Analog Devices LINEAR LT8491 Manual page 77

APPLICATIONS INFORMATION
• Following the calculation of R
R and C according to the following formulas:
DACO2 DACO
0.833 •R
FBOUT1
=
R
DACO2
⎛
R
⎝ ⎜
FBOUT2
R –R
FBOUT2
0.833 • 234,684 • 20k
=
⎛
20k •14.2 •
⎝ ⎜
= 107,556Ω
Choose R = 107k which is the closest standard
DACO2
value resistor.
= 0.2 •R
R
DACO1 DACO2
= 0.2 •107,556Ω
= 21,511Ω
Choose R = 21.5k which is the closest standard
DACO1
value resistor.
1
C =
DACO
500 •R
DACO1
1
=
500 • 21,511
= 93nF
• Using the standard value resistors calculated above, the
V , N1 and N2 checking equations yield the following:
X3
V = 14.31V
X3
N1 = 1.22
N2 = 0.804
• To find a resistor combination that yields V
to the desired 14.2V, R
higher standard value and the above calculations are
repeated.
, solve for R
FBOUT1
•R
FBOUT2
Ω
1.241
⎞
• V • –
⎠ ⎟
S2
1.211
FBOUT1
1.241
⎞
– 20k – 234,684
⎠ ⎟
1.211
Ω
F
F
X3
is increased to the next
FBOUT2
For more information
, • Iterations of the previous step are performed that
DACO1
include adjustments to R
until the following standard value feedback resistors
were chosen:
R = 274k
FBOUT1
R = 23.2k
FBOUT2
R = 26.1k
DACO1
Ω
R = 124k
DACO2
C = 0.082µF
DACO
where:
V = 14.27V
X3
N1 = 1.22
N2 = 0.805
• With the output feedback network determined, use
V and solve for the input resistor feedback network
MAX
according to the following formulas:
= 100k •
R
FBIN1
= 100k •
= 97,865Ω
The closest standard value for R
= 2.75 •
R
DACI2
closer
= 2.75 •
= 5726Ω
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LT8491
, R
FBOUT1 DAC01
⎡ ⎤
⎛ ⎞
4.47V
1+
⎢ ⎥
⎝ ⎜ ⎠ ⎟
V – 6V
⎢ ⎥
MAX
Ω
⎢ ⎥
⎛ ⎞
5.593V
1+
⎢ ⎥
⎝ ⎜ ⎠ ⎟
V – 6V
⎣ ⎦
MAX
⎡ 4.47V ⎤
⎛ ⎞
1+
⎝ ⎜ ⎠ ⎟
⎢ ⎥
53V – 6V
⎢ ⎥ Ω
5.593V
⎛ ⎞
⎢ ⎥
1+
⎝ ⎜ ⎠ ⎟
⎢ ⎥
53V – 6V
⎣ ⎦
is 97.6k.
FBIN1
⎛ ⎞
R
FBIN1
Ω
⎝ ⎜ ⎠ ⎟
V – 6V
MAX
97,865
⎛ ⎞
Ω
⎝ ⎜ ⎠ ⎟
53V – 6V
and R
DAC02
Rev. 0
77