DOI : 10.17577/IJERTV5IS030131
- Open Access
- Total Downloads : 206
- Authors : Jyoti Devi, Dr. Abhro Mukherjee, Amiya Ranjan Behera
- Paper ID : IJERTV5IS030131
- Volume & Issue : Volume 05, Issue 03 (March 2016)
- DOI : http://dx.doi.org/10.17577/IJERTV5IS030131
- Published (First Online): 15-03-2016
- ISSN (Online) : 2278-0181
- Publisher Name : IJERT
- License:
This work is licensed under a Creative Commons Attribution 4.0 International License
Embedded Controller Design for Efficient Lighting System
Jyoti Devi, Dr. Abhro Mukherjee, Amiya Ranjan Behera
Department of Electrical & Electronics Manipal Institute of Technology,
Manipal National Institute of Science & Technology, Berhampur
AbstractLight Emitting Diode is the future of lighting. LED is a semiconductor device and its nonlinear. For driving LED we need a driver circuit. Hence LEDs are nonlinear devices and boost driver circuit is unstable, its difficult to achieve the desired output. So here we have taken these as a problem statement and tried to linearize and stabilize the system using a proper controller. This paper demonstrates how to stabilize the system. In this paper a controller is designed by considering LEDs are nonlinear (not the linearized form of LED is taken). The steps how to stabilize the nonlinear system is shown in detail, and it is also showing the obtained results.
Keywords Efficient Lighting, LED, Driver Circuit, Lighting Control, Linearization, Non-linear system.
-
INTRODUCTION
Nowa days LEDs are replacing every lamp because of long life, good CRI, low energy consumption and free of pollution. Hence it is a semiconductor device, frequently switching does not affect the lamp life. The proposed system contains set of high power LEDs and a 12V dc supply.
The light output from LED depends on the average current flows through the LED. Because of manufacturing fault the forward voltage drop of LEDs varies, so the current flowing through each LED will differ. And due to different currents flowing through LEDs the lumen output from LEDs will vary. This leads to non-uniform distribution of light, which is not acceptable to users. Therefore the system requires a Driver circuit which will make sure that the current flowing through all LEDs are same. The design of driver circuit depends on mostly the LED forward voltage drop, available input voltage. After designing driver circuit the current through LED doesnt meet the required current, because of non-linearity. The V-I characteristic of LED is non-linear. Hence the system becomes unstable. Therefore the system needs a controller to overcome this problem.
In this paper the controller construction is discussed briefly for a non-linear dynamic system. As LED is a non-linear
cancels out the systems non-linearities and the other controls the resulting linear system.
-
DRIVER CIRCUIT DESIGN
×
×
×
×
The proposed system contains 36 identical LEDs, they are arranged in 6 parallel strings and each string contains 6 LEDs connected in series. The forward voltage drop of each LED is varies from 2.7V to 3.2V. The typical forward voltage drop is 2.85V. The maximum output voltage will be 3.2 6 = 19.2 V. The input voltage is varies from 10V to 14V. The minimum input voltage is 10V. When the output voltage is greater than input voltage, it requires Boost driver circuit design. In Boost circuit design the output voltage is minimum 50% greater than input voltage. The required current is 700 mA through each LED, there are six strings therefore the load current is 700 6
= 4.2 mA .The circuit shown below is Boost driver circuit.
Fig. 1. Boost Driver Circuit
-
DRIVER CIRCUIT DESIGN Input Voltage (VS ) = 10 V
Output Voltage (VO ) =19.2 V
Required output current (ILOAD) = 4.2 A Switching frequency (FS ) = 25 KHz
For calculating L and C value, it is required to calculate Duty Cycle (D) and Inductor current (IL).
device, the control design process in the case of linear systems is different from non-linear systems. Feedback Linearization Method is used to construct a state-feedback control strategy.
Vs(min)
V
V
DutyCycle(D) = 1
o(max)
10
= 1 19.2 = 0.4791 (1)
The basic idea of the feedback linearization approach is to use a control consisting of two components; one component
ILoad 4.2
IL = (1 D) = 0.4791 = 8.0629A (2)
s(min) × ×
s(min) × ×
V D T
InductanceL =
IL
Load × ×
Load × ×
I D T
CapacitanceC =
= 1.188 × 104H (3)
= 1.3973 × 105F (4)
The above set of equations is continuously differentiable. Using Taylors series expansion method the state space matrix can be written in the following way
IF × Rd
Rl+Rs 0 1 X 1
0
0
-
CONTROLLER DESIGN
which leads to unacceptable output current and voltage. The lumen output of LED is directly proportional to the current
which leads to unacceptable output current and voltage. The lumen output of LED is directly proportional to the current
Because of non-linear devices are present in the circuit,
X =
L
Is
C.V t
VC
× e V t
1 + L .U X2 0
Y =
Y =
V t
0 1
V t
0 1
V t
V t
1
X2
1
X2
+
+
.U
.U
0 Is × e VC X 0
0
0
0
0
flow through the LED. So it is necessary to design controller which make sure that the stable desired current will flow
At equilibrium point (0, 0) the state space matrix will become,
X2
X2
through LEDs.
-
LINEARIZATION OF DIFFERENTIAL EQUATIONS OF
X =
Rl+Rs
L
0
0
Is
C.V t
X1 +
1
L .U
0
NON-LINEAR SYSTEM
0 Is X1
0
0
0
The replacement of a nonlinear system model by its linear approximation is called linearization. The need for lineariza-
Where,
Y = V t
1
X2 +
0 .U
tion is that the dynamical behavior of many nonlinear system models can be well approximated within some range of variables by linear system models. The derivation of dynamic
AON =
Rl+Rs L
0
0
Is
CON =
CON =
V t
1
V t
1
C.V t
1
, BON = L ,
0
system model by nonlinear differential equation;
0 Is
0
0
0
0
0
When Switch is closed; According to Kirchhoffs Voltage and
0
0
When Switch is closed; According to Kirchhoffs Voltage and
0
0
&DON =
&DON =
When switch is opened; According to Kirchhoffs laws the
Fig. 2. Equivalent Boost Converter circuit for the switch closed
Current laws following equations can be obtained.
VS = VL + ILRL + ILRS (5)
Where RL, RS and RC are the internal resistances of Inductor, Switch and Capacitor respectively and whose values are 0.1,
0.001 and 0.01 ohm respectively.
VL = VS IL(RL + RS ) (6)
Fig. 3. Equivalent circuit for switch opened
following equation can be written
VS = VL + VC + ILRL + ICRC (13)
VL = VS VC ILRL (IL IO )RC (14)
dt L
dt L
S C L L C C S V t
S C L L C C S V t
dil = 1 (V V I (R + R ) + R I .e VC ) (15)
IL = IC + IO, IC = IL IO (16)
diL 1
dVC = 1 I
IS VC
.e
(17)
=
dt L
(VS IL(RL + RS )) (7)
V t
L
L
dt C C
L
L
IC = IO (8)
At equilibrium point (0, 0) the state space matrix
dvC
C = Idt
dvC
C = Idt
VC
× e V t
VC
× e V t
(9)
(9)
X =
X =
L
1
C
L
1
C
Rl+Rc
1 + Is.Rc X1
X2
X2
1
+
L
0
.U
+
L
0
.U
S
X2
S
X2
L.V t
I s
C.V t
L.V t
I s
C.V t
dVC
= IS e VC
(10)
0 Is X1
+ 0 .U
V t
dt C
VC
V t
V t
Y =
Y =
Where,
0 1 X2 0
Rl+Rc
Rl+Rc
1 + Is.Rc
1 + Is.Rc
1
1
T heoutputcurrentI0 = IS × e V t (11)
T heoutputvoltageV0
= V
C
(12)
T heoutputvoltageV0
= V
C
(12)
AOFF =
L
1
C
L
L.V t
I s
C.V t
, BOFF =
L
0
,
AOFF =
L
1
C
L
L.V t
I s
C.V t
, BOFF =
L
0
,
T heoutputvoltageV0
= V
C
(12)
AOFF =
L
1
C
L
L.V t
I s
C.V t
, BOFF =
L
0
,
T heoutputvoltageV0
= V
C
(12)
AOFF =
L
1
C
L
L.V t
I s
C.V t
, BOFF =
L
0
,
COFF =
COFF =
V t
1
V t
1
&DOFF =
&DOFF =
0 Is l 0l
0
0
0
0
X = 8417.508Z + 4633459372Z
1
1
2
2
1
1
The average of two states can be done in this manner A =
X1l = (8417.508)Z2 + (4633459372)Z1l
AON
. D + AOFF
. (1- D)
X2 313774709.8Z1)
Where D is the duty cycle. Similarly the average of B, C and D can be done. By substituting all the values the state space matrix of the system can be obtained as shown in below.
Differentiate Z(t) with respect to time gives
Z = T .F (x) T .G(x)atX = T 1(Z)
x x
l
l
c
c
l
l
s
s
(
(
+
+
s
s
c )(1 D)
c )(1 D)
A =
A =
(R +R )(1D)(R +R )D 1 I .R
L
L
L.V t
L
L
L.V t
l Z1 = (0).f1 + (0.3187 × 108).f2
,
8
,
8
1 (1 D) Is
= (0.3187 × 10 ).(37277.17X1 550484.6742X2)
C
1 l
0 Is l
C.V t
0l
= Z2
B = L
0
, C = 0
V t &D =
1 0
Z2
= (1.188 × 104).f1
(1.7543 × 103).f2
By substituting all the values
A =
A =
l
l
889.630 4047.392
889.630 4047.392
37.277 × 103 55.0484 × 104l
, B =
8417.508
l
l
0 ,
= (0.3187 × 108).(889.6304X1 4047.397X2) (1.7543 × 103).(37277.17X1 550484.6742X2)
0 7.6923
C = 0 1
&D =
0l
0
0
= 965.715X2 65.5521X1
= (3.0765 × 1011)Z1 (551785.3262)Z2
-
CONTROLLER FORM:
The objective of this is to constructing state-feedback stabi- lizing controller for dynamic system model. To construct such
T
× 4 ×
× 4 ×
.G(x) = (1.188 10 ) (8417.508)u
x
= u
a controller, it is advantageous to work with an equivalent system model rather than with the original nonlinear one. Thus the nonlinear system model reduced to linear system model.
Z =
Z2 +
l
l
(3.0765 × 1011)Z1 (551785.3262)Z2
0l
u
u
The Controllability matrix QC = [ B; AB ]
QC =
QC =
0.0001 × 108 0.0749 × 108l
0 3.1378 × 108
0 3.1378 × 108
Let v = ((3.0765 × 1011)Z1 (551785.3262)Z2) + u
-
LINEAR STATE-FEEDBACK CONTROL:
The linear state-feedback control law, for a system modeled
The linear state-feedback control law, for a system modeled
The linear state-feedback control law, for a system modeled
The linear state-feedback control law, for a system modeled
l
l
T1 = the last row of Inverse of the Controllability matrix
by X = AX + Bu, is the feedback of a linear combination of
QC 1
0.1188 × 103 0.0028 × 103
1 H × l l
1 H × l l
=
=
0 0.3187 × 108
all the state variables and has the form u = -KX, where K is a constant matrix.
The closed loop system then is X = (A – BK)X
T = 0 0.3187 108 X1
X2
×
×
l
l
T1 = (0.3187 108)X2 The goal is now construct a state variable transformation Z=T(X) for which there is a inverse X= T1(Z).
The poles of the closed loop system are the roots of the characteristic equation
|SI (A BK)| = 0
|
|
The roots of the closed-loop characteristic equation |SI (A
T (X) = T1 Lf.T1
T T
BK) = 0 are in desirable locations in the complex plane.
The Transfer Function of linearized system is
2.414 × 109
1 1
Lf.T1 = .f1 + .f2
G(S) = 2 5 8
x1
x2
S + (5.514 × 10 )S + (6.406 × 10
= (0).f1 + (0.3187 × 108).f2
= (0.3187 × 108).(37277.1706X1 550484.6742X2)
(0.3187 × 108)X2
(0.3187 × 108)X2
= (1.188 × 104)X1 (1.7543 × 103)X2
T (X) =
(1.188 × 104)X1 (1.7543 × 103)X2l
×
×
× ×
× ×
Let (0.3187 108) X2 = Z1 and (1.188 104)X1 – (1.7543 103) X2 = Z2
Then,
Fig. 4. Closed loop System
X2 = 313774709.8Z1
(1.188×104)X1 (1.7543 × 103)X2 = Z2
The characteristics equation of closed loop is 1 + K G(S) = 0
S2 + (5.514105)S + 6.406 × 108 + K(2.414 × 109) = 0
Rouths stability criterion can be used to find the range of K at which the system reamain stable. For the value of K greater than -0.2653, the system is stable. By tuning the K value in simulation the desired output can obtained. Here the K value is 0.7351 at which the output current meets the desired current output i.e; 4.2 A. Let the desired Eigen values be 1, 2 then
(S 1)(S 2) = 1 + KG(S) S2 + S(1 + 2) + 12 =
S2 + S(5.514 × 105) + (6.406 × 108) + K(2.414 × 109)
Substitute K = 0.7351;
S2 + S(1 + 2)+ 12 = S2 + S(5.514 ×105)+ 2415131400
By comparing co-efficients 1, 2 values can be found
1 = 4415.355
2 = 546984.645
Fig. 5. Simulation of Nonlinear System
|SI (A BK)|;
(A BK) = 889.630 4047.392 1
1
1
37277.17 550484.6742
8417.508K1 8417.508K2
1
1
0 0
= 889.63 8417.508K1 4047.392 8417.508K2
37277.17 550484.6742
1
1
|SI (A BK)|
= S + 889.63 + 8417.508K1 4047.392 + 8417.508K2
37277.17 S + 550484.6742
= (S + 889.63 + 8417.508K1)(S + 550484.6742)
(4047.392 + 8417.508K2)(37277.17)
Compare the above polynomial with the closed loop system characteristic equation
551374.3046 + 8417.508K1 = 5.514 × 105
K1 = 3.0526 × 103K2 = 5.61023
-
-
RESULT:
The Simuation Model of nonlinear System:
dil = (Rl + Rs)D (Rl + Rc)(1 D) .i
Fig. 6. The plot of unstable Output Current
Fig. 7. The Closed loop System
dt L L
c
c
isRc Vc 1 1
+ ( e V t )(1 D).V + U CV t L L
dt
dt
L
L
L
L
C.V t
C.V t
V t
V t
c
c
dVc = 1 (1 D)i
+ Is .e Vc V
Substitute all the values, then the equations are
diL = 889.63i
dt L
1515.6639VC
+ 8417.508U
Fig. 8. The Plot of Stable Output Current
VC = 37277.17i
dt L
550780.6117VC
To stabilize it, the feedbacks have been taken from state variables and multiplied by appropriate gain then given to the
The output of the system can be obsereved clearly as unstable. system as input. The block diagram of the closed loop system
is shown below.
1
1
V = (3.0765 × 10 1)Z1 (551785.3262)Z2 + U WhereZ1 = (0.3187 × 108)X2,
Z2 = (1.188 × 104)X1(1.7543 × 103)X2
v = K1.iL K2.VC
U = (65.5552)iL 1954.08773VC
Here it can be clearly observed that the system became stable even though there are non-linear elements in system using State Feedback Linearization Control Method.
-
CONCLUSION:
To reduce energy consumption and to improve effi- ciency,other conventional lamps can be replaced by LEDs. Number of LEDs are used in a fixture to provide high light output. If the LED voltage is greater than the input available voltage then a Boost Driver circuit is required to drive current through LEDs. But the boost convertor is an unstable converter, so that eventually the system becomes unstable and nonlinear. The unstable nonlinear system can be stabilized using Linearizing State-Feedback Controller. In this paper the unstable current through LED is stabilized using Linearizing State- Feedback Control technique. Simulation result and mathematical results indicates the same. Further this can be taken forward for practical implementation in lighting industry.
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