- #1
Yasir
- 9
- 0
Hi,
I am planning to run a DC motor using a super capacitor. The specification of the motor are:
I have finalized this super capacitor from maxwell for the above operation http://www.maxwell.com/products/ultracapacitors/docs/160vmodule_ds_3000246-5.pdf
I have planned to charge this capacitor to 120V and then discharge till 90V DC.
I have used the formula:
t=[C*(V0-V1)/I] where,
t = Discharge time (sec)
C = Capacitance (F)
V0 = Initial Voltage (V)
V1 = Final Voltage (V)
I = Current (A)
Thus applying the values to the formula:
t ={5.8(120-90)/15} (120 - 90 is the voltage range for the motor, 15 is req. current)
={5.8(30)/15} (5.8 F is the capacitance of the Super cap mentioned in datasheet)
=11.6 sec.
11.6 sec which is more than what i require.
I got the above formula from Elna Supercapacitor, Illinois Capacitor Inc and Maxwell (Page 2) http://www.maxwell.com/products/ultracapacitors/docs/applicationnote_maxwelltestprocedures.pdf so m pretty sure about the formula.
So guys please help me here and let me know whether the calculations and the product i have finalized for running the motor for 10 seconds (atleast) is suitable or not. What problems i may face during this experiment. I have planned to use AC 110V supply with transformer/rectifier to charge the capacitor to 120 V. Do i need to add any other equipment between motor and supercap or a direct connection is sufficient.
Thanks in advance. :)
I am planning to run a DC motor using a super capacitor. The specification of the motor are:
- Voltage : 90V - 120 V DC
- Current : 15 A
- Duration : Should run for atleast 10 seconds.
I have finalized this super capacitor from maxwell for the above operation http://www.maxwell.com/products/ultracapacitors/docs/160vmodule_ds_3000246-5.pdf
I have planned to charge this capacitor to 120V and then discharge till 90V DC.
I have used the formula:
t=[C*(V0-V1)/I] where,
t = Discharge time (sec)
C = Capacitance (F)
V0 = Initial Voltage (V)
V1 = Final Voltage (V)
I = Current (A)
Thus applying the values to the formula:
t ={5.8(120-90)/15} (120 - 90 is the voltage range for the motor, 15 is req. current)
={5.8(30)/15} (5.8 F is the capacitance of the Super cap mentioned in datasheet)
=11.6 sec.
11.6 sec which is more than what i require.
I got the above formula from Elna Supercapacitor, Illinois Capacitor Inc and Maxwell (Page 2) http://www.maxwell.com/products/ultracapacitors/docs/applicationnote_maxwelltestprocedures.pdf so m pretty sure about the formula.
So guys please help me here and let me know whether the calculations and the product i have finalized for running the motor for 10 seconds (atleast) is suitable or not. What problems i may face during this experiment. I have planned to use AC 110V supply with transformer/rectifier to charge the capacitor to 120 V. Do i need to add any other equipment between motor and supercap or a direct connection is sufficient.
Thanks in advance. :)