Hello everybody, I'm using Arduino 2009 with Adafruit motor shield to drive a stepper motor that runs at 4.8v 1.6A.
As the online docs says, I've piggyback one of the two L293D on the board.
If I power arduino and the motor shield through the usb input everything works fine,
but using a 12v lead battery to power them doesn't works because the output coming off the shield is 12v.
How can I lower the output to 4.8v needed by the motor?
Thanks
Luciano
Help with Motor Shiled and 12V battery
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Please be positive and constructive with your questions and comments.
- adafruit_support_mike
- Posts: 67454
- Joined: Thu Feb 11, 2010 2:51 pm
Re: Help with Motor Shiled and 12V battery
Basically, you want a voltage regulator.. something that takes a higher voltage as input and produces a steady lower voltage as output.
The easy way is to use a linear regulator like the LM7805. That takes any input up to about 30v and produces 5v output. The trouble is that the LM7805 is basically a self-tuning resistor, so all the voltage you lose gets wasted as heat. In this case, you'd generate 7W of heat for every 5W that reached your motor.
The efficient solution is more complicated. That involves a 'switching regulator' that basically opens and shuts a switch really fast. It's pretty much the same idea as PWM, just faster, smoothed out better, and with better control of the output.
Some "low dropout" voltage regulators use switching circuits, but it can be hard to tell from the datasheets. You can get dedicated switching controllers and switching regulators, but you usually need to build a circuit around them. If you're going to do that, you might as well look at pre-built circuits like this one:
http://www.mouser.com/ProductDetail/Mur ... %2f%2fo%3d
That one costs about $12.50, but will drop your battery voltage to 5v at 2A with a simple drop-in circuit.
The easy way is to use a linear regulator like the LM7805. That takes any input up to about 30v and produces 5v output. The trouble is that the LM7805 is basically a self-tuning resistor, so all the voltage you lose gets wasted as heat. In this case, you'd generate 7W of heat for every 5W that reached your motor.
The efficient solution is more complicated. That involves a 'switching regulator' that basically opens and shuts a switch really fast. It's pretty much the same idea as PWM, just faster, smoothed out better, and with better control of the output.
Some "low dropout" voltage regulators use switching circuits, but it can be hard to tell from the datasheets. You can get dedicated switching controllers and switching regulators, but you usually need to build a circuit around them. If you're going to do that, you might as well look at pre-built circuits like this one:
http://www.mouser.com/ProductDetail/Mur ... %2f%2fo%3d
That one costs about $12.50, but will drop your battery voltage to 5v at 2A with a simple drop-in circuit.
- arctic_eddie
- Posts: 233
- Joined: Tue Feb 28, 2012 6:01 pm
Re: Help with Motor Shiled and 12V battery
Here's one for less money but it will take a few weeks to get here. Many of us in the astronomy world have used these to power cameras at 4.2V on telescopes. You can adjust them for the desired output.
http://www.ebay.com/itm/1PCS-LM2596-DC- ... 1088690509?
http://www.ebay.com/itm/1PCS-LM2596-DC- ... 1088690509?
Please be positive and constructive with your questions and comments.