Super capacitor for running DC motor

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SUMMARY

This discussion focuses on using a Maxwell super capacitor to power a DC motor with specifications of 90V - 120V DC and 15A current for a minimum of 10 seconds. The user calculated a discharge time of 11.6 seconds using the formula t=[C*(V0-V1)/I], confirming the suitability of the chosen super capacitor with a capacitance of 5.8F. Additionally, the user plans to charge the capacitor using a 110V AC supply with a transformer and rectifier. Recommendations include using a reverse biased Schottky diode to protect the motor from inductive kickback.

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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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