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Author Topic: URGENT-battery pack help needed  (Read 736 times)

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Offline aruna1Topic starter

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URGENT-battery pack help needed
« on: March 22, 2010, 06:16:34 PM »
hi guys,i'm working on a robot that uses 2 pololu metal gear motors as drivers.these are 6v motors.i use a 12v 1500mAh battery pack(1.2v cells x 10).i use 7806 to convert 12v to 6v for motors.my question is as my motors drain high curr ent,battey voltage drops to 8v quickly,after this point regulator dont work properly as they need atleast 8v input to operate.so i have to recharge the battery.bt batter still have 6v,which i cant use.it seems pretty waste of battery power.any idea how to solve this and use batteries maximamly?  thanks
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Offline billhowl

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Re: URGENT-battery pack help needed
« Reply #1 on: March 22, 2010, 07:33:56 PM »
For battery pack(1.2v cells x 10) 8V is lowest it can go, if it go any lower it may damage the battery. BTW 7806 is not efficient and wastes more power stepping down the voltage than it actually ends up delivering to the target device!

You can use switching regulator like the BEC
 


http://shop.ebay.com/i.html?_nkw=5A+BEC&_sacat=0&_dmpt=Radio_Control_Parts_Accessories&_odkw=BEC&_osacat=0&_trksid=p3286.c0.m270.l1313
http://www.dimensionengineering.com/ParkBEC6v.htm
« Last Edit: March 22, 2010, 10:06:00 PM by billhowl »

Offline aruna1Topic starter

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Re: URGENT-battery pack help needed
« Reply #2 on: March 22, 2010, 09:46:48 PM »
For battery pack(1.2v cells x 10) 8V is lowest it can go, if it go any lower it may damage the battery. BTW 7806 is not efficience and wastes more power stepping down the voltage than it actually ends up delivering to the target device!

You can use switching regulator like the BEC
 


http://shop.ebay.com/i.html?_nkw=5A+BEC&_sacat=0&_dmpt=Radio_Control_Parts_Accessories&_odkw=BEC&_osacat=0&_trksid=p3286.c0.m270.l1313
http://www.dimensionengineering.com/ParkBEC6v.htm

hi,i dont have enough time to buy it now.can you say how much voltage battery will have once it fully charged? 15v maybe?
thanks
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Offline billhowl

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Re: URGENT-battery pack help needed
« Reply #3 on: March 22, 2010, 10:02:45 PM »
For battery pack(1.2v cells x 10) fully charged will be about 14.5V.


Quote
How do I tell if I need a switching regulator?
As a general rule of thumb, if your linear voltage regulation solution is wasting less than 0.5 watts of power, a switching regulator would be overkill for your project. If your linear regulator is wasting several watts of power, you most certainly want to replace it with a switcher! Here is how to calculate power losses:

The equation for wasted power in a linear regulator is:

Power wasted = (Input voltage – output voltage) * load current

For example, let’s say you have a 12V lead-acid battery and you want to power a microcontroller that draws 5mA, and an ultrasonic rangefinder that draws 50mA. Both the microcontroller and the ultrasonic rangefinder run off of 5V. You use an LM7805 (a very common linear regulator) to get the voltage down to 5V from 12V.

Power wasted = (12V – 5V) * (0.050A + 0.005A) = 0.385W

0.385W is not too bad for power losses. The LM7805 can handle this without a big heatsink. You could get more battery life if you used a switching regulator, but in this case the power consumption is so low that the battery life will be very long anyway.

Now let’s expand on this example, and add two servos that draw an average of 0.375A each, and also run off of the 5V supply. How much power is wasted in a linear regulator now?

Power wasted = (12V – 5V) * (0.050A + 0.005A + 0.375A + 0.375A) = 5.635W

5.6 Watts is a lot of waste heat! Without a large heatsink the LM7805 would get so hot it would desolder itself or melt your breadboard or defeat Iceman. Even with the heatsink, 5.6W is also a lot of life to suck out of your battery for no reason. A switching regulator such as a DE-SW050 would be very useful in this case, and would reduce power losses to around 0.5W.


 


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