12V power supply mini breadboard project. How did I do?

Have questions about a SparkFun product or board? This is the place to be.

Moderator: phalanx

12V power supply mini breadboard project. How did I do?

Postby toddthefrog » Tue Feb 23, 2010 10:09 am

Today is literally day 0 of learning how to solder, touching resistors, capacitors, everything. I bought an Arduino, 12V stepper motor and the Easy Driver 4.3. I needed a semi-permanent way to feed 12 volts to the steppers.

This is based on the SparkFun tutorial here

Image

Image

Image

I have the 3.3v linear regulator to supply current to my LED so I'll know when the thing is turned on.

The red and black wires sticking straight up indicate which line is + and gnd out.

One thing I haven't figure out yet and hopefully someone can help me with is if my wall wart is 12v / 650 mah and each stepper is supposed to get 330 mah max will I use a resistor to downgrade the mah's? I don't understand how that works yet. Also do you suggest I place a 100uf cap on the positive out?

What do you guys think?

Thanks!!

Todd
toddthefrog
 
Posts: 5
Joined: Tue Feb 23, 2010 9:45 am

Postby waltr » Tue Feb 23, 2010 10:18 am

No resistor needed.
Your wall wart will have a 12V potential across its output but its the load that draws current. Connect a resistor of 10 k Ohm across the wall wart output and use Ohm's law to calculate the current. The voltage on the resistor is 12V and the current flowing through the resistor is 1.2mA. Now change the resistor to 100 Ohm, the voltage on the resistor is the same but the current through this resistor is now 120mA.
You can keep decreasing the resistor value and the current will go up until the wall wart can not supply it. Then the output voltage of the wall wart will drop and it may even get hot.
Make sense?
waltr
 
Posts: 2651
Joined: Tue Sep 08, 2009 12:07 pm
Location: Philadelphia, USA

Postby toddthefrog » Tue Feb 23, 2010 10:23 am

So if I'm understanding you correctly the max that the wall wart can output is 650 mah but if the stepper only NEEDS 330 mah than I can safely power 2 steppers. If I put 3 steppers on the wall wart the max it will supply is around 220 mah per stepper, if that much mah is required. Does that mean that each stepper has some kind of limiter that will not allow more than 330 mah's and no matter what I hook it up to, it will only "absorb" it's max? Is that why an LED needs a resistor because it has no such built in limiter aka resistor?

Thanks so much for taking the time to explain,

Todd
toddthefrog
 
Posts: 5
Joined: Tue Feb 23, 2010 9:45 am

Postby gmarsh » Tue Feb 23, 2010 10:43 am

Your 3.3V regulator probably needs capacitors on its input and output to work correctly.

Other than that, damn good soldering job if it's your first time!
gmarsh
 
Posts: 91
Joined: Thu Jul 19, 2007 5:49 am
Location: Halifax, NS

Postby Grimfox » Tue Feb 23, 2010 11:01 am

toddthefrog wrote:So if I'm understanding you correctly the max that the wall wart can output is 650 mah but if the stepper only NEEDS 330 mah than I can safely power 2 steppers. If I put 3 steppers on the wall wart the max it will supply is around 220 mah per stepper, if that much mah is required. Does that mean that each stepper has some kind of limiter that will not allow more than 330 mah's and no matter what I hook it up to, it will only "absorb" it's max? Is that why an LED needs a resistor because it has no such built in limiter aka resistor?

Thanks so much for taking the time to explain,

Todd


330mAH * 2 = 660mAH slightly more than your wallwort can supply. It's only 10mA so you should be ok, but if both motors are pulling that level of power continuously then the wallwort will heat up. So be aware. Also if you are powering the audrino and LED off that same supply they need current too. So 650mAH is not as much as it first seems.

The motors themselves have a built-in resistance that is roughly based on the resistance in the windings but is also affected by the magnetic field that they create. The LED is effectively just a short piece of wire but only allows current to travel in a single direction (the D stands for diode) If you reverse you LED it wont turn on. This is part of the reason why you needed the resistor. The other part of that is to drop the voltage that the LED does not.

welcome to the wonderful world of electronics.
Grimfox
 
Posts: 123
Joined: Tue Oct 27, 2009 9:16 am

Postby toddthefrog » Tue Feb 23, 2010 11:03 am

gmarsh wrote:Your 3.3V regulator probably needs capacitors on its input and output to work correctly.

Other than that, damn good soldering job if it's your first time!


I saw they used caps in the tutorial but i figured I wouldn't need them because I'm only powering the LED from it so it's not getting an intensive workout or anything. The capacitors only stabilize the current correct?

Thanks for the compliment, that's my first soldering job! I'm so excited to finally be doing this. I've been thinking about trying electronics for a while now.

Todd
toddthefrog
 
Posts: 5
Joined: Tue Feb 23, 2010 9:45 am

Postby toddthefrog » Tue Feb 23, 2010 11:09 am

Grimfox wrote:
toddthefrog wrote:So if I'm understanding you correctly the max that the wall wart can output is 650 mah but if the stepper only NEEDS 330 mah than I can safely power 2 steppers. If I put 3 steppers on the wall wart the max it will supply is around 220 mah per stepper, if that much mah is required. Does that mean that each stepper has some kind of limiter that will not allow more than 330 mah's and no matter what I hook it up to, it will only "absorb" it's max? Is that why an LED needs a resistor because it has no such built in limiter aka resistor?

Thanks so much for taking the time to explain,

Todd


330mAH * 2 = 660mAH slightly more than your wallwort can supply. It's only 10mA so you should be ok, but if both motors are pulling that level of power continuously then the wallwort will heat up. So be aware. Also if you are powering the audrino and LED off that same supply they need current too. So 650mAH is not as much as it first seems.

The motors themselves have a built-in resistance that is roughly based on the resistance in the windings but is also affected by the magnetic field that they create. The LED is effectively just a short piece of wire but only allows current to travel in a single direction (the D stands for diode) If you reverse you LED it wont turn on. This is part of the reason why you needed the resistor. The other part of that is to drop the voltage that the LED does not.

welcome to the wonderful world of electronics.


I'll be powering the arduino from the built in DC in.

OK now I'm a tid bit confused because I thought the resistor only effects the current and not the voltage. So you're telling me a resistor effects both? If I have a 12V (like I do on this board) in and need to power a 2.2 volt part (the LED) I can just plop a resistor on there and not have to use the linear regulator?
toddthefrog
 
Posts: 5
Joined: Tue Feb 23, 2010 9:45 am

Postby edward.ford » Wed Feb 24, 2010 2:21 pm

Great looking soldering you've got going there! My first solder job didn't look half that good.

Just wanted to point out that an ATX power supplies can provide a clean 12v for your project. If you have one laying around, it might be worth a look. I used one to power my steppers before investing in a decent 24v PS.

Remember that you have to jump one of the connectors (usually the green one) to ground in order for the PS to turn on.
edward.ford
 
Posts: 22
Joined: Mon Nov 23, 2009 9:03 am
Location: Dixon, IL

Postby TheDirty » Wed Feb 24, 2010 2:33 pm

I thought I posted in here, but I guess I didn't. You need the capacitors with the LDO or the regulator will be unstable, not just noise in the power, but oscillate.

Seems like a lot of power to run through single connections on a breadboard. You might want to double up the wiring.
Mark Higgins
TheDirty
 
Posts: 657
Joined: Sun Jan 04, 2009 8:06 am
Location: Toronto, ON, Canada

Postby MichaelN » Wed Feb 24, 2010 2:58 pm

I know I'm being pedantic, but....

"mAh" is a measure of capacity/charge, not a measure of current. Normally used to specify battery capacity.

If you're talking about current, it should be "A", "mA" or "Amps" instead.
MichaelN
 
Posts: 1188
Joined: Fri May 08, 2009 6:14 pm
Location: Australia

Postby toddthefrog » Wed Feb 24, 2010 11:06 pm

MichaelN wrote:I know I'm being pedantic, but....

"mAh" is a measure of capacity/charge, not a measure of current. Normally used to specify battery capacity.

If you're talking about current, it should be "A", "mA" or "Amps" instead.


Thanks for correcting me! I want to say things the right way.

Thanks everyone for your help and insight! I appreciate it.

Todd
toddthefrog
 
Posts: 5
Joined: Tue Feb 23, 2010 9:45 am

Re: 12V power supply mini breadboard project. How did I do?

Postby markaren1 » Fri Feb 26, 2010 1:00 am

you do need 2 x 100nF caps

input to gnd
and out to gnd. The regulator may oscillate otherwise. This is normal good practice.

-Mark
markaren1
 
Posts: 366
Joined: Sat Dec 19, 2009 12:16 pm

Re: 12V power supply mini breadboard project. How did I do?

Postby markaren1 » Fri Feb 26, 2010 1:01 am

you do need 2 x 100nF caps

input to gnd
and
out to gnd

The regulator may oscillate otherwise which may cause all manner of interesting problems.

This is normal good practice.

-Mark
markaren1
 
Posts: 366
Joined: Sat Dec 19, 2009 12:16 pm

Re:

Postby Grimfox » Fri Feb 26, 2010 3:25 pm

toddthefrog wrote:OK now I'm a tid bit confused because I thought the resistor only effects the current and not the voltage. So you're telling me a resistor effects both? If I have a 12V (like I do on this board) in and need to power a 2.2 volt part (the LED) I can just plop a resistor on there and not have to use the linear regulator?


Current Resistance and Voltage are all related though Ohm's law V(voltage) = I(current) x R(Resistance)

Therefore if you change the resistance, current, or voltage the other two components of the equation will also be affected. Adding your resistor restricts the flow of current and drops some voltage. As a general case if you have 1 volt dropping across a 1ohm resistor there would be 1 amp passing through your resistor. If you double the resistance then you have 1 volt/2ohms = .5 amps.

Take some time and read up on Ohm's Law and Kirchoff's Current and Voltage Laws. (wikipedia might help) They will help you understand some of the effects of what you are doing. I took this class as a freshman I think you will be able to pull some useful information from the website. The lecture and powerpoints data should be useful you might be able to find some practice problems within the homework but it looks like a lot of that has been moved to a restricted location online.

http://www2.tech.purdue.edu/ecet/courses/ecet107/

Everything in exam 1 and 2 is pretty good stuff. exam 3 is still good but gets more difficult.

One more key note, Power(W) = I x V So before you go dumping 1amp through a 1ohm resistor note that you'll be using 1Watt of power and that little resistor (more than likely limited to 1/4W) will burst into flames.

(yay new forums)
Grimfox
 
Posts: 123
Joined: Tue Oct 27, 2009 9:16 am


Return to SparkFun Product Questions

Who is online

Users browsing this forum: No registered users and 4 guests