# Charging Circuit

Discussion in 'The Projects Forum' started by DeanF, Mar 10, 2010.

1. ### DeanF Thread Starter New Member

Mar 9, 2010
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0
I am building a project that required that a 12v battery be used to power a motor. The problem is that I want to apply 12v to move an object but want the object to return to original position upon removing the power. The motor is a electric window motor for a car which I assume draws about 2-3 amps. I can find that out for sure,,, but you get the idea, I'm sure. Thanks.

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
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If it's like a small car with wheels, can you use a spring to pull it back? You can't expect the motor to do anything but stop when power is cut.

3. ### retched AAC Fanatic!

Dec 5, 2009
5,201
313
You may want to use a spring return actuator for your movement/return project.

That will allow you to apply the 12v and move the object, and when you cut power, the spring returns it to the start position.

4. ### DeanF Thread Starter New Member

Mar 9, 2010
2
0
I guess I should clarify my problem. I want to charge a source when the charging voltage is removed that the charged source will power the motor to return to the starting point. This is not where a spring application can be used easily. Thank you.

5. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
Use a rechargeable battery of some type to be determined by your distance to travel and the power requirements of the motor to move your load, including the battery of course.

You might consider supercapacitors, which can be charged quickly, but they have a very different discharge curve from a battery, and would be considerably larger than a battery for the same amount of power stored.

6. ### ifixit Distinguished Member

Nov 20, 2008
639
110
1. How much current does the motor draw (exactly) during this return action?
2. How long does it take to return?
3. If a capacitor is used to power the motor then it will discharge to lower voltages as the current is taken by the motor. At what voltage does the motor stop due to lack of voltage.
Formulas...
• Time = (Capacitance x (Vt - Vi)) / Current.
• Capacitance = Current x Time / (Vi - Vt)
• Example...
• Ans to (1) = 2 Amps.
• Ans to (2) = 2 Seconds.
• Ans to (3) motor stops with 8 or less Volts on it.
• Capacitor needed = 2 A x 2 s / (12V - 8V) = 1F
The above doesn't take into account the motor slowing down as the voltage drops, and a one Farad cap is huge. Likely cost alot as well.

A mechanical solution is likely the best, but as SgtWookie suggests, a 12V Ni-cad pack from an old battery powered drill/driver might work well enough. You can charge it from the other battery that gets removed.

Good Luck

7. ### hwy101 Active Member

May 23, 2009
91
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If I was building this project, I would probably be using a windshield wiper motor, they always return to their original starting point, but keep in kind these motors require a constant power as well as a switched source, that's the tricky part.
But that should work with Wookie's battery idea.

8. ### Bernard AAC Fanatic!

Aug 7, 2008
4,243
420
You have a 12 V battery, motor & object; object to be moved from point A to point B.
Does 12 V battery stay at point A?
Does 12 V battery move with object & stay at point B?
Just one motor?