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參數資料
| 型號： | MCM20027IBMN |
| 廠商： | MOTOROLA INC |
| 元件分類： | 圖像傳感器 |
| 英文描述： | Color SXGA Digital Image Sensor 1280 x 1024 pixel progressive scan solid state image sensor with integrated CDS/PGA/ADC, digital programming, control |
| 中文描述： | IMAGE SENSOR-CMOS, 10fps, 0.20-2.80V, SQUARE, SURFACE MOUNT |
| 封裝： | CERAMIC, LCC-48 |
| 文件頁數： | 25/74頁 |
| 文件大?。?/td> | 973K |
| 代理商： | MCM20027IBMN |
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Revision 8.0 - 28 November 2001 : MCM20027
MOTOROLA
25
MOTOROLA
SEMICONDUCTOR TECHNICAL DATA
ImageMOS
To ensure that Strobe signal fires, the integration time
must be large enough to ensure that all rows are inte-
grating simulanteously for at least 2 Row periods (
T
row
)
(see
“Frame Rate and Integration Time Control” on
page 13
)
where
T
row
= (vcw
d
+ shs
d
+ shr
d
+ 19)
To accomplish this - ensure that the Integration time
(
cint
d
) greater than 2 Row periods (
T
row
) larger than
the active Window of Interest Row depth.
Min. Integration time =T
intmin
=
(cint
min
+1)*
T
row
cint
min
=wrd
d
+x where x > 2
where
wrd
d
is the Window Of Interest Row depth.
T
strobe1
=
T
row
T
strobe2
=
T
intmin
- (wrd
d
+1)*
T
row
EXAMPLE:
Below you will find an example of how to
ensure that the strobe signal will fire and to determine
the length of the STROBE signal in default mode:
(Refer to
Figure 16, on page 24
for timing analysis)
Goal (For example purpose):
Strobe Signal that lasts for at least 250us, which is the
length of a typical strobe/flash event.
Assumptions:
1) Active Window of Interest = 1280 x 1024
ie.
(wcw
d
)=1279
(
wrd
d
)=1023
2) Virtual Column Width (
vcw
d
)= 1290
3) Virtual Row Depth (
vrd
d
) = 1034
4) Sample & hold time (
shs
d
) = 10
5) Sample & hold time (
shr
d
) = 10
6) MCLK = 13 Mhz
Variables:
Integration Time (
cint
min
) is the main variable used to control
the time of the Strobe signals.
T
intmin
=
(cint
min
+1)*
T
row
Calculations:
Row Time =Trow = (vcwd + shsd + shrd + 19)
= (1290 + 10 + 10 + 19) / 13.5e6
= 98.44
μ
s
T
intmin
=
(cint
min
+1)*
T
row
cint
min
=wrd
d
+x where x > 2
Let
cint
min
=
wrd
d
+x where x > 2
= 1023 + 4 = 1029
where x=4
Therefore,
T
intmin
= 101.39
T
strobe1
= 98.44
μ
s
T
strobe2
=
T
intmin
- (wrd
d
+2)*
T
row
= 3 *
T
row
= 295us
Results
:
NOTE!! Refer to
Figure 16, on page 24
for timing
analysis
Signal
Value
T
row
98us
T
int
101ms
T
strobe1
98us
T
strobe2
295us
T
frame
202ms
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