參數(shù)資料
型號: MC14551B
廠商: ON SEMICONDUCTOR
英文描述: Quad 2-Channel Analog Multiplexer/Demultiplexer
中文描述: 四2通道模擬多路復用器/解復用器
文件頁數(shù): 7/9頁
文件大?。?/td> 250K
代理商: MC14551B
MOTOROLA CMOS LOGIC DATA
7
MC14551B
APPLICATIONS INFORMATION
Figure A illustrates use of the on–chip level converter
detailed in Figure 2. The 0–to–5 volt Digital Control signal is
used to directly control a 9 Vp–p analog signal.
The digital control logic levels are determined by VDD and
VSS. The VDD voltage is the logic high voltage; the VSS volt-
age is logic low. For the example, VDD = + 5 V = logic high at
the control inputs; VSS = GND = 0 V = logic low.
The maximum analog signal level is determined by VDD
and VEE. The VDD voltage determines the maximum recom-
mended peak above VSS. The VEE voltage determines the
maximum swing below VSS. For the example, VDD – VSS
= 5 volt maximum swing above VSS; VSS – VEE = 5 volt
maximum swing below VSS. The example shows a
±
4.5 volt
signal which allows a 1/2 volt margin at each peak. If voltage
transients above VDD and/or below VEE are anticipated on
the analog channels, external diodes (Dx) are recommended
as shown in Figure B. These diodes should be small signal
types able to absorb the maximum anticipated current
surges during clipping.
The absolute maximum potential difference between VDD
and VEE is 18.0 volts. Most parameters are specified up to
15 volts which is the recommended maximum difference be-
tween VDD and VEE.
Balanced supplies are not required. However, VSS must
be greater than or equal to VEE. For example, VDD =
+ 10 volts, VSS = + 5 volts, and VEE = – 3 volts is acceptable.
See the table below.
Figure A. Application Example
EXTERNAL
CMOS
DIGITAL
CIRCUITRY
9 Vp–p
ANALOG SIGNAL
0–TO–5 V DIGITAL
CONTROL SIGNAL
VDD
VSS
VEE
SWITCH
I/O
COMMON
O/I
CONTROL
MC14551B
–5 V
+
5 V
9 Vp–p
ANALOG SIGNAL
+4.5 V
–4.5 V
GND
VDD
VDD
VEE
VEE
Dx
Dx
Dx
Dx
SWITCH
I/O
COMMON
O/I
Figure B. External Schottky or Germanium Clipping Diodes
+
5 V
VDD
VSS
VEE
Logic High/Logic Low
Maximum Analog Signal Range
+ 5
0
0
+ 5/0
+ 5 to 0 = 5 Vp–p
+ 5
0
– 5
+ 5/0
+ 5 to – 5 = 10 Vp–p
+ 10
– 5
+ 10/ + 5
+ 10 to – 5 = 15 Vp–p
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