- #1

Yasir

- 9

- 0

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. :)