An Electrolytic Capacitor-Less LED Driver using Harmonics … · 2016. 5. 19. · Krunal H. Patel1...

IJSRD - International Journal for Scientific Research & Development| Vol. 4, Issue 03, 2016 | ISSN (online): 2321-0613

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An Electrolytic Capacitor-Less LED Driver using Harmonics Injection

Technique Krunal H. Patel1 Dhiren A. Luhar2 Prof. Chetan D. Upadhyay3

1,2M.E. Student 3Assistant Professor 1,2,3L.D. College of Engineering, Ahmedabad-380015

Abstract— An LED lighting system requires a driver for to

the power conversion from ac voltage to dc regulated current

output. The electrolytic capacitor is use in the LED driver.

But life of electrolytic capacitor is less then LED. Here we

use film capacitor instead of electrolytic capacitors. But it’s

Maximum capacitance per Unit volume is lower Then

electrolytic capacitor. The output current in above driver is

pulsating. In this paper we use harmonics injection

Technique for make output current constant.

Key words: Flyback Converter, Peak to Average Ratio,

Harmonics Injection Technique

I. INTRODUCTION

Nowadays LEDs are widely used in lighting area. These are

niche applications compared with general illumination. An

LED is a p-n junction semiconductor which emits light

spontaneously directly from an external electric field

(electroluminescence effect). Typical lifetimes are 25,000 to

50,000 hours, but heat and current settings can extend or

shorten this time significantly. An LED lighting system use

driver to handle the power conversion from ac voltage to dc

regulated current output. In driver large capacitance is

Require for balance the energy difference between the input

pulsating power and the output DC power delivered to the

LED load. Electrolytic capacitors are use for the energy

storage capacitor in LED lamps due to their high energy

density and low cost. The estimated lifespan of the LED

devices can last up to at least 50,000 hours but electrolytic

capacitors typically last only up to approximately 10,000

hours. The reliability of the electrolytic capacitors include

their sensitivity to their operating temperature, ripple current

and internal equivalent series resistance (ESR). [1] Table I

shows a comparison on the properties of three types of

capacitors. Properties compared at the ambient temperature

of 50oC, available range of values and maximum

capacitance per unit volume (μF/cm3) at the same voltage

rating. Among the three types of capacitors, electrolytic

capacitors give the highest value available in a single

package. The lifetime of the electrolytic capacitors will be

shortened by half if the operating temperature is increased

by 10oC. [2]

Capacitor Lifetime

(hours)

Available

Range

Maximum

capacitance

per Unit

volume

(µF/cm3)

Electrolytic <10000 1 µF-12 mF 200

Polyester

Film >100000 10 pF-80 µF 30

Ceramic >100000 10 pF-10 µF 5

Table 1: Comparison among several types of capacitors

Thus, for high ambient temperature, the lifetime of

the electrolytic capacitor becomes the critical factor that

determines the lifetime of the entire application. polyester

film capacitor and Ceramic capacitor are the best choice in

terms of the lifetime. But their capacitance per unit volume

and range are lesser then Electrolytic capacitor. So the

pulsating current brings flicker with twice the line

frequency[3]. The peak to average ration of output current

should be 1. Here we use harmonics injection Technique for

make output current constant and make this ratio near to 1.

We can set duty cycle of flyback converter for Harmonics

injection. In order to inject the third and fifth harmonics into

the input current, the function of the duty cycle in a half-line

cycle is derived.[5]

II. RELATIONSHIP BETWEEN VOLTAGE RIPPLE AND STORAGE

CAPACITANCE

Assume that the input power factor is unity. The input

voltage and input current is defined as Vin (t) = Vm sinωt [1]

Iin (t) = Im sin ωt [2]

The instantaneous input power can be derived as

Pin (t) = Vin (t) Iin (t) = (1 cos 2 )

2

m mV I t [3]

It can be obtained from (3) that the average input

power is

Pin 2

m mV I

[4]

Fig. 1: Key waveform of the converter (power factor =

unity)

Assume the efficiency of the converter is 100%. It

means that the average input power is equal to the output

power Po. These results in

Po 2

m mV I

[5]

It can be seen that when Pin > Po, CB is charged,

and VC increases, and when Pin < Po, CB is discharged, and

VC decreases.

Substitution of (5) into (3) yields

Pin (t) = Po (1 − cos 2ωt) [6]

From the (6) and Fig. 1, we can see that Pin is equal

to Po at Tline/8 and 3Tline/8. It can be seen that CB is charged

from Tline/8 to 3Tline/8, and VC increases. Consequently, VC

gets its minimum and maximum values at Tline/8 and

3Tline/8, respectively.

An Electrolytic Capacitor-Less LED Driver using Harmonics Injection Technique

(IJSRD/Vol. 4/Issue 03/2016/270)


The energy charging CB from Tline/8 and 3Tline/8 is

line

line

3 /8

/8

[ ]

T

in

T

PoE P t Po dt

[7]

ΔE can also be expressed as

ΔE = 1

2

CB*(V 2Cmax− V 2Cmin) [8]

From (7) and (8), we have

CB =2 2

max min

2

* ( ) C C

Po

V V

=

max min

2

* *[( ) / 2] C C C

Po

V V V

[9]

From this equation we know that if ΔVC is

intentionally increased, the value of CB will be significantly

reduced.[4] Thus, film capacitors can be used instead of

electrolytic capacitors.

III. PEAK-TO-AVERAGE RATIO OF OUTPUT CURRENT

Fig.1 shows that the waveform of the output current follows

the shape of a sine square function. Its peak-to-average ratio

is 2. Therefore, it is necessary to reduce the peak-to-average

ratio of the output current to guarantee LED’s safety.[5]

A. Harmonics Injection Into Input Current to Reduce Peak-

to-Average Ratio of Output Current

The optimum output current for driving the LED is a pure dc

current, its peak-to-average ratio is 1. In such a situation, the

output power will be purely dc, and correspondingly, the

input power must be purely dc due to the absence of the

storage capacitor. So, the input current will be

( )( ) * sin

o o o

in

V V Ii t

Vin t Vm t

[10]

According to this the input current tends to infinity

at the zero-crossings of the input voltage. It is show in Fig.

2.

Fig. 2: Key waveform when Pin=Po

There is a conflict that if the input power factor is

unity, the peak-to-average ratio of the output current is 2,

and if the peak-to- average ratio of the output current is to

be reduced to 1, the input power factor will be zero.

Therefore, the trade-offs must be balanced. This is

achievable by injecting some odd harmonics into the input

current to reduce the peak-to-average ratio of the output

current, while maintaining the input power factor to be

higher than 0.9 as required by the regulation standards.

B. Relationship between Injected Third and Fifth

Harmonics and Peak-to-Average Ratio of Output Current

When both the third and fifth harmonics are injected, the

input current is derive by

iin1+3+5(t) = I (sin ωt + I*3 sin 3ωt + I*

5 sin 5ωt) [11]

And the corresponding output current io1+3+5 is

1 3 5

1 3 5

* *

3 5

( ) ( )( )

sin (sin sin 3 sin 5 )

in in

o

o

m

o

V t I ti t

V

V It t i t I t

V

[12]

The average value of io1+3+5 in a half-line cycle is

1 3 5 1 3 5

0

1( )

2

m

o o

o

V Ii i t

V

[13]

It can be seen from (13) that the average value of

io1+3+5 in a half-line cycle is determined by the fundamental

component of the input current and has nothing to do with

the injected Harmonics. From (12) and (13), the normalized

output current with the base of Io1+3+5 is derived as

* 1 3 5

1 3 5

1 3 5

2 * * * * 2 * 4

3 5 3 5 5

( )( )

2 sin [(1 3 5 ) 4( 5 ) sin 16 sin ]

o

o

o

i ti t

i

t I I I I t I t

[14]

Fig. 3: Waveform of i∗o1+3+5

As shown in Fig. 3, it is necessary to find the

turning points of i∗o1+3+5 by using (14). After that, calculate

the minimum peak-to-average ratio of the output current of

the three cases within the corresponding range of I∗3 and I*5.

Case I Case II Case I

I*3 0.465 0.44 0.389

I*5 0.135 0.092 0.087

Min(i∗o1+3+5|peak) 1.340 1.382 1.396

Table 2: Minimum peak to average ratio of the output

current and the corresponding I*3 and I*

5 in three case.

When we determine the range of I*3 and I*

5, the

input power factor should be ensured to be higher than 0.9.

From Table the minimum peak-to-average ratio is 1.34,

which occurs in case I.

IV. CALCULATION FOR FLYBACK CONVERTER

Fig. 4: Flyback converter

Electrolytic capacitor less flyback converter is shown in

Fig.4. Here L0 and C0 is high frequency harmonics filter in

the secondary side. The switching frequency is much higher

than the line frequency, the input voltage can be treated to

be constant in a switching period. Here switching frequency

is 200kHz and line frequency is 50Hz. The flyback

converter is designed in discontinuous current mode (DCM)




so that a high power factor is automatically achieved. Two

mode of operation.

A. Mode 1:

When the switch Q turns on, the primary current ip increases

linearly from zero, and it reaches the peak value when the

switch turns off. The peak and average value of ip in a

switching cycle are

sin( )

in y m L y

p pk

p s p s

v D V w t Di t

L f L f

[15]

2sin1

( ) ( )2 2

m L y

p av p pk y

p s

V w t Di t i t D

L f

[16]

From (16) and Fig. 4, it can be obtained that the

average input current in a switching cycle is 2

sin( )

2

m L y

in

p s

V w t Di t

L f

[17]

It can be seen from (17) that if Dy is kept constant

in a halfline cycle, the input current will be proportional to

the input voltage, and so, PFC can be automatically

achieved.

B. Mode 2:

When Q turns off, the secondary diode conducts, and the

energy stored in the primary side of the transformer is

transferred to the secondary side. The peak value of the

secondary current is then

sin( ) ( )

m L y

s pk p pk

p s

nV w t Di t ni t

L f

[18]

Thus, the secondary current ts is will decrease

linearly. The reset time tr for is to fall to zero is

0 0 0

sin sins s pk s m L y m L y

r

p s s

L i L nV w t D V w t Dt

V V L f nV f

[19]

The corresponding duty cycle is

0

sinm L yr

r

s

V w t DtD

T nV

[20]

In a switching cycle, the average value of the

secondary current can be expressed as 2 2 2

0

sin1( ) ( )

2 2

m L y

s av s pk r

p s

V w tDi t i t D

L V f

[21]

Fig. 5: waveforms of the primary and secondary currents of

flyback converter

Thus, the average value of output current in a half-

line cycle can be derived as 2 2

0

00

1( )

4

m y

s av L

p s

V DI i t d w t

L V f

[22]

According to (17), if the duty cycle is kept constant

at Dy 0, the fundamental component of the input current is 2

12

m y

p s

V DI

L f

[23]

If the third and fifth harmonics are injected into the

input current, substitution of (17) and (23) into (11), leads to 2

1 3 5

2

0 * *

3 5

sin ( )

2

(sin sin 3 sin 5 )2

m L y

p s

m y

L L L

p s

V w tD t

L f

V Dw t I w t I w t

L f

[24]

Equation (24) can be rewritten as

1 3 5

* 4 * * 2 * *

0 5 3 5 3 5

( )

16 sin (4 20 ) sin (1 3 5 )

y

y L L

D t

D I w t I I w t I I

[25]

Equation (25) means that we can vary the duty

cycle in a half-line cycle to inject the third and fifth

harmonics into iin . Substituting I∗3 = 0.465 and I∗ 5 = 0.135

into (25), the duty cycle is 4 2

1 3 5 0( ) 2 .16 sin 4.56 sin 3.07

y y L LD t D w t w t

[26]

The function of Dy1+3+5 is too complicated to be

implemented. The accuracy of (26) strongly depends on ωLt0

,For the sake of simplicity,(26) can be rewritten as

0( ) (1 sin )

y fit y LD t D a k w t

[27]

Where k is determines the input power factor and

ak determines the output power. k is find by the set power

factor to 0.9. and a is find by the take output current

constant.

Substitution of k=-0.61 and a=2.02, equation (27)

became

0( ) 2.02(1 0.61 sin )

y fit y LD t D w t

[28]

So the duty cycle can be easily implemented by

simply sampling the input voltage.

V. SIMULATION AND RESULTS

Data form the prefer paper. [5]

Parameters values

Input voltage Vin 198-264VAC/50Hz

Output voltage Vo 25VDC

Output current 0.35A

the transformer: n 3

Primary inductance: Lp 270μH

Primary filter capacitor Co 0.1 μF

Input filter inductance Lin 100μH

Output filter capacitor Co 0.47 μF

Output filter inductance Lo 30 μH

Table 1: Data form the prefer paper. [5]

A. Simulation Model (Harmonics Injection-Sine Block) In

Matlab

Fig. 6: Simulation Model using Sine Block




Fig. 7: input voltage and current using sine block

Fig. 8: output voltage and output current and secondary

current using sine block

B. Simulation Model (Function of Duty Cycle) in Matlab

Fig. 9: simulation model using duty cycle function

Fig. 10: input voltage and current using duty cycle function

Fig. 11: output voltage and output current and secondary

current using duty cycle function

VI. CONCLUSION

These papers represent the harmonics injection using

function of duty cycle. By injecting the third and fifth

harmonics into the input current the peak-to-average ratio of

the output current is reduced. In both simulation model with

direct sine block and duty cycle function the output and

input waveforms are similar.The experimental results

showed the effectiveness of the proposed electrolytic

capacitor-less ac–dc LED driver.

APPENDIX: LIST OF SYMBOLS

Vm peak input voltage

I m peak input current

w angular frequency

Pin input power

Po optput power

VC voltage of CB storage capacitor

VC max maximum voltage values of CB

VC min minimum voltage values of CB

I*3 the normalized amplitude of the third

harmonics

I∗5 the normalized amplitude of the fifth

harmonics

1 3 5

( )y

D t

duty cycle with third & fifth harmonics

injection

( )y fit

D t

fitting function of the dutycycle

(harmonics injection)

( )

p pki t

primary peak current

( )

p avi t

primary average current

( )

s pki t secondary peak current

( )s av

i t

secondary average current

IO output current

Lp primary inductance

REFERENCES

[1] Shelf-Life Evaluation of Aluminuin Electrodytic

Capacitcirs(IEEE Trans. Components, Hybrids, and

Manufacturing Technology)

[2] Current Source Ballast for High Power Lighting

Emitting Diodes without Electrolytic Capacitor (Y. X.




Qin, 2Henry S. H. Chung, Senior Member, IEEE, 3D.

Y. Lin, and 4S. Y. R. Hui, Fellow, IEEE)

[3] A Flicker-free Electrolytic Capacitor-less AC-DC LED

Driver(Shu Wang, Xinbo Ruan, Kai Yao, Zhihong Ye)

[4] Means of Eliminating Electrolytic Capacitor in AC/DC

Power Supplies for LED Lightings(Linlin Gu, Xinbo

Ruan, Senior Member, IEEE, Ming Xu, Senior

Member, IEEE, and Kai Yao)

[5] A Method of Reducing the Peak-to-Average Ratio of

LED Current for Electrolytic Capacitor-Less AC–DC

Drivers(Beibei Wang, Xinbo Ruan, Senior Member,

IEEE, Kai Yao, and Ming Xu, Senior Member, IEEE)

An Electrolytic Capacitor-Less LED Driver using Harmonics … · 2016. 5. 19. · Krunal H. Patel1...

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