Manual revision 001
Section 2: Description of SBE 39
plus
-IM
SBE 39
plus
-IM
14
Example 1:
Ten 39
plus
-IMs
with pressure
are deployed on a mooring, set up to
sample autonomously
every 5 sec
(720 samples/hour), with full power transmit setting (
SetEnableFullPwrTX=Y
), and
average is requested by computer
every hour
. It is set to
#iiOutputFormat=1
, resulting in conservatively 80 characters/sample (varies, depending on
sample number). How long can it be deployed?
Sampling current for 39
plus
-IM
with pressure and full power
= 0.84 mA-sec/sample = 0.00084 A-sec/sample
In 1 hour, sampling = 720 samples * 0.00084 A-sec/sample =
0.6048 A-sec/hour
Transmit current/sample = 0.032 mA-sec/character = 0.000032 A-sec/character
0.000032 A * 80 characters = 0.00256 A-sec/sample
In 1 hour, transmit = 1 query/hour * 0.00256 A-sec/sample =
0.00256 A-sec/hour
Awake c
urrent (awake but not sampling) = 800 μA = 0.8 mA = 0.0008 A
Awake current per query = 0.0008 A * 1 sec/IM to be queried * 9 other IMs on mooring = 0.0072 A-sec/query
In 1 hour, awake = 1 query/hour * 0.0072 A-sec/query =
0.0072 A-sec/hour
Quiescent curre
nt = 60 μA = 0.06 mA = 0.00006 A
In 1 hour, quiescent = 0.00006 A * (3600 sec/hour – 720 * 0.28 sec/sample – 9 sec awake) =
0.2034 A-sec/hour
In 1 hour, 39
plus
-IM takes 720 samples and transmits average to computer:
Current consumption / hour = 0.6048 + 0.00256 + 0.0072 + 0.2034
=
0.8180 Amp-sec/hour
Capacity = (4.0 Amp-hours * 3600 sec/hour) / (0.8180 Amp-sec/hour) = 17603 hours = 733 days =
2.0 years
Total number of samples = 17603 hours * 720 samples/hour =
12,674,000 samples
However, memory can only hold approximately 5,500,000 samples; 39plus-IM will not overwrite data in memory.
Example 2:
Same as Example 1, but
average is requested by computer every minute
. How long can it be deployed?
Sampling current for 39
plus
-IM
with pressure and full power
= 0.84 mA-sec/sample = 0.00084 A-sec/sample
In 1 hour, sampling = 720 samples * 0.00084 A-sec/sample =
0.6048 A-sec/hour
Transmit current/sample = 0.032 mA-sec/character = 0.000032 A-sec/character
0.000032 A * 80 characters = 0.00256 A-sec/sample
In 1 hour, transmit = 60 query/hour * 0.00256 A-sec/sample =
0.1536 A-sec/hour
Awake current (awake but not sampling) = 800 μA = 0.8 mA = 0.0008 A
Awake current per query = 0.0008 A * 1 sec/IM to be queried * 9 other IMs on mooring = 0.0072 A-sec/query
In 1 hour, awake = 60 query/hour * 0.0072 A-sec/query =
0.432 A-sec/hour
Quiescen
t current = 60 μA = 0.06 mA = 0.00006 A
In 1 hour, quiescent = 0.00006 A * (3600 sec/hour – 720 * 0.28 sec/sample – 60 x 9 sec awake) =
0.1715 A-sec/hour
In 1 hour, 39
plus
-IM takes 720 samples and transmits average to computer 60 times:
Current consumption / hour = 0.6048 + 0.1536 + 0.432 + 0.1715
=
1.3619 Amp-sec/hour
Capacity = (4.0 Amp-hours * 3600 sec/hour) / (1.3619 Amp-sec/hour) = 10573 hours = 440 days =
1.2 years
Total number of samples = 10573 hours * 720 samples/hour =
7,612,000 samples
However, memory can only hold approximately 5,500,000 samples; 39plus-IM will not overwrite data in memory.
Example 3:
39
plus
-IM
without pressure
is deployed on a mooring, with full power transmit setting
(
SetEnableFullPwrTX=Y
), and is
poll sampled
every minute (60 samples/hour). It is set to
#iiOutputFormat=1
,
resulting in conservatively 80 characters/sample.
PwrOff
is sent immediately after each sample. How long can it be
deployed?
Polled Sampling current for 39
plus
-IM
without pressure, with full power
= 1.9 mA-sec/sample = 0.0019 A-sec/sample
In 1 hour, sampling = 60 samples * 0.0019 A-sec/sample =
0.1140 A-sec/hour
Transmit current/sample = 0.032 mA-sec/character = 0.000032 A-sec/character
0.000032 A * 80 characters = 0.00256 A-sec/sample
In 1 hour, transmit = 60 query/hour * 0.00256 A-sec/sample =
0.1536 A-sec/hour
Awake current (awake but not sampling) = 800 μA = 0.8 mA = 0.0008 A
Awake current per query = 0.0008 A * 1 sec/IM to be queried * 9 other IMs on mooring = 0.0072 A-sec/query
In 1 hour, awake = 60 query/hour * 0.0072 A-sec/query =
0.432 A-sec/hour
Quiescent current = 60 μA = 0.06 mA = 0.00006 A
In 1 hour, quiescent = 0.00006 A * (3600 sec/hour – 60 * 0.28 sec/sample – 60 * 9 sec awake) =
0.1826 A-sec/hour
In 1 hour, 39
plus
-IM takes 60 samples and transmits each sample to computer:
Current consumption / hour = 0.1140 + 0.1536 + 0.432 + 0.1826
=
0.8822 Amp-sec/hour
Capacity = (4.0 Amp-hours * 3600 sec/hour) / (0.8822 Amp-sec/hour) = 16322 hours = 680 days =
1.8 years
Total number of samples = 16322 hours * 60 samples/hour =
979,000 samples
This shows that polled sampling is a very inefficient use of power.
