where:
24 = 24V output
12 = Input voltage
Total Current @ 24V = Sum of 24V loads powered by the 12/24V
boost converter (in mA).
0.88 = Efficiency
The 5232 DC Input supply current is calculated as follows:
5232 DC Input Current = (5 x Total Current @ 5V) / (12 x 0.85)
= (5 x 0.670) / (12 x 0.85)) = 328
mA
where:
5 = 5V power supply
12 = Input voltage
Total Current @ 5V = Sum of 5V loads powered by the 5232 (in mA).
0.85 = Efficiency
The total current required from the 12V battery for this application is:
455mA + 328mA =
783mA.
It should be noted that in a battery-powered application such as this there are several opportunities to
reduce the power consumption. There will be significant power savings when the LEDs are disabled.
Switching VLOOP off under program control, when not required by the application will reduce the
power required.
4.2.5.3 SCADAPack P4B
As illustrated in section 4.2.1-Recommended AC Power Supply Configuration the 5232 controller
board may be powering a 5601 I/O module as well as other 5000 Series modules.
The following table shows current requirements of the controller board and the 5601 I/O modules:
5V Current (mA) 11-30V Current (mA)
5232 Controller Board 500 0
5606 I/O Module 30 to 600mA
See specifications
for details.
12
In the following example, a SCADAPack P4B equipped with analog outputs is powered using the
Recommended AC setup such that 24V will be available across the DC PWR terminals on connector
P3 to power the analog circuit on the integrated I/O board. This 24V must be available for both
analog outputs and only 4 of the 8 analog inputs. Assume all 12 digital output relays will be on with
the LEDs enabled. The minimum current requirement for each module on the SCADAPack P4 is
given in the table below.
SCADAPack 32 Controller Hardware Manual
May 24, 2006
16
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