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Powering DC motors with Batteries

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enridp

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Hi !
I'm looking for the best way for powering a DC motor with batteries.
The motor is 24V and has a consumption of 100mA to 300mA.

I was thinking in using 6 AA batteries (9V) and boost them to 25V with a DC-DC converter like MC34063.

But calculating the components with this page:

**broken link removed**

I found that from 9 to 25V I have a max Iout of 240mA (Imax supported is 1.5A for MC3403), so the efficiency is:
[(25*0.24)/(9*1.5)] = 44% !!

Am I doing something wrong? I thought that MC3403 was a very efficient DC-DC converter.

Does somebody know a better way for powering this motor?

Thanks !!
Enrique.
 
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If we take your 24 VDC motor that draws lets say 200 mA under normal load the power consumed is about 4.8 Watts. That is to do a certain amount of work. If the same motor were wound and designed for 12 volt service it would draw 400 mA to do the same amount of work (4.8 Watts) and if it were wound and designed for 48 volts it would draw 100 mA to do the same amount of work. So what we have is if we half the voltage we double the current and if we double the voltage we half the current. The bottom line is 4.8 Watts of power.

DC to DC converters work the same way. If for example I want to go from 12 volts in to 24 volts out the current in for a given load will be twice the current out for that load. Then we can worry about the efficiency of the converter. For the converter to do its thing it will have loss. A good example would be heat loss. You can only get so much and there is a trade off.

If I were you I would consider two 12 volt SLA batteries in series with adequate current to run your motor. With a 200 to 300 mA load at 24 volts your AA batteries in series will be very short lived. I would also fuse the motor as in a stall the current draw will be real high.

Ron
 
Or better still, use a more sensible motor - a 24V motor sounds pretty unsuitable - although I suspect it's probably a FAR higher current motor than he suggests, and he's only measuring it's idle current.
 
What is the application for your motor. Is it in a portable equipment? if it is not, why not to consider a 24v power supply using your home electricity outlet?
 
I will try to explain it better (sorry for my english):

This is the motor:
**broken link removed**

I'm using this motor because is the only that I have with worm gears, I don't have many options in this.
I have measured the idle current an it's about 100mA without load, and braking the gears (simulating load) is about 200-300 mA.
It's a portable equipment.
I can power the motor with 12V or less, but then is too slow, the motor says 24V in it's case, and the optimal speed is with about 30V.

These are my calculations:
24[V]*0.3[A] = 7[W] = 7[V*A]
The motor turns a plate in about 12 sec. waits about 5-10 seconds and turns the plate again.
So, every turn needs:
7[W]*12 = 84[J] = 84[V*A*s] = 0,024 [V*A*h] = 24[V*mA*h]

With 1 AA battery of 1.5[V] and 2500[mAh] we have:
1.5[V]*2500[mAh] = 3750[V*mA*h]
and therefore:
NTurns = 3750/24 = 156

With 1 AA battery we can turn the plate 156 times.
But I need 24V, not 1.5V.
Boosting 1.5 to 24 seems very utopian, so I was thinking in using 6 AA batteris for boosting from 9V to 25V.
(also with 6 AA batteries we have 963 turns).

But the problem is that MC34063 looks very inefficient (44%), and with very low Iout (from 9 to 25 we have an output of 240mA).

So I'm looking for a better way for powering this motor.
The easy option is to use 16 AA batteries but is more expensive that the whole project :(

Do you know a better option?

PS: SLA batteries are too heavy and too big, and also are very expensive and difficult to find here in Argentina
 
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For 100 mA, you can use 2 of 9Volts cell, even that's closer to 24V.
 
But calculating the components with this page:
**broken link removed**
I found that from 9 to 25V I have a max Iout of 240mA (Imax supported is 1.5A for MC3403), so the efficiency is:
[(25*0.24)/(9*1.5)] = 44% !!

The efficiency might be that bad, but the average current taken by a boost converter will not be the peak current, so the efficiency will be better than that.

Boost conversion is never 100% efficient, but you might get 80%

The biggest problem that you have is that the capacity of batteries is less when you draw more current. You are looking for 7 W or so of power. That is more than you can ever take from one AA cell. If you have 6 AA cells, the power is over 1 W, and their capacity is less at that power.

https://data.energizer.com/PDFs/l91.pdf shows the expensive lithium batteries against standard alkali AA cells.
 
I've replied two times but nothing was posted, I know that I'm new and my posts are moderated, but why they are not approved? :(
 
Sometimes it just takes awhile. You should be fine now.

Ron
 
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