MC14051B, MC14052B, MC14053B
APPLICATIONS INFORMATION
Figure A illustrates use of the on−chip level converter
detailed in Figures 2, 3, and 4. The 0−to−5 V Digital Control
signal is used to directly control a 9 V analog signal.
peak. If voltage transients above V and/or below V are
DD EE
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.
p−p
The digital control logic levels are determined by V
DD
and V . The V voltage is the logic high voltage; the V
SS
DD
SS
voltage is logic low. For the example, V = + 5 V = logic
The absolute maximum potential difference between
DD
high at the control inputs; V = GND = 0 V = logic low.
V
and V is 18.0 V. Most parameters are specified up to
DD EE
SS
The maximum analog signal level is determined by V
15 V which is the recommended maximum difference
between V and V
DD
and V . The V
voltage determines the maximum
.
EE
EE
DD
DD
recommended peak above
V
.
SS
The
V
voltage
Balanced supplies are not required. However, V must
EE
SS
determines the maximum swing below V . For the
be greater than or equal to V . For example, V = + 10
SS
EE
DD
example, V
− V = 5 V maximum swing above V
;
V, V = + 5 V, and V – 3 V is acceptable. See the Table
DD
SS
SS
SS EE
V
− V = 5 V maximum swing below V . The example
below.
SS
EE
SS
shows a± 4.5 V signal which allows a 1/2 volt margin at each
+5 V
−5 V
V
DD
V
SS
V
EE
+4.5 V
9 V
SWITCH
I/O
p−p
+5 V
9 V
ANALOG SIGNAL
COMMON
O/I
p−p
GND
MC14051B
MC14052B
MC14053B
ANALOG SIGNAL
EXTERNAL
CMOS
DIGITAL
−4.5 V
0−TO−5 V DIGITAL
CONTROL SIGNALS
INHIBIT,
A, B, C
CIRCUITRY
Figure A. Application Example
V
DD
V
DD
D
D
D
X
X
X
ANALOG
I/O
COMMON
O/I
D
X
V
EE
V
EE
Figure B. External Germanium or Schottky Clipping Diodes
POSSIBLE SUPPLY CONNECTIONS
Control Inputs
Logic High/Logic Low
In Volts
V
V
V
EE
In Volts
Maximum Analog Signal Range
In Volts
DD
SS
In Volts
In Volts
+ 8
0
0
– 8
+ 8/0
+ 5/0
+ 8 to – 8 = 16 V
p–p
+ 5
– 12
0
+ 5 to – 12 = 17 V
p–p
+ 5
0
+ 5/0
+ 5 to 0 = 5 V
p–p
+ 5
0
– 5
+ 5/0
+ 5 to – 5 = 10 V
+ 10 to – 5 = 15 V
p–p
+ 10
+ 5
– 5
+ 10/ + 5
p–p
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