CHAPTER 15 VFD CONTROLLER/DRIVER
332
User’s Manual U11302EJ4V0UM
15.10.3 Display type in which a segment spans two or more grids (display mode 2: DSPM05 = 1)
The calculation method for the total power dissipation in the case of the display example in Figure 15-26 is
described below.
Example
Assume the following conditions:
V
DD
= 5 V
±
10%, 5.0 MHz oscillation
Supply current (I
DD
) = 21.6 mA
Display output: 23 segments
×
7 patterns (cut width = 1/16: when DIMS1 to DIMS3 = 000B)
Maximum current at the display output pin is 15 mA.
Display output voltage (V
OD
) = V
DD
– 2 V (voltage drop of 2 V)
Fluorescent display control voltage (V
LOAD
) = –35 V
Mask option pull-down resistor = 25 k
Ω
By placing the above conditions in calculation <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:
(V
DD
– V
OD
)
×
Total current value of each grid
×
Digit width (1 – Cut width) =
No. of grids + 1
2 V
×
15 mA
×
9 grids
×
(1 –
1
)
= 31.6 mW
7 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
×
9 grids
×
(1 –
1
)
= 62.5 mW
25 k
Ω
7 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 – 0.5 V – (–35 V))
2
×
22 dots
×
(1 –
1
)
= 152.8 mW
25 k
Ω
7 grids + 1
16
Total power dissipation = <1> + <2> + <3> = 118.8 + 31.6 + 62.5 + 152.8 = 365.7 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.
Figure 15-25 shows the grid driving timing, and Figure 15-26 shows a display example and display data of a
display type in which a segment spans two or more grids.