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AD8387JSVZ bảng dữ liệu(PDF) 13 Page - Analog Devices

tên linh kiện AD8387JSVZ
Giải thích chi tiết về linh kiện  High Performance, 12-Bit, 12-Channel Decimating, LCD DecDriver
PDF  16 Pages
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nhà sản xuất  AD [Analog Devices]
Trang chủ  http://www.analog.com
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AD8387JSVZ bảng dữ liệu(HTML) 13 Page - Analog Devices

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AD8387
Rev. 0 | Page 13 of 16
THEORY OF OPERATION
TRANSFER FUNCTION AND ANALOG OUTPUT
VOLTAGE
The DecDriver has two regions of operation where the video
output voltages are either above or below the reference voltage
VRL. The transfer function defines the video output voltage as
the function of the digital input code as:
VOUTN(n) = VIDx(n) = VRL + VFS × (1 − n/4095),
for INV = HIGH
VOUTP(n) = VIDx(n) = VRL − VFS × (1 − n/4095),
for INV = LOW
where n is the input code.
VFS = 2 × (VRH − VRL)
A number of internal limits define the usable range of the video
output voltages, VIDx, as shown in Figure 17.
VIDx – VOLTS
AVCC
(VRL + VFS)
VRL
(VRL – VFS)
AGND
0
VIDx vs. INPUT CODE
INPUT CODE
VOUTP
VOUTN
4095
≥1.3V
0
≤ VFS
≤ 5.25V
0
≤ VFS ≤ 5.25V
≥1.3V
5.25V
≤ VRL
≤ (AVCC – 4)
11V
≤ AVCC
≤ 18V
INTERNAL LIMITS AND
USABLE VOLTAGE RANGES
Figure 17. AD8387 Transfer Function and Usable Voltage Ranges
ACCURACY
To best correlate transfer function errors to image artifacts, the
overall accuracy of the DecDriver is defined by three
parameters, VDE , VCME, and ΔVDE.
VDE, the differential error voltage, measures the difference
between the rms value of a channel and the ideal rms value of
that channel. The defining expression is
VOUTN(n) VOUTP(n)
n
VDE(n)
1
VFS
2409
⎡⎤
⎛⎞
⎣⎦
=−
⎜⎟
⎝⎠
5
×
VCME, the common-mode error voltage, measures ½ the dc
bias of a channel. The defining expression is
+
=
VRL
n
VOUTP
n
VOUTN
n
VCME
2
)
(
)
(
2
1
)
(
ΔVDE measures the maximum VDE mismatch between
channels. The defining equation is
ΔVDE = max{VDE(n)(0 − 11)} − min{VDE(n)(0 − 11)}
ΔV measures the maximum mismatch between channels. The
defining expression is
ΔV(n) = max{ΔVN(n), ΔVP(n)}
where:
ΔVN(n) = max{VOUTN(n)(0 − 11)} − min{VOUTN(n)(0 − 11)}
ΔVP(n) = max{VOUTP(n)(0 − 11)} − min{VOUTP(n)(0 − 11)}



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