value.
9.11 Choosing the Switching MOSFET
The MSL3080 uses an external logic level MOSFET to drive the boost converter. Choose a MOSFET that can pass at least twice the
peak inductor current, and that minimizes simultaneously both the MOSFET on-resistance, R
DSon
, and gate charge for fast switching
speed. Make sure that the MOSFET drain-source voltage rating is above the maximum un-optimized boost output voltage, with some
extra margin for voltage overshoot due to excess circuit board stray inductance and output rectier recovery artefacts. Assure that the
MOSFET is heat sunk to withstand the worst-case power dissipation while maintaining die temperature within the MOSFET ratings.
9.12 Choosing the Output Rectier
The output rectier passes the inductor current to the output capacitor and load during the switching off-time. Due to the high boost
regulator switching frequency use a Schottky rectier. Use a Schottky diode that has a current rating sufcient to supply the load
current, and a voltage rating higher than the maximum boost regulator output voltage. Schottky rectiers have very low on voltage
and fast switching speed, however at high voltage and temperatures Schottky leakage current can be signicant. Make sure that the
rectier power dissipation is within the rectier specications. Place the MOSFET and rectier close together and as close to the output
capacitor(s) as possible to reduce circuit board radiated emissions.
9.13 Loop Compensation
Use a series RC network from COMP to FB to compensate the MSL3080 regulation loop (Figure 8.1 on page 14). The regulation loop
dynamics are sensitive to output capacitor and inductor values. To begin, determine the right-half-plane zero frequency:
where R
LOAD
is the minimum equivalent load resistor, or
The output capacitance and type of capacitor affect the regulation loop and method of compensation. In the case of ceramic capacitors
the zero caused by the equivalent series resistance (ESR) is at such a high frequency that it is not of consequence. In the case of
electrolytic or tantalum capacitors the ESR is signicant, so must be considered when compensating the regulation loop. Determine the
ESR zero frequency by the equation:
where C
OUT
is the value of the output capacitor, and ESR is the Equivalent Series Resistance of the output capacitor. Assure that the loop
crossover frequency is at least 1/5th of the ESR zero frequency.
Place the MOSFET and rectifier close together and as close to the output capacitor(s) as possible to reduce circuit board
radiated emissions.
LOOP COMPENSATION
Use a series RC network from COMP to FB to compensate the MSL3040/41 regulation loop (Figure 3 on page 12). The
regulation loop dynamics are sensitive to output capacitor and inductor values. To begin, determine the right-half-plane
zero frequency:
L
R
V
V
f
LOAD
OUT
IN
RHPZ
2
2
,
where R
LOAD
is the minimum equivalent load resistor, or
)( MAXOUT
OUT
LOAD
I
V
R
.
The output capacitance and type of capacitor affect the regulation loop and method of compensation. In the case of
ceramic capacitors the zero caused by the equivalent series resistance (ESR) is at such a high frequency that it is not of
consequence. In the case of electrolytic or tantalum capacitors the ESR is significant, so must be considered when
compensating the regulation loop. Determine the ESR zero frequency by the equation:
OUT
ESRZ
CESR
f
2
1
where C
OUT
is the value of the output capacitor, and ESR is the Equivalent Series Resistance of the output capacitor.
Assure that the loop crossover frequency is at least 1/5
th
of the ESR zero frequency.
Next determine the desired crossover frequency as 1/5
th
of the lower of the ESR zero f
ESRZ
, the right-half-plane zero f
RHPZ
or the switching frequency f
SW
. The crossover frequency equation is:
OUTLOADCS
LOAD
TOP
COMP
C
CRR
R
R
R
f
2
1
11
,
where f
C
is the crossover frequency, R
TOP
is the top side voltage divider resistor (from the output voltage to FB), R
COMP
is
the resistor of the series RC compensation network. Rearranging the factors of this equation yields the solution for R
COMP
as:
OUTCCSTOPCOMP
C f RR R
21 1 .
These equations are accurate if the compensation zero (formed by the compensation resistor R
COMP
and the
compensation capacitor C
COMP
) happens at a lower frequency than crossover. Therefore the next step is to choose the
compensation capacitor such that the compensation zero is 1/5
th
of the crossover frequency, or:
COMPCOMP
C
COMPZ
CR
f
f
2
1
5
.
Solving for C
COMP
:
CCOMP
COMP
fR
C
2
5
.
Page 20 of 22
© Atmel Inc., 2011. All rights reserved.
Place the MOSFET and rectifier close together and as close to the output capacitor(s) as possible to reduce circuit board
radiated emissions.
LOOP COMPENSATION
Use a series RC network from COMP to FB to compensate the MSL3040/41 regulation loop (Figure 3 on page 12). The
regulation loop dynamics are sensitive to output capacitor and inductor values. To begin, determine the right-half-plane
zero frequency:
L
R
V
V
f
LOAD
OUT
IN
RHPZ
2
2
,
where R
LOAD
is the minimum equivalent load resistor, or
)(
MAXOUT
OUT
LOAD
I
V
R
.
The output capacitance and type of capacitor affect the regulation loop and method of compensation. In the case of
ceramic capacitors the zero caused by the equivalent series resistance (ESR) is at such a high frequency that it is not of
consequence. In the case of electrolytic or tantalum capacitors the ESR is significant, so must be considered when
compensating the regulation loop. Determine the ESR zero frequency by the equation:
OUT
ESRZ
CESR
f
2
1
where C
OUT
is the value of the output capacitor, and ESR is the Equivalent Series Resistance of the output capacitor.
Assure that the loop crossover frequency is at least 1/5
th
of the ESR zero frequency.
Next determine the desired crossover frequency as 1/5
th
of the lower of the ESR zero f
ESRZ
, the right-half-plane zero f
RHPZ
or the switching frequency f
SW
. The crossover frequency equation is:
OUTLOADCS
LOAD
TOP
COMP
C
CRR
R
R
R
f
2
1
11
,
where f
C
is the crossover frequency, R
TOP
is the top side voltage divider resistor (from the output voltage to FB), R
COMP
is
the resistor of the series RC compensation network. Rearranging the factors of this equation yields the solution for R
COMP
as:
OUTCCSTOPCOMP
C f RR R
21 1 .
These equations are accurate if the compensation zero (formed by the compensation resistor R
COMP
and the
compensation capacitor C
COMP
) happens at a lower frequency than crossover. Therefore the next step is to choose the
compensation capacitor such that the compensation zero is 1/5
th
of the crossover frequency, or:
COMPCOMP
C
COMPZ
CR
f
f
2
1
5
.
Solving for C
COMP
:
CCOMP
COMP
fR
C
2
5
.
Page 20 of 22
© Atmel Inc., 2011. All rights reserved.
Place the MOSFET and rectifier close together and as close to the output capacitor(s) as possible to reduce circuit board
radiated emissions.
LOOP COMPENSATION
Use a series RC network from COMP to FB to compensate the MSL3040/41 regulation loop (Figure 3 on page 12). The
regulation loop dynamics are sensitive to output capacitor and inductor values. To begin, determine the right-half-plane
zero frequency:
L
R
V
V
f
LOAD
OUT
IN
RHPZ
2
2
,
where R
LOAD
is the minimum equivalent load resistor, or
)( MAXOUT
OUT
LOAD
I
V
R
.
The output capacitance and type of capacitor affect the regulation loop and method of compensation. In the case of
ceramic capacitors the zero caused by the equivalent series resistance (ESR) is at such a high frequency that it is not of
consequence. In the case of electrolytic or tantalum capacitors the ESR is significant, so must be considered when
compensating the regulation loop. Determine the ESR zero frequency by the equation:
OUT
ESRZ
CESR
f
2
1
where C
OUT
is the value of the output capacitor, and ESR is the Equivalent Series Resistance of the output capacitor.
Assure that the loop crossover frequency is at least 1/5
th
of the ESR zero frequency.
Next determine the desired crossover frequency as 1/5
th
of the lower of the ESR zero f
ESRZ
, the right-half-plane zero f
RHPZ
or the switching frequency f
SW
. The crossover frequency equation is:
OUTLOADCS
LOAD
TOP
COMP
C
CRR
R
R
R
f
2
1
11
,
where f
C
is the crossover frequency, R
TOP
is the top side voltage divider resistor (from the output voltage to FB), R
COMP
is
the resistor of the series RC compensation network. Rearranging the factors of this equation yields the solution for R
COMP
as:
OUTCCSTOPCOMP
C f RR R
21 1 .
These equations are accurate if the compensation zero (formed by the compensation resistor R
COMP
and the
compensation capacitor C
COMP
) happens at a lower frequency than crossover. Therefore the next step is to choose the
compensation capacitor such that the compensation zero is 1/5
th
of the crossover frequency, or:
COMPCOMP
C
COMPZ
CR
f
f
2
1
5
.
Solving for C
COMP
:
CCOMP
COMP
fR
C
2
5
.
Page 20 of 22
© Atmel Inc., 2011. All rights reserved.
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x