Super capacitor for running DC motor

AI Thread Summary
A user is planning to run a DC motor with specifications of 90V-120V and 15A using a super capacitor from Maxwell, which has a capacitance of 5.8F. They calculated that the capacitor can discharge for 11.6 seconds, exceeding their requirement of at least 10 seconds. To charge the capacitor to 120V, they will rectify 110V AC, needing approximately 85V AC for proper charging. Suggestions include adding a reverse biased Schottky diode across the motor for protection against inductive kick. The user is open to recommendations for alternative super capacitor products or modifications to their setup.
Yasir
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Hi,

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. :)
 
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C = Q / V
Q = I * t
∴ t = V * C / I
and so: dt = dv * C / I
Your 11.6 sec is correct.

To charge the capacitor to 120V DC you will need to rectify 120V / Sqr(2) = 85V AC

To protect the motor insulation, you might consider a reverse biased Schottky diode across the motor to catch any inductive kick produced during switching.
 
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Thanks for the reply Baluncore...
Guys all kind of suggestions are welcome...anything you want to add any modifications or any other super cap product cheaper and smaller than what i have suggested in the first post ...even if you think whatever i have posted is right, please let me know...
Thanks a lot again..
 
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