From ARVP
This project covers the implementation of the power systems on Bearacuda, the power consumption of major components, fusing and voltage regulation. Bearacuda is powered by two 18.5 V Lithium Polymer batteries; one to power the thrusters and one to power the internal electronics and external servos.
Project Members:
- Brodi Roberts
- Darrel Ross
Overview
Bearacuda is powered by two LiPo batteries. The first battery powers the six thrusters through three Roboteq motor controllers, which are in turn controlled by the embedded computer. The second battery is connected to a Vicor switching buck regulator, which takes the battery input voltage and converts it to 12 V regulated DC, which is in turn used to power the internal electronics. The embedded computer draws power from a PicoPSU ATX converter, which converts the 12 V DC into the necessary ATX voltages. Any devices requiring 5 V DC (including external servos) draw it from a 5 V switching regulator on the mainboard.
Batteries:
- Lithium Polymer:
http://thunderpowerrc.com/html/extreme.html
Charge Capacity: 5Ah
Weight: 613g
Voltage: 18.5V
Notes: dimensions 42x48x160 [mm], charge available from same site for $130, purchase of a LiPo charging bag is recommended
Information on the handling, storage, and transportation of LiPo Batteries is available in the Primer Document
Must handle with caution.
Voltage Regulator
- Takes a regulated 12V in and distributes it to an ATX power array (3.3V 5V -/+12V)
Battery Charger
Notes: Can charge, discharge, and balance. 1 to 6 LiPo Cells charging capability.
The battery charger purchased requires a 12V/10A DC supply to charge from and has been purchased.
Other Components
- LipoSac: the Liposac is to be used whenever the LiPo batteries are being charged. This is a saftey precaution and is designed to help contain any potential fires, which are possible when charging lithium polymer batteries.
Bearacuda Power Consumption:
Embedded Computer: 40-50 W
MC68332: 0.455 W
Sensor array: ~5 W
Servos: 2*3A*5V/0.8 = 37.5 W
dsPIC: .250A*5V/0.8 = 1.6 W
P = 50W + 0.455W + 5W + 37.5W + 1.6W = 94.56 W Take vicor regulator to have 0.85 efficiency, then worst case power consumption is 94.56/0.85 = 111.24 W
Depreciated Information
New Voltage Regulator:
- A new voltage regulator must be acquired, as the 100 W Vicor regulator we currently have will not be able to provide power for all the electronics once the embedded computer is installed. A 150 W regulator would probably be sufficient but to maintain a safe margin, a 200 W supply would be preferred. A 200 W Vicor is being ordered for testing purposes but it would be preferable if we could develop our own as well. Anybody interested in this project should contact Darrel. - DarrelRoss 07:58, 21 September 2008 (MDT)
- A new 200 W voltage regulator has been acquired from Vicor. The voltage regulator was redesigned to include the large regulator and large header pins were used as the team found that the old board's header pins tended to bend during competition due to the large cables. - ChrisWoloschuk
