CHAPTER 15 VFD CONTROLLER/DRIVER
327
User’s Manual U11302EJ4V0UM
By placing the above conditions in calculations <1> to <3>, the total dissipation can be worked out.
<1> CPU power dissipation: 5.5 V
×
21.6 mA = 118.8 mW
<2> Output pin power dissipation:
Grid
(V
DD
– V
OD
)
×
Total current value of each grid
×
Digit width (1 – Cut width) =
No. of grids + 1
2 V
×
15 mA
×
11 grids
×
(1 –
1
)
= 25.8 mW
11 grids + 1
16
Segment (V
DD
– V
OD
)
×
Total segment current value of illuminated dots
×
Digit width (1 – Cut width) =
No. of grids + 1
0.4 V
×
3 mA
×
31 dots
×
(1 –
1
)
= 2.9 mW
11 grids + 1
16
<3> Pull-down resistor power dissipation:
Grid
(V
OD
– V
LOAD
)
2
×
No. of grids
×
Digit width (1 – Cut width) =
Pull-down resistor value
No. of grids + 1
(5.5 V – 2 V – (–35 V))
2
×
11 grids
×
(1 –
1
)
= 50.9 mW
25 k
Ω
11 grids + 1
16
Segment
(V
OD
– V
LOAD
)
2
×
No. of illuminated dots
×
Digit width (1 – Cut width) =
Pull-down resistor value No. of grids + 1
(5.5 V – 0.4 V – (–35 V))
2
×
31 dots
×
(1 –
1
)
= 155.8 mW
25 k
Ω
11 grids + 1
16
Total power dissipation = <1> + <2> + <3> = 118.8 + 25.8 + 2.9 + 50.9 + 155.8 = 354.2 mW
In this example, the power dissipation problem is cleared because the total power dissipation does not exceed
the allowable total power dissipation rating shown in Figure 15-22.
