Buck high-brightness LED driver based on the ST1S10
step-down DC-DC converter voltage regulator
Introduction
High-brightness LEDs are becoming a prominent source of light because of their long life,
ruggedness, design flexibility, small size and energy efficiency. LEDs are now available in
higher and higher wattages per package (1 W, 3 W and 5 W) with currents up to 1.5 A. At
these current levels, the traditional means of limiting current with a resistor is not sufficiently
accurate nor efficient. Today, single-dice, white HBLEDs capable of delivering up to
90 lm/W of light are available. A typical 1 W white LED delivers an optical efficiency of
30 lm/W, whereas a typical 60 W light bulb delivers 15 lm/W.
It is known that the brightness of an LED is proportional to the forward current, so the best
way to supply LEDs is to control the forward current to get good matching of the output light.
LED manufacturers specify the characteristics (such as lumens, beam pattern) of their
devices at a specified forward current (I
This application note describes how to implement a constant current control to drive highbrightness LEDs by a step-down DC-DC converter voltage regulator. A switching regulator
is the best choice for driving HBLEDs when high efficiency and low power dissipation are
required.
), not at a specific forward voltage (VF).
F
The circuit uses the ST1S10 high-efficiency buck converter configured to drive a single
HBLED in constant current mode.
The ST1S10 is a step-down monolithic power switching regulator which needs few external
passive components and it is capable of delivering 3 A. An internal oscillator fixes the
switching oscillation at 900 kHz, and it is possible to synchronize the switching frequency
with an external clock from 400 kHz to 1.2 MHz.
This application note includes a schematic diagram, bill of material (BOM), and test data.
When designing a power supply for a white high-brightness LED, the main requirements are
efficiency, size and cost of the complete solution.
A standard buck converter is the best choice for providing a constant current because only
the buck converter among the switching topologies has an average inductor current that is
equal to the average load current. For this reason, the conversion of a constant voltage into
constant current is much easier.
LEDs are current-driven devices whose brightness is proportional to their forward current.
Forward current can be controlled in two ways: voltage mode and current mode. The first
method uses the LED V-I curve to determine what voltage has to be applied to the LED in
order to generate the desired forward current. This is typically accomplished by applying a
voltage source and using a ballast resistor as shown in Figure 1. This method has two
serious drawbacks. The first is that every change in LED forward voltage creates a change
in LED current. The second problem is the power lost across the ballast resistor which
reduces the efficiency.
Figure 1.Constant voltage control
Equation 1
⎛
⎜
1VV×+×=
+=
FBOUT
⎜
⎝
LEDs are PN junction devices with a steep I - V curve. For this reason, driving an LED with
a voltage source can lead to large swings of forward current in response to even a small
change in forward voltage. In general, to meet the needs of a driver for an HBLED, the
current output must be in the ±5% to ±20% range.
The best way to drive the LEDs is to control the forward current so that it eliminates changes
in current due to variations in forward voltage, which translates into a constant LED
brightness. Figure 2 illustrates the configuration of a typical buck converter driver circuit.
The value of current-sense resistor (R
SENSE
feedback voltage that the buck converter requires. Multiple LEDs should be connected in a
series configuration to keep an identical current flowing in each LED.
4/20
⎞
R
H
⎟
⎟
R
L
⎠
RIVn
FBFMAX_F
) depends on the desired LED current and the
AN2754Background
Figure 2.Constant current control
VOUT
Voltage Regulator
FB
R
SENSE
Equation 2
V
FB
I =
F
R
FB
Accuracy and efficiency are the two main goals of the current sensing even if they are in
direct conflict. The higher the sense voltage is, the higher the signal-noise ratio, but the
higher the power dissipated on R
To reduce the power dissipated in the series resistance,
SENSE
.
Figure 3 shows a simple method of
amplifying the current sense signal by using a single supply op-amp. This method allows the
user to select the current sense resistor R
while setting the average value of I
with the gain of the op-amp.
F
according to the desired power dissipation
SENSE
Figure 3.Constant current control with V
Voltage Regulator
VOUT
FB
Equation 3
I
=
F
SENSE
5/20
amplification
SENSE
+
OUT
-
RFBRSENSERIN
V
FB
⎛
R
FB
⎜
1R
+⋅
⎜
R
IN
⎝
⎞
⎟
⎟
⎠
Buck topology switching power supplyAN2754
2 Buck topology switching power supply
The buck topology switching power supply is an efficient voltage regulator which produces
an output voltage always less than or equal to the source voltage in the same polarity. The
first step of conversion is to generate a chopped version of input source. A single-pole
double-throw (SPDT) switch is connected as shown in
Figure 4.Buck topology
The switch output voltage is equal to the converter input voltage when the switch is in
position 1 and equal to zero when the switch is position 2. The position is varied periodically
at a frequency of 1/T, where T represents the switching cycle period. The ratio of the on-time
to the period is referred to as the duty cycle D. So the switch output is a rectangular
waveform having amplitude equal to the source voltage, frequency equal to 1/T and duty
cycle equal to D. By inserting a low-pass filter between the (SPDT) switch and the load, a
basic buck topology is formed. The DC value of switch output voltage is simply the source
voltage multiplied by the duty cycle. The L-C filter cutoff is selected to pass the desired lowfrequency components of the switch output but also to attenuate the high-frequency
switching harmonics.
Figure 4.
A power stage can operate in continuous or discontinuous inductor current mode.
Continuous inductor current mode is characterized by current flowing continuously in the
inductor during the entire switching cycle in steady-state operation. In discontinuous mode
the inductor current drops to zero for a portion of the switching cycle. In this section we will
derive the voltage conversion relationship for the continuous conduction mode buck power
stage. In continuous conduction mode, the power stage assumes two states per switching
cycle.
The ON state is when the high-side switch is ON and the low-side switch is OFF.
Figure 5.Buck converter circuit while the switch is in position 1
During the ON state the voltage applied on the inductor is given by:
Equation 4
VVsv−=
outL
6/20
AN2754Buck topology switching power supply
TnT (
We can find the inductor current by integrating the inductor voltage waveform.
Equation 5
VV
−
()()
L
nTitnTi
+=+
outS
t
⋅
L
Figure 6.Inductor current waveform
iL
VV
−
I
OUT
()()
L
nTitnTi
+=+
OUTS
''t
⋅
L
() ()
L
V
OUT
DTnTiti
''''t
⋅−+=
L
t’t’’
nT+D
n+1)T
Equation 6
VV
−
maxL
()
nTiI
+=
outS
DT
⋅
L
The inductor current increase during the ON state is given by:
Equation 7
VV
−
outS
i
=∆
L
DT
⋅
L
The OFF state is when the high side is OFF and the low side ON.
Figure 7.Buck converter circuit while the switch is in position 2
t
7/20
Buck topology switching power supplyAN2754
[
]
The inductor voltage during the OFF state is given by:
Equation 8
VV−=
outL
The inductor current during the OFF state is given by:
Equation 9
V
()()
L
DTnTiti
L
out
t
⋅−+=
The inductor current decrease during the OFF state is given by:
Equation 10
V
out
()
i
L
L
TD1
−⋅=∆
The volt-time product of each switch state must be equal in steady-state operation, so the
current increase during the ON state and the current decrease during the OFF state must be
equal.
Equation 11
DT
V
outoutS
()
L
TD1
−⋅=⋅
VV
−
L
From the above equation we obtain the continuous conduction mode buck voltage
conversion relationship.
Equation 12
VsDV
out
×=
To guarantee continuous mode, the following equation must be satisfied:
Equation 13
LOADLavg
2
∆
i
L
≥=
II
The following relationship provides the minimum value of the inductance which is necessary
to guarantee the fixed ripple current in continuous mode:
Equation 14
L⋅
()
=
MIN
FI
⋅∆
VoutIESRRVin
−⋅+−
LavgLDSONMAX
Vout
Vin
MAXSWL
The function of the output capacitor is both filtering the AC current and providing the charge
that is necessary to supply the load during the transients. Constant current drivers are free
of load transient by design. For this reason, the capacitor is only needed to obtain a lower
current ripple amplitude across the LEDs. The value of the output capacitor is chosen to
8/20
AN2754Buck topology switching power supply
reduce the ripple current on the LEDs branch. To calculate the ripple current that flows
through the LEDs it is necessary to estimate the impedances of the branches of both the
LED and output capacitor (Zo, Zc). In this procedure let us suppose that the triangular shape
of the ripple current on the inductor is approximately sinusoidal.
Figure 8.Output impedence
Equation 15
RFBrDZ
O
+=
Equation 16
ESRZ
C
+=
1
CoutFsw2
⋅⋅π⋅
The following equation can be used to estimate the impedance of the output capacitor which
guarantees the desired ripple current on the LEDs for a given inductor ripple current:
Equation 17
i
∆
Z×
C
LED
=
∆−∆
Z
ii
LEDL
O
9/20
Design exampleAN2754
3 Design example
This section outlines a step-by-step procedure for the design of a constant current control by
means of a switching step-down voltage regulator. The aim is to maintain both high
efficiency and good accuracy. The following design procedure is helpful in selecting the
component values of the application.
3.1 Design parameters
LED manufacturers generally recommend values for ∆IF ranging from ±5% to ±20% of IF .
The higher LED ripple allows the use of smaller inductors and smaller output capacitors.
The advantages of higher ripple current are reductions in the solution size and cost. Lower
ripple current requires more inductor output and more capacitor output. The advantages of a
lower ripple current are reductions in heating of the LED itself and a greater range of the
average LED current before the current limit is reached. The application is designed to
supply up to four HBLEDs. The LED used in the application is a Lumides LUXEON III
Emitter LXHL-PW09 with a typical forward voltage of 3.7 V at 700 mA.
Ta bl e 1 provides a summary of the specifications of a particular application.
Table 1.Performance specification summary
Input source 4 AA batteriesVin6 V
White LED LUXEON III Emitter LXHL-PW09
LED forward voltageV
= Average inductor currentI
I
LEDavg
Ripple current on the inductor (%ILavg)∆I
Ripple current on the LED branch (%∆I
3.2 Power stage selection
For the power stage we use the ST1S10 which is a general-purpose voltage regulator stepdown DC-DC converter which has been optimized for high-efficiency small-sized equipment.
A high switching frequency (900 kHz) allows the use of tiny surface-mount components.
The synchronous rectification is implemented in order to obtain efficiency higher than 90%.
The ST1S10 provides up to 3 A over an input voltage range from 2.5 V to 16 V.
The minimum input voltage to maintain regulation, depending on the load current and output
voltage, can be calculated as:
LED
SymbolValueUnit
3.7 at 700 mAV
1A
60%
10%
) ∆I
L
LED
F
LED
= % I
= % I
LED
LED
Equation 18
V
V+⋅+=
minin
10/20
OUT
D
MAX
()
RRI
maxON_DSLLED
AN2754Design example
where R
inductor and V
is the maximum PMOS switch-on resistance, RL is the DC resistance of the
DS (on)
is the nominal output voltage.
OUT
3.3 Current sense
Amplifying the sensed voltage is a way to reduce the power loss in the current sense
resistor. The operational amplifier selected for this application must be able to work with a
common mode input voltage close to zero. The selected device is the TS951, a rail-to-rail
BiCMOS operational amplifier. The value of the current sense resistor is determined by two
factors: power dissipation on R
R
In order to keep the power dissipation for the current sensing at a minimum value, a good
choice for the sense voltage with a forward current of 1 A is 100 mV.
The ratio between feedback voltage and sense voltage gives the value of the gain of the
amplification stage:
Equation 19
reduces power dissipation but the detection of the feedback signal is more difficult.
SENSE
and the threshold level for amplifier input. Smaller
SENSE
V
Gain×=
Gain=
FB
RI
SENSELED
8.0
=
×
8
1.01
3.4 Inductor selection
The buck power stage is designed to operate in continuous mode for load current greater
than 30% of full load. We choose an inductor value producing a maximum peak-peak ripple
current equal to sixty percent of the maximum load current. This limits the RMS current in
the output filter capacitor and, as a second order effect, keeps the core losses in the
inductor reasonable.
By using a single HBLED in conjunction with the chosen current sense resistor the output
voltage is given by the following equation:
Equation 20
Let us set the value of the ripple current equal to 60% of the average current:
Equation 21
The minimum value of the output current to guarantee continuous mode is given by:
Equation 22
V9.3V1.0V8.3VVV
SENSEFOUT
LEDAVGL
LCCMminLED
=+=+=
A6.016.0I6.0I
=×=×=∆
mA3002/6.02/II
==∆=
11/20
Design exampleAN2754
[
]
[
]
The minimum value of inductance which guarantees a ripple current of 300 mA can be
calculated using
Equation 23
Equation 14:
L⋅
()
=
MIN
FI
⋅∆
VoutIESRRVin
−⋅+−
LavgLDSONMAX
Vout
Vin
MAXSWL
3.5 Output capacitor selection
The target tolerance for the LED ripple current is 10% of the forward current. In this
particular example the forward current is 1 A and the ripple current onLEDs branch is
100 mA. In current-mode converters, the load consists of the dynamic resistance of the
diodes, rD and the operating point resistance V
LED manufacturers and for those do not, it must be determined by examining the slope of
the I-V curve that is provided in all LED datasheets.
The LUXEON III Emitter LXHL-PW09 datasheet gives a typical value for the dynamic
resistance rD of 0.8 Ω at 700 mA. Given the ripple current on the inductor with the equation
below, it is possible to calculate the Z
branch equal to 10% of I
:
F
impedance to guarantee a ripple current on LEDs
C
. Typical values for rD are provided by
O/ILED
Equation 24
Z
=18.0
C
A ceramic capacitor is used and the required capacitance is selected based on the
impedance at 900 kHz.
Equation 25
3.6 Input capacitor selection
Because of the pulsating input current nature of the buck converter, a low ESR input
capacitor is required. A good input voltage filtering is important for minimizing the
interference with other circuits caused by high input voltage spikes. The following equation is
used to calculate the input ripple voltage due to capacitance and ESR:
Equation 26
iZ
∆⋅
Z
=
C
()
()
[]
()
LEDO
ii
∆−∆
LEDL
05.01.08.0
×+
05.03.0
−
()
IinRIinVinpp
ESR
Ω=
fESRZ2/1C×−×π×=
SWC
3
11090001.018.02/1C
=⋅×−×π×=
D
⋅+⋅=∆
CFsw
⋅
A 4.7 µF input capacitor is sufficient for effective input voltage filtering.
12/20
AN2754Description of the board
4 Description of the board
The evaluation board is configured as constant current supply. Current regulation is
accomplished by regulating the voltage across a current sense resistor
4.1 Input/output connection
The following table describes the input/output connections.
Table 2.Input/output connections
Reference
designator
J1LED cathodeOutput to cathode of LED
J2LED anodeOutput to anode of LED
J3Supply/sync
J4Enable
NameDescription
VIN_SW: Power input supply voltage to be tied to VIN_A. (VIN_SW max=18 V)
VIN_A: Analog input supply voltage to be tied to VIN_SW. (VIN_ A max=18 V)
SYNC: Synchronization and frequency select. Connect SYNC to GND for 900 kHz
switching frequency or connect to an external clock from 400 kHz to 1.2 MHz.
Use this connector to enable and disable DC-DC converter. Connect a jumper
between the ON pin and the center pin to enable the supply. Connect a jumper
between the OFF pin and the center pin to disable the supply. If this pin is left open,
the EVM does not operate correctly. This pin is also used for PWM dimming control
of the LED current.
13/20
Schematic and bill of material (1 A LED current)AN2754
A
5 Schematic and bill of material (1 A LED current)
Figure 9.Schematic - LED current 1 A
CIN A
100nF
1
6
8
4
Vin a
Vin sw
Pgnd
gnd
SYN
VIN
EN
GND
J3
1
2
3
4
5
Supply
J4
1
2
3
CIN SW
4.7uF
Vin
Vin
Vin
Table 3.Bill of material
QuantityReferencePart/valuePCB footprint
2Cs,C
1C
1C
1C322 nFSM/C_0805
1D312 VSM/D_1406
U1
Vin_a
Vin_sw
PGND
AGND
ST1S 10
IN
SW
3
Vf b
C3
22nF
Vin
Inh
Sync
2
5
DZ2
2.4V
A100 nFSM/C_0805
IN
SW4.7 µFSM/C_1210
OUT
COUT
2.2uF
Rz1
330
R2 5. 6k
Dz
12V
RFB
Cs
100nF
6.8k
VIN
+
-
U3
TS951ILT
RIN
1k
2.2 µFSM/C_1210
4.7uH
L1
7
1
2
1
2
Rsense
0.1
J2
J1
LED AN ODE
LED CATHOD E
1J1LED cathodeSIP/TM/L.200/2
1J2LED anodeSIP/TM/L.200/2
1J3SupplySIP/TM/L.500/5
1J4InhSIP/TM/L.300/3
1L14.7 µFSM/L_2220
1RFB6.8 kΩSM/R_0805
1RIN1 kΩSM/R_0805
1R
SENSE
1Rz330SM/R_0805
1R25.6 kΩSM/R_0805
1U1ST1S10VFDFPN 8L 4X4
1U2TS951ILTSOT23-5
14/20
0.1SM/R_0805
AN2754Board layout
6 Board layout
Board layout is critical for all switched-mode power supplies. Figure 5, 6 and 7 show the
board layout for the ST1S10 constant current control evaluation board. It is essential to keep
the high switching current circulating paths as small as possible to reduce radiation and
resonance problems.
In general the following rules should be applied:
●Traces should be as short as possible
●L1, C
●Especially the connection from the IC pin SW and the inductor must be kept short
●Ground areas should be as large as possible. If a 2-layer PCB is used, one layer should
be assigned as the ground layer and good connectivity between both layers should be
observed
●C
Figure 10. Assembly layer
, and C
IN
should be placed close to pin (VS) of the chip; CIN directly to pin 4
IN
should be placed as close together as possible
OUT
15/20
Board layoutAN2754
Figure 11. Top layer
Figure 12. Bottom layer
16/20
AN2754Typical application waveforms
7 Typical application waveforms
7.1 Startup
Figure 13. Startup
Conditions:
V
: 6V; I
IN
: 1A; L=4.7uH; C
LED
=2.2uF.
OUT
Ch3: Inductor current
Ch4: LED current
7.2 Switching waveform
Figure 14. Steady-state operation
Conditions:
V
: 6V; I
IN
: 1A; L=4.7uH; C
LED
=2.2uF.
OUT
CH1:SW
Ch3: Inductor current
Ch4: LED current
17/20
Typical application waveformsAN2754
7.3 PWM dimming using the enable function
Figure 15. PWM dimming
Conditions:
V
: 7 V;
IN
I
: 1 A; L=4.7 µH;
OUT
C
=2.2 µF, Freq=100 Hz;
OUT
duty cycle 50%.
CH1: SW; CH2: Enable Ch3: inductor current Ch4:LED
18/20
AN2754Revision history
8 Revision history
Table 4.Document revision history
DateRevisionChanges
28-Aug-20081Initial release
19/20
AN2754
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