TDA8359 查看數據表（PDF） - Philips Electronics

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TDA8359

Full bridge vertical deflection output circuit in LVDMOS

Philips Electronics 

Philips Semiconductors

Full bridge vertical deflection output circuit

in LVDMOS

Product specification

TDA8359J

Flyback supply voltage calculation

If the flyback time is known, the required flyback supply

voltage can be calculated by the simplified formula:

VFB = Icoil(p–p) × R-1----c-–--o--e-i-l--–+--t--F-R-B---⁄-M-x-

where:

x = -R----c---oL---i-lc--+-o--i--lR-----M--

The flyback supply voltage calculated this way is

approximately 5% to 10% higher than required.

Calculation of the power dissipation of the vertical

output stage

The IC total power dissipation is given by the formula:

Ptot = Psup − PL

The power to be supplied is given by the formula:

Psup = VP × I--c---o---i-l--(2-p---e---a---k--) + VP × 0.015 [A] + 0.3 [W]

In this formula 0.3 [W] represents the average value of the

losses in the flyback supply.

The average external load power dissipation in the

deflection coil and the measuring resistor is given by the

formula:

PL = -(--I--c--o---i--l-(--p3---e---a--k---)--)--2- × (Rcoil + RM)

Example

Table 1 Application values

SYMBOL

Icoil(peak)

Icoil(p-p)

Lcoil

Rcoil

RM

fvert

tFB

VALUE

1.2

2.4

5

6

0.6

50

640

UNIT

A

A

mH

Ω

Ω

Hz

µs

Table 2 Calculated values

SYMBOL

VP

RM + Rcoil (hot)

tvert

x

VFB

Psup

PL

Ptot

VALUE

14

7.8

0.02

0.000641

30

8.91

3.74

5.17

UNIT

V

Ω

s

V

W

W

W

Heatsink calculation

The value of the heatsink can be calculated in a standard

way with a method based on average temperatures. The

required thermal resistance of the heatsink is determined

by the maximum die temperature of 150 °C. In general we

recommend to design for an average die temperature

not exceeding 130 °C.

EXAMPLE

Measured or given values: Ptot = 6 W; Tamb(max) = 40 °C;

Tj = 120 °C; Rth(j-c) = 4 K/W; Rth(c-h) = 2 K/W.

The required heatsink thermal resistance is given by:

Rth(h – a) = T----j---–-P----Tt--o-a-t--m----b- – (Rth(j – c) + Rth(c – h))

When we use the values given we find:

Rth(h – a) = 1----2---0---6--–-----4---0-- – (4 + 2) = 7 K/W

The heatsink temperature will be:

Th = Tamb + (Rth(h-a) × Ptot) = 40 + (7 × 6) = 82 °C

2002 Jan 21

11

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