HIN202CPZ Intersil, HIN202CPZ Datasheet - Page 10

• Manufacturer Part Number: HIN202CPZ
• Description: IC TXRX RS-232 5V 16-PDIP
• Manufacturer: Intersil
• Type: Transceiverr

Specifications of HIN202CPZ

• Number Of Drivers/receivers: 2/2
• Protocol: RS232
• Voltage - Supply: 4.5 V ~ 5.5 V
• Mounting Type: Through Hole
• Package / Case: 16-DIP (0.300", 7.62mm)
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

Typical Performance Curves
Test Circuits (HIN202)
Application Information
The HIN2XX may be used for all RS-232 data terminal and
communication links. It is particularly useful in applications
where ±12V power supplies are not available for
conventional RS-232 interface circuits. The applications
presented represent typical interface configurations.
A simple duplex RS-232 port with CTS/RTS handshaking is
illustrated in Figure 9. Fixed output signals such as DTR
(data terminal ready) and DSRS (data signaling rate select)
is generated by driving them through a 5kW resistor
connected to V+.
±30V INPUT
3kΩ
12
10
8
6
4
2
0
0.1µF
3.0
RS-232
0.1µF
OUTPUT
0.1µF
C1
T2
C2
FIGURE 5. V- SUPPLY VOLTAGE vs V
C3
0.1µF C4
FIGURE 7. GENERAL TEST CIRCUIT
+
+
-
-
+
3.5
+
-
-
1
2
3
4
5
6
7
8
C1+
V+
C1-
C2+
C2-
V-
T2
R2
4.0
OUT
IN
10
R1
R2
T1
HIN202, HIN206, HIN207, HIN208, HIN211, HIN213
GND
R1
V
T1
T2
V
OUT
OUT
OUT
4.5
CC
CC
IN
IN
IN
10
16
15
14
13
12
11
9
5.0
RS-232 ±30V INPUT
TTL/CMOS OUTPUT
TTL/CMOS INPUT
TTL/CMOS INPUT
TTL/CMOS OUTPUT
+5.5V INPUT
+4.5V TO
T1 OUTPUT
CC
5.5
3kΩ
0.1µF
6.0
In applications requiring four RS-232 inputs and outputs
(Figure 10), note that each circuit requires two charge pump
capacitors (C1 and C2) but can share common reservoir
capacitors (C3 and C4). The benefit of sharing common
reservoir capacitors is the elimination of two capacitors and
the reduction of the charge pump source impedance which
effectively increases the output swing of the transmitters.
FIGURE 8. POWER-OFF SOURCE RESISTANCE
12
10
8
6
4
2
0
0
FIGURE 6. V+, V- OUTPUT VOLTAGE vs LOAD
T
TRANSMITTER OUTPUTS
OPEN CIRCUIT
A
= 25°C
CONFIGURATION
5
V- (V
CC
R
OUT
10
V
= 4V)
1
2
3
4
5
6
7
8
IN
C1+
V+
C1-
C2+
C2-
V-
T2
R2
= V
= ±2V
OUT
IN
IN
|I
15
LOAD
/I
R1
R2
T1
T2
GND
R1
| (mA)
T1
T2
V
OUT
OUT
OUT
OUT
CC
T1
A
IN
IN
IN
20
OUT
16
15
14
13
12
11
10
9
V+ (V
25
CC
V+ (V
V- (V
= 5V)
CC
CC
30
= 4V)
= 5V)
35