MIL-M-38510-121 Microcircuits, Linear Monolithic And Multichip, Silicon 12 Bit Digital-To-Analog ConvertersThis specification covers the detail requirements for monolithic silicon, bipolar, interface, memory core drivers. 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 For this product, the requirements of MIL-M-38510 have been superseded by MIL-PRF-38535, see 6.3

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Subgroup

Symbol

Test no.

Adapter pin numbers  1/

Energized relays

Measured pin

Equation   2/

Oevice

Types

Limits

Unit

4

6

7 thru 18

22 thru 37

No.

Value

Unit

Min

Max

3

TA  = -55°C

MCE

110

111

112

113

114

115

116

117

118

119

120

121

+15 V

"

" " " " " " " " " " " " " " " " " " " " " "

-15 V

"

" " " " " " " " " " " " " " " " " " " " " "

0111 1111 1111

1000 0000 0000

0011 1111 1111

0100 0000 0000

0001 1111 1111

0010 0000 0000

0000 1111 1111

0001 0000 0000

0000 0111 1111

0000 1000 0000

0000 0011 1111

0000 0100 0000

0000 0001 1111

0000 0010 0000

0000 0000 1111

0000 0001 0000

0000 0000 0111

0000 0000 1000

0000 0000 0011

0000 0000 0100

0000 0000 0001

0000 0000 0010

0000 0000 0000

0000 0000 0001

1000 0000 0000 1111

0111 1111 1111 1111

1100 0000 0000 1111

1011 1111 1111 1111

1110 0000 0000 1111

1101 1111 1111 1111

1111 0000 0000 1111

1110 1111 1111 1111

1111 1000 0000 1111

1111 0111 1111 1111

1111 1100 0000 1111

1111 1011 1111 1111

1111 1110 0000 1111

1111 1101 1111 1111

1111 1111 0000 1111

1111 1110 1111 1111

1111 1111 1000 1111

1111 1111 0111 1111

1111 1111 1100 1111

1111 1111 1011 1111

1111 1111 1110 1111

1111 1111 1101 1111

1111 1111 1111 1111

1111 1111 1110 1111

None

"

" " " " " " " " " " " " " " " " " " " " " "

21

"

" " " " " " " " " " " " " " " " " " " " " "

E95

E96

E97

E98

E99

E100

E101

E102

E103

E104

E105

E106

E107

E108

E109

E110

E111

E112

E113

E114

E115

E116

E117

E118

V

"

" " " " " " " " " " " " " " " " " " " " " "

For each set of major carry codes N and N + 1

Calculate the associated error in LSB's as

Follows:

MCE = 4.1 (EA - EB)

where

EB = measured value from E95,E97,E99,etc. EA = measured value from E96,E98,E100,etc.

03, 04

"

" " " " " " " " " " " " " " " " " " " " " "

-1.0

"

" " " " " " " " " " " " " " " " " " " " " "

1.0

"

" " " " " " " " " " " " " " " " " " " " " "

LSB

"

" " " " " " " " " " " " " " " " " " " " " "

4

TA  = 25°C

1ZS

122

"

"

0000 0000 0000

1111 1111 1111 1111

K1

19

130

1ZS = (130) (4096)/(127  - 130)

03, 04

-2

2

LSB

AE

123

"

"

"

"

1111 1111 1111

0000 0000 0000

0000 0000 0000 1111

1111 1111 1111 1111

None

21

"

E119

E120

V

"

AE  = 0.1 (E119  - E120)

"

-0.20

0.20

%

BPOE

124

"

"

0000 0000 0000

1111 1111 1111 1111

K2, K3

"

E121

"

BPOE  = (-2.05) (E121)

"

-8

8

LSB

BPAE

125

"

"

1111 1111 1111

0000 0000 0000 1111

K2

"

E122

"

BPAE  = 0.05 (E122)

"

-0.20

0.20

%

BZE

126

"

"

1000 0000 0000

0111 1111 1111 1111

K2

"

E123

"

BZE  = (-2.05) (E123)

"

-6.0

6.0

LSB

LE1

LE15

127

128

129

130

131

132

133

134

135

136

137

138

139

140

141

"

"

" " " " " " " " " " " " "

"

"

" " " " " " " " " " " " "

0001 0000 0000

0010 0000 0000

0011 0000 0000

0100 0000 0000

0101 0000 0000

0110 0000 0000

0111 0000 0000

1000 0000 0000

1001 0000 0000

1010 0000 0000

1011 0000 0000

1100 0000 0000

1101 0000 0000

1110 0000 0000

1111 0000 0000

1110 1111 1111 1111

1101           "

1100           "

1011           "

1010           "

1001           "

1000           "

0111           "

0110           "

0101           "

0100           "

0011           "

0010           "

0001           "

0000           "

None

"

" " " " " " " " " " " " "

"

"

" " " " " " " " " " " " "

E124

E125

E126

E127

E128

E129

E130

E131

E132

E133

E134

E135

E136

E137

E138

"

"

" " " " " " " " " " " " "

For each input code N calculate the

Corresponding linearity error in LSB.

LE = 4.1((E119-E120)(N/4095) + E120  -EA)

where

EA = measured value E124  through E146

N = OUT input digital code word in decimal form

Thus for 12 bits N = (A1)(2 exp 11) + (A2) (2 exp 10 )+ … +(A12)(2 exp 0)

A1, A2, A12 are 0 or 1.

E119and E120  correspond to the error amplifier Outputs at full scale and zero scale respectively.

"

"

" " " " " " " " " " " " "

" "

-0.5

"

" " " " " " " " " " " " "

" "

+0.5

"

" " " " " " " " " " " " "

" "

"

"

" " " " " " " " " " " " "

" "

A. Record the OUT code-word, (XXXX 0000 0000), with the most positive error.            2/ B. Record the OUT code-word, (YYYY 0000 0000), with the most negative error.          2/