MC14551B
http://onsemi.com
8
APPLICATIONS INFORMATION
Figure A illustrates use of the on−chip level converter
detailed in Figure 2. The 0−to−5.0 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
= + 5.0 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
= 5.0 V maximum swing above V
SS
; V
SS
– V
EE
= 5.0 V
maximum swing below V
SS
. The example shows a ± 4.5 V
signal which allows a 1/2 V margin at each peak. If voltage
transients above V
DD
and/or below V
EE
are anticipated on the
analog channels, external diodes (D
x
) 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 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
= + 5.0 V, and V
EE
= – 3.0 V is acceptable. See the table below.
Figure A. Application Example
EXTERNAL
CMOS
DIGITAL
CIRCUITRY
9 V
p-p
ANALOG SIGNAL
0-TO-5 V DIGITAL
CONTROL SIGNAL
V
DD
V
SS
V
EE
SWITCH
I/O
COMMON
O/I
CONTROL
MC14551B
-5 V+ 5 V
9 V
p-p
ANALOG SIGNAL
+ 4.5 V
- 4.5 V
GND
V
DD
V
DD
V
EE
V
EE
D
x
D
x
D
x
D
x
SWITCH
I/O
COMMON
O/I
Figure B. External Schottky or Germanium Clipping Diodes
+ 5 V
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 + 10/ + 5 + 10 to – 5 = 15 V
p–p
