R
×
5RL
16
TYPICAL APPLICATION
In R
×
5RL Series, a constant voltage can be obtained
without using Capacitors C1 and C2. However, when the
wire connected to Vin is long, use Capacitor C1. Output
noise can be reduced by using Capacitor C2.
Insert Capacitors C1 and C2 with the capacitance of
0.1µF to 2.0µF between Input/Output Pins and GND Pin
with minimum wiring.
As shown in the circuit diagram, a dual power sup-
ply circuit can be constructed by using two R
×
5RL
Series.
This circuit diagram shows a dual power supply
circuit with an output of 3V and an output of 5V.
When the minimum output current of IC2 is larger
than I
SS
of IC1, Resistor R is unnecessary. Diode D is
a protection diode for the case where V
OUT
2 becomes
larger than V
OUT
1.
• VOLTAGE BOOST CIRCUIT
• DUAL POWER SUPPLY CIRCUIT
C2
C1
V
IN
GND
GND
GND
R
×
5RL
SERIES
V
OUT
V
OUT
V
IN
+
+
C2
C1
V
IN
GND
R
×
5RL
SERIES
V
OUT
V
OUT
V
IN
R1
R2
I
SS
+
+
C1
C2
V
IN
GND
GND
GND
R
×
5RL20A
R
×
5RL30A
V
OUT
V
OUT
1
5V
V
OUT
2
3V
V
IN
C3
GND
V
OUT
V
IN
R
I
SS
IC1
IC2
D
+
+
+
The output voltage can be obtained by the follow-
ing formula :
V
OUT
=Vreg · (1+R2/R1) + I
SS
R · 2
Since the quiescent current of R
×
5RE Series is so
small that the resistances of R1 and R2 can be set as
large as several hundreds k
Ω
and therefore the sup-
ply current of “Voltage Boost Circuit” itself can be
reduced.
Furthermore, since R
×
5RL Series are operated by
a constant voltage, the supply current of “Voltage
Boost Circuit” is not substantially affected by the
input voltage.
*
1
APPLICATION CIRCUITS
