ADR5040BRTZ Analog Devices Inc, ADR5040BRTZ Datasheet - Page 10
![IC, SM, SHUNT, VOLT REF, 2.048V](/photos/19/9/190903/42334026_sml.jpg)
ADR5040BRTZ
Manufacturer Part Number
ADR5040BRTZ
Description
IC, SM, SHUNT, VOLT REF, 2.048V
Manufacturer
Analog Devices Inc
Datasheet
1.ADR5041ARTZ-R2.pdf
(16 pages)
Specifications of ADR5040BRTZ
Topology
Shunt
Reference Voltage
2.048V
Reference Voltage Tolerance
2.048mV
Temperature Coefficient
75ppm/°C
Voltage Reference Case Style
SOT-23
No. Of Pins
3
Operating Temperature Range
-40°C
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Part Number:
ADR5040BRTZ
Manufacturer:
ADI/亚德诺
Quantity:
20 000
Company:
Part Number:
ADR5040BRTZ-REEL7
Manufacturer:
AD
Quantity:
3 000
Part Number:
ADR5040BRTZ-REEL7
Manufacturer:
ADI/亚德诺
Quantity:
20 000
ADR5040/ADR5041/ADR5043/ADR5044/ADR5045
TERMINOLOGY
Temperature Coefficient
The change in output voltage with respect to operating temperature
changes. It is normalized by an output voltage of 25°C. This
parameter is expressed in ppm/°C and is determined by the
following equation:
where:
V
V
V
OUT
OUT
OUT
(25°C) = V
(T
(T
TCV
1
2
) = V
) = V
OUT
OUT
OUT
⎡
⎢ ⎣
ppm
OUT
°
at Temperature 1.
at Temperature 2.
C
at 25°C.
⎤
⎥ ⎦
=
V
V
O
OUT
UT
(
25
( )
T
2
°
C
−
)
V
×
OUT
(
T
2
( )
−
T
T
1
1
)
×
10
6
(1)
Rev. A | Page 10 of 16
Thermal Hysteresis
The change in output voltage after the device is cycled through
temperatures ranging from +25°C to −40°C, then to +125°C, and
back to +25°C. This is common in precision reference and is
caused by thermal-mechanical package stress. Changes in envi-
ronmental storage temperature, board mounting temperature, and
the operating temperature are some of the factors that can
contribute to thermal hysteresis. The following equation
expresses a typical value from a sample of parts put through
such a cycle:
where:
V
V
−40°C, then to +125°C, and back to +25°C.
OUT
OUT_TC
(25°C) = V
V
V
OUT
OUT
= V
_
_
HYS
HYS
OUT
[
at 25°C after a temperature cycle from +25°C to
=
ppm
OUT
V
OUT
at 25°C.
]
=
(
25
V
OUT
°
C
(
V
)
25
−
OUT
V
°
OUT
C
(
25
)
−
_
°
V
TC
C
OUT
)
_
TC
×
10
6
(2)