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