R⋅5RL
TYPICAL APPLICATION
VIN | R×5RL | VOUT | |
VIN | SERIES | VOUT | |
+ | + | ||
| |||
C1 |
| C2 |
GND
GND 
GND
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.
APPLICATION CIRCUITS
• VOLTAGE BOOST CIRCUIT
VIN | R×5RL | VOUT |
VIN | SERIES | VOUT |
|
| |
| GND | + |
| C2 R1 |
The output voltage can be obtained by the follow-
ing formula :
*1
VOUT=Vreg · (1+R2/R1) + ISS R · 2
Since the quiescent current of R⋅5RE Series is so small that the resistances of R1 and R2 can be set as
+
C1
ISS
R2
large as several hundreds kΩ and therefore the supply 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.
• DUAL POWER SUPPLY CIRCUIT
IC1
| VIN | VOUT |
| VOUT1 |
VIN | R×5RL20A |
|
| |
|
|
| + | 5V |
|
| C1 | D | |
|
|
| ||
| GND |
|
| ISS |
| IC2 |
|
|
|
| VIN | VOUT |
| VOUT2 |
| R×5RL30A |
|
| 3V |
| + |
| + | |
| C3 | R | ||
| C2 |
| ||
| GND |
|
|
|
GND |
|
|
| GND |
As shown in the circuit diagram, a dual power supply 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 ISS of IC1, Resistor R is unnecessary. Diode D is a protection diode for the case where VOUT2 becomes larger than VOUT1.
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