SECTION 3: PRODUCT OVERVIEW
Entire Contents Copyright
2014 by Adaptive Power Systems, Inc. (APS) • All Rights Reserved • No reproduction without written authorization from APS.
42L Series Modular DC Load Operation Manual
Page 18 of 88
Minimum
Tr
Use the following formulas to calculate the minimum transition time for a given slew rate on
a 42L0860 module:
Maximum current range for this module is 60A so 30% of 60 = 18. The minimum required
slew rate can be calculated as follows:
𝑇𝑇
min =
18
𝑠𝑠𝑠𝑠
𝑟𝑟𝑟𝑠
(
𝐴
/
𝑠
)
∗
(
90%−10%
)
100%
µ
s
Which is equivalent to:
𝑇𝑇
min =
18
𝑠𝑠𝑠𝑠
𝑟𝑟𝑟𝑠
(
𝐴
/
𝑠
)
∗
0.8
µ
s
For a programmed slew rate of 2.5A/s, this results in:
𝑇𝑇
min =
18
2
.
5
∗
0.8
µ
s = 5.76
µ
s
Example 1:
Assume high current level (CCH) = 12A and low current level (CCL) = 0A. For a 42L0860, a
16A delta change in current represents less than 30% of full scale current this load module
(< 18A). If the programmed slew rate is set to 2.5A/s, the expected transition time would be:
𝑇𝑇
=
0
.
8∗
(
12−
0
)
2
.
5
µ
s = 3.84
µ
s
However, we determined that
Tr min
for a slew rate of 2.5A/s is at least 5.76
µ
s so the
actual transition time will be limited to no less than 5.76
µ
s.
Maximum
Tr
Use the following formula to calculate the maximum transition time for a given slew rate:
𝑇𝑇
max =
60
(
𝑀𝑟𝑀
.
𝐶𝐶𝑟𝑟𝑠𝐶𝑟
)
𝑠𝑠𝑠𝑠
𝑟𝑟𝑟𝑠
(
𝐴
/
𝑠
)
∗
0.8
µ
s
For a slew rate of 0.1A/s, this results in:
𝑇𝑇
max =
60
2
.
5
∗
0.8
µ
s = 19.2
µ
s
Example 2:
Assume high current level (CCH) = 40A and low current level (CCL) = 0A. Since 40A
represents more than 30% of the current range for the module used (40 > 18). If the slew
rate is set to 0.1A/s, the expected transition time would be:
𝑇𝑇
=
0
.
8∗
(
40−
0
)
2
.
5
µ
s = 12.8
µ
s
Since
Tr max
for a slew rate of 2.5A/s is 19.2
µ
s so the actual transition time will be lesser
of these two values or 12.8
µ
s.
