MIL-M-38510-114 Microcircuits, Linear, Bi-Fet Operational Amplifiers, Monolithic SiliconThis specification covers the detail requirements for monolithic silicon, BI-FET operational amplifier microcircuit. Two product assurance classes and a choice of case outlines and lead finishes are provided and are reflected in the complete part number. For this product, the requirements of MIL-M-38510 have been superseded by MIL-PRF-38535, see 6.3.

Pages 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23

Subgroup

Symbol

MlL- STD-

883

method

Test no.

Notes

Adapter pin numbers

Energized relays

Measured pin

Equation

1/

Device type

Test limits

Unit

1

2

3

4

No.

value

Units

Min

Max

1

TA =

+25°C

vlO

4001

" " "

1

2

3

4

2/

" " "

35 v

5 v

20 v

5 v

-5 v

-35 v

-20 v

-5 v

Open

" " "

-15 v

15 v

0 v

0 v

None

" " "

5

" " "

E1

E2

E3

E4

v

" " "

vlO = E1 vlO = E2 vlO = E3

vlO = E4

01,02,03

-5

5

mv

" " "

04,05,06

-2

2

+llB

4001

" " " "

5

"

5 v

-35 v

"

15 v

K1, K8

"

E5

"

+llB = 10,000 (E2 - E5)

All

-100

3500

pA

" " " "

6

"

"

5 v

5 v

-25 v

-25 v

"

"

10 v

10 v

None

K1

"

"

E6

E7

"

"

+llB = 200 (E6 - E7)

"

"

-100

300

7

8

"

"

20 v

35 v

-20 v

-5 v

"

"

0 v

-15 v

K1

K1

"

"

E8

E9

"

"

+llB = 200 (E3 - E8)

+llB = 200 (E1 - E9)

"

"

-100

-100

100

100

-llB

"

" " "

"

9

"

5 v

-35 v

"

15 v

K2, K8

"

E10

"

-llB = 10,000 (E10 - E2)

All

-100

3500

"

" " "

"

10

"

"

5 v

5 v

-25 v

-25 v

"

"

10 v

10 v

None

K2

"

"

E11

E12

"

"

-llB = 200 (E12 - E11)

"

"

-100

300

11

12

"

"

20 v

35 v

-20 v

-5 v

"

"

0 v

-15 v

K2

K2

"

"

E13

E14

"

"

-llB = 200 (E13 - E3)

-llB = 200 (E14 - E1)

"

"

-100

-100

100

100

llO

"

13

"

Calculate value using data from tests 3, 7, and 11

llO = 200(2E3 - E8 - E13)

"

-20

20

pA

+PSRR

-PSRR

4003

4003

14

15

10 v

20 v

-20 v

-10 v

Open

Open

0 v

0 v

None

None

5

5

E15

E16

v

v

4

+PSRR = 20 logl 10  /(E3 - E15) l

4

-PSRR = 20 logl 10  /(E3  - E16) l

"

"

85

85

dB

CMR

4003

16

3/

Calculate value using data from tests 1 and 2

CMR = 20 logl 3 x 104/(E1  - E2) l

"

85

dB

vlO

ADJ(+)

vlO

ADJ(-)

17

18

20 v

20 v

-20 v

-20 v

Open

" "

"

0 v

0 v

K7

K6, K7

5

5

E17

E18

v

v

vlO ADJ(+) = (E3 - E17)

vlO ADJ(-) = (E3 - E18)

"

"

8

-8

mv

mv

lOS(+)

lOS(-)

3011

19

20

5/

5/

15 v

15 v

-15 v

-15 v

"

"

-10 v

10 v

None

None

6

6

l1

l2

mA

mA

lOS(+) = l1

lOS(-) = l2

"

"

-50

50

mA

mA

lCC

3005

21

15 v

-15 v

"

0 v

None

2

l3

mA

lCC = l3

01,04

4

mA

02,03,

05,06

7

2

TA =

+125°C

vlO

4001

" " "

22

23

24

25

2/

" " "

35 v

5 v

20 v

5 v

-5 v

-35 v

-20 v

-5 v

Open

" " "

-15 v

15 v

0 v

0 v

None

" " "

5

" " "

E19

E20

E21

E22

v

" " "

vlO = E19 vlO = E20 vlO = E21

vlO = E22

01,02,03

-7

7

mv

" " "

04,05,06

-2.5

2.5

vlO/

T

"

26

6/

vlO / T = [vlO (Test 24) - vlO (Test 3)] / 100°C

01,02,03

-30

30

∝v/

°C

04,05,06

-10

10

+llB

4001

" " " "

27

2/

5 v

-35 v

Open

15 v

K1, K8

5

E23

v

+llB = 10 (E20 - E23)

All

-10

60

nA

28

"

"

5 v

5 v

-25 v

-25 v

"

"

10 v

10 v

None

K1, K8

"

"

E24

E25

"

"

+llB = 10 (E24 - E25)

"

"

-10

50

"

"

29

30

"

"

20 v

35 v

-20 v

-5 v

"

"

0 v

-15 v

K1, K8

K1, K8

"

"

E26

E27

"

"

+llB = 10 (E21 - E26)

+llB = 10 (E19 - E27)

"

"

-10

-10

50

50

"

"

-llB

"

" " "

"

31

32

"

" "

5 v

5 v

5 v

-35 v

-25 v

-25 v

"

" "

15 v

10 v

10 v

K2, K8

None

K2, K8

"

" "

E28

E29

E30

"

" "

-llB = 10 (E28 - E20)

-llB = 10 (E30 - E29)

All

" "

-10

-10

60

50

"

" "

33

34

"

"

20 v

35 v

-20 v

-5 v

"

"

0 v

-15 v

K2, K8

K2, K8

"

"

E31

E32

"

"

-llB = 10 (E31 - E21)

-llB = 10 (E32 - E19)

"

"

-10

-10

50

50

"

"