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Designing a high-voltage generator circuit for HDTV

Posted: 16 Feb 2004     Print Version  Bookmark and Share

Keywords:POWER 

/ARTICLES/2004FEB/B/2004FEB16_AMD_POW_TA.PDF

By Zhuang Wanchun

Shui Gonghao

HDTV Laboratory of Digital

R&D Center

TV Business Unit

Skyworth Group

HDTV features support for

three kinds of video compres-

sion formats: DTV signals such

as PAL line-double/field-

double, PAL60 and PAL75;

HDTV image format such as

1080i@50Hz, 1080i@60Hz

and 720P; and VGA format for

PC such as VGA 640_480 and

VGA 800_600.

Scanning circuits which

meet this requirement usually

have broad working frequency

spectrum, with horizontal fre-

quencies (fh) ranging from

28kHz to 48kHz and vertical

frequencies (fv) ranging from

50Hz to 75Hz. Increased hori-

zontal frequency will double

the active power of the tube.

This increment, along with the

inherent high-voltage active

power which is as high as 60W

for the 32-inch CRT, 16:9

Toshiba tube, will significantly

boost the operation current of

horizontal output tube in the

horizontal output stage. This

will generate row deflection

and high-voltage and will incur

design difficulties like heat and

high power driving.

Hence, separating the high-

voltage generator circuit from

horizontal output stage has be-

come a prior solution for de-

signers. One of the advantages

of this is that the load package,

traditionally combining row

deflection coil and high-volt-

age stage, is separated into two

independent units, which re-

duces the distributed power

and improves the reliability of

the product.

Another advantage is that

separating high-voltage from

horizontal output stage can

dramatically remove the corre-

lation between high-voltage

and horizontal current, which

is vital for multi-frequency

scanning circuit. Designers can

develop specific high-voltage

stage independently according

to different application criteria

and employ feedback control

circuit to ensure that the dis-

play system maintain a highly

stablehigh-voltageintheanode

of the CRT under different scan

patterns and image contents.

Inourexample,inductorLis

charged by power supply B+

when electric switch S is

switched on, with the current

going in the direction of the

solid arrow. The current will

increase and reach its maxi-

mum value of ILM during the

period Ton, and the power of L

is calculated as:

PC =

1

2

CVcm

2

(t=2) (1)

Capacitor C is charged by L--

which acts as a power supply--

when S is switched off, and the

internal current of L will drop

over time, reaching zero at the

moment of T2. The power of C

is calculated as:

PL = 1

2

LIm

2

(t=1) (2)

Atthispoint,thetopofChas

cumulated positive electrons

and negative electrons at the

bottom. C will in turn act as a

power supply and charges L

when L's current reaches zero.

This time the current flow has

an opposite direction, shown as

the broken line in Figure 1. As

the charging current increases,

the voltage on C will reduce

gradually and reach zero at the

moment 3. We define Tr as the

flyback pulse width, which is

equal to (3-1), and the rela-

tionship between Tr and L and

C is

Tr = LC (3)

During the periods 3 and

4, the energy accumulated in

the inductor L will be released

again. However, this time the

induced electromotive force of

L is positive at the bottom and

negative at the top. This will

power on diode D and transfer

the energy back into B+

. The

diode acts as a damper in this

situation.

Thehigh-voltagediodeDH is

only powered on during the

flybackpulseduration,andload

RL onlyconsumespowerduring

theoffperiodofS.Accordingto

the law of conservation of en-

Designing a high-voltage generator circuit for HDTV

ergy, the energy accumulated

duringon-periodofSisequalto

that of off-period. And we can

arrive at this equation:

PL = PC + P0 (4)

P0 in the equation is the en-

ergy absorbed by load RL. From

Equation 4 we can deduce the

formula of flyback pulse volt-

age when P0 = 0.

If P0=0, it will follow that

PL=PC, and from Equation 1

and Equation 2 we get:

1 LIm

2

= 1 CVcm

2

C 2 (5)

so:

Vcm

2 =

L

C

Im

2

since:

Im = 1

L

B+Ton

the following holds:

Vcm

2

=

L 1

C L

B+Ton =

1

(B+Ton)2

LC( )2

(6)

or:

Vcm =

1

LC

B+Ton =

B+Ton

Tr

From Equation 6 we can

know that there is a propor-

tional relationship between the

width of flyback pulse and the

voltage in power supply B+

and

thedurationofon-periodTon.If

we apply the square wave of

horizontal frequency, TH, on S

as the actuating signal, we can

get Ton=Toff= 1

/2TH, and Equa-

tion 6 can be simplified to

T S

D C L

FBT

D

HV

R

B

on

Toff

TH

+ L

H

Figure 1: Designers can develop a specific high-voltage stage and employ feedback control circuit to maintain stability.

IL

1

V

0

0

t

t

L

2 3 4

Figure 2: Wave forms of current and voltage of inductor L.

Vcm = 1

2

B+TH

Tr

(7)

or

B+ =

2VcmTr

TH

(8)

Figure 3 is a high-voltage

generator circuit implementa-

tion for Equation 7, in which

S2, D2 and L2 combine to form

a buck power supply B+

--its out-

put voltage is controlled by the

PWM duty cycle. On the other

hand, switch S1 will generate

flyback pulse Vcm, under the

control of horizontal frequency

square wave. From Equation7

we can know that the flyback

pulse's amplitude has a propor-

tional relationship with B+

when TH is constant, whereas

high voltage VH is equal to

nVcm (n is the turn ratio of the

high-voltage transformer).

Stableflybackpulse'sampli-

tude is a premise for the high-

voltage to maintain stability.

According to the law of conser-

vation of energy, the amplitude

of flyback pulse will decrease as

thepowerofthesecondaryload

increase, with constant B+

. In a

practical implementation,

when B+

equal to 100V and

CRT's anode current is equal to

zero, measured value for Vcm

should be 1000Vp-p. The am-

plitude of flyback pulse will

shrink to 850Vp-p when the

anode current increase to

1.5mA.

The working mechanism for

the circuit shown in Figure3 is

that high-voltage will drop as

the anode current increases.

High-voltage is sampled by R5

to increase the PWM duty cycle

of the output of power manage-

ment IC TL494, which will pro-

long the on-period of S2 and

boost the voltage of B+

, com-

pensating the amplitude of Vcm

to the original level and keep-

PWM

ZD1 R1 D2 CL

R3

D3

S3

S1

S2

L

L2

D

HV

R4

R5

+12V

PWMTL494

1:n

D1

+24V

+200V

R2

A

TH

Cr

L1

Vcm

TH

0

0

0

0

0

0

0

H

Figure 3: High-voltage generator circuit has an output voltage controlled by the PWM duty cycle.

+12V

L1 C8

C3

R16

R13

C4

Q8 D2

C17

R19

R20 R11

R14 C6 D1

D3 C13 R15

Q5

Q1

R17

C9 C15

C5

R2

13R5

R6

Q3

C1

R4

TL494

9

R7

Q2 D5

S1

S2

R10

2

R8

C10

C7R1

D6

S3

H-PUS

L

7

C18

8

7

6

5

4

3

2

1

9

10

11

12

13

14

15

16

+

Figure 4: Sample & control circuit of high-voltage generator.

ing high-voltage stable.

Figure 4 is a sampling and

controlling circuit, which will

keep the high-voltage of anode

stable.Thesamplingvoltagein-

puts at S2 will produce a PWM

control signal output at S1.

ASIC TL494 is employed to

control the PWM and consists

of two internal sample amplifi-

ers, a sawtooth wave generator,

a comparator and an output

stage which can be designed

into an emitter-follower or in-

verter. Voltage sampled by cw

goes through a low-pass filter

formed by C15, R13 and C9 to get

rid of the clutter involved.

Then it is amplified by the

error amplifier formed by the

lead 1, 2 and 3 of the TL494,

with the gain decided by the

feedbacknetworkformedbyR1,

C2 and C1. Adjusting the value

ofR1 andC2 canchangethesen-

sitivity of PWM control. R4 and

C3 form a timing circuit for the

internal sawtooth wave genera-

tor, which oscillates in phase

with horizontal synchroniza-

tion pulse. A maximum PWM

duty cycle can be obtained and

high-voltage can be controlled

by TL494's pin 4. Pin 4 also can

control R17 and C16 to provide

a soft start-up. S3 is an over-

voltage controller.

Another error amplifier, the

flyback monitoring circuit

formed by pin 14, 15 and 16 of

TL494 and Q5, is controlled by

the horizontal flyback pulse

produced by the row deflection

output circuit. Row pulse will

die away when there is an error

in row deflection circuit, which

will cut-off Q5 and lift up the

voltage in pin 16 of TL494,

stoppingtheinternalPWMand

remove high-voltage.





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