HIGH RIPPLE-REJECTION AND LOW DROPOUT CMOS VOLTAGE REGULATOR
Rev.5.0_00
S-1112/1122 Series
Electrical Characteristics
Table 7
(Ta = 25°C unless otherwise specified)
Test
Item
Symbol
VOUT(E)
Conditions
Min.
Typ.
Max.
Unit
Circuit
VOUT(S) VOUT(S) VOUT(S)
Output voltage*1
V
IN = VOUT(S) + 1.0 V, IOUT = 30 mA
V
1
× 0.99
150*5
× 1.01
Output current*2
IOUT
Vdrop
VIN ≥ VOUT(S) + 1.0 V
OUT = 100 mA
mA
V
3
1
Dropout voltage*3
I
1.5 V ≤ VOUT(S) ≤ 1.6 V
1.7 V ≤ VOUT(S) ≤ 1.8 V
1.9 V ≤ VOUT(S) ≤ 2.3 V
2.4 V ≤ VOUT(S) ≤ 2.7 V
2.8 V ≤ VOUT(S) ≤ 5.5 V
0.32
0.28
0.25
0.20
0.19
0.55
0.47
0.35
0.29
0.26
∆VOUT1
∆VIN•VOUT
V
I
V
OUT(S) + 0.5 V ≤ VIN ≤ 6.5 V,
OUT = 30 mA
IN = VOUT(S) + 1.0 V,
1.0 mA ≤ IOUT ≤ 80 mA
IN = VOUT(S) + 1.0 V, IOUT = 10 mA,
−40°C ≤ Ta ≤ 85°C
IN = VOUT(S) + 1.0 V, ON/OFF pin = ON,
no load
IN = VOUT(S) + 1.0 V, ON/OFF pin = OFF,
no load
Line regulation
Load regulation
0.05
0.2
40
% / V
mV
∆VOUT2
12
∆VOUT
∆Ta•VOUT
Output voltage
V
ppm/
°C
100
50
temperature coefficient*4
Current consumption
during operation
Current consumption
during shutdown
Input voltage
Shutdown pin
input voltage “H”
Shutdown pin
input voltage “L”
Shutdown pin
input current “H”
Shutdown pin
input current “L”
V
ISS1
90
µA
2
V
ISS2
VIN
0.1
1.0
6.5
2.0
1.5
V
VSH
V
V
V
IN = VOUT(S) + 1.0 V, RL = 1.0 kΩ
4
VSL
ISH
IN = VOUT(S) + 1.0 V, RL = 1.0 kΩ
IN = 6.5 V, VON/OFF = 6.5 V
−0.1
−0.1
0.3
0.1
0.1
µA
ISL
V
IN = 6.5 V, VON/OFF = 0 V
V
IN = VOUT(S) + 1.0 V, f = 1.0 kHz,
∆Vrip = 0.5 Vrms, IOUT = 30 mA
IN = VOUT(S) + 1.0 V, ON/OFF pin = ON,
RR
Ripple rejection
80
200
dB
5
3
V
Short-circuit current
Ishort
mA
V
OUT = 0 V
*1. VOUT(S): Specified output voltage
VOUT(E): Actual output voltage at the fixed load
The output voltage when fixing IOUT(= 30 mA) and inputting VOUT(S) + 1.0 V
*2. The output current at which the output voltage becomes 95% of VOUT(E) after gradually increasing the output current.
*3. Vdrop = VIN1 − (VOUT3 × 0.98)
VOUT3 is the output voltage when VIN = VOUT(S) + 1.0 V and IOUT = 100 mA.
VIN1 is the input voltage at which the output voltage becomes 98% of VOUT3 after gradually decreasing the input voltage.
*4. The change in temperature [mV/°C] is calculated using the following equation.
∆VOUT
∆Ta
∆VOUT
∆Ta • VOUT
*2
*3
[
mV/°C
]
*1 = VOUT(S)
[
V
]
×
ppm/°C ÷1000
[ ]
*1. The change in temperature of the output voltage
*2. Specified output voltage
*3. Output voltage temperature coefficient
*5. The output current can be at least this value.
Due to restrictions on the package power dissipation, this value may not be satisfied. Attention should be paid to the power
dissipation of the package when the output current is large.
This specification is guaranteed by design.
Seiko Instruments Inc.
9