TDA8359J Datasheet, PDF (11/20 Pages) NXP Semiconductors – Full bridge vertical deflection output circuit in LVDMOS

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:
V_FB = I_{coil(p–p)} × ^R-1----c-_–--^o--e-ⁱ-^l--_–⁺--_t--_F-R-_B---_⁄-^M-_x-
where:
x = -R----_c---_o^L---_i-_lc--₊-^o--ⁱ--^l_R-----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:
P_tot = P_sup − P_L
The power to be supplied is given by the formula:
P_sup = V_P × ^I--c---^o---ⁱ-^l--⁽_2-^p---^e---^a---^k--⁾ + V_P × 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:
P_L = -⁽--I--^c--^o---ⁱ--^l-⁽--^p_3---^e---^a--^k---⁾--⁾--²- × (R_coil + R_M)
Example
Table 1 Application values
SYMBOL
I_coil(peak)
I_coil(p-p)
L_coil
R_coil
R_M
f_vert
t_FB
VALUE
1.2
2.4
5
6
0.6
50
640
UNIT
A
A
mH
Ω
Ω
Hz
µs
Table 2 Calculated values
SYMBOL
V_P
R_M + R_coil (hot)
t_vert
x
V_FB
P_sup
P_L
P_tot
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: P_tot = 6 W; T_amb(max) = 40 °C;
T_j = 120 °C; R_th(j-c) = 4 K/W; R_th(c-h) = 2 K/W.
The required heatsink thermal resistance is given by:
R_{th(h – a)} = ^T----j---^–-P----T_t--_o-^a-_t--^m----^b- – (R_{th(j – c)} + R_{th(c – h)})
When we use the values given we find:
R_{th(h – a)} = ^{1----2---0---6--–}-----4---0-- – (4 + 2) = 7 K/W
The heatsink temperature will be:
T_h = T_amb + (R_th(h-a) × P_tot) = 40 + (7 × 6) = 82 °C
2002 Jan 21
11