Power two Arduinos and a linear actuator from a single wall wart

Hi all.

I'm looking to power two Arduinos and a linear actuator from a single wall wart, as in the title, using two jacks linked together and to the motor shield. The linear actuator draws at most 150 mA, and I understand that Arduinos call for about 250 mA each. I'm powering a MAX486 from one of my Arduinos and one LED from the other, so I can't imagine that calls for much, so by my reckoning I need about 650 mA. The wall wart is 12V and 1 A. Is that sufficient for my needs? Is there any problem with running all three of the same one? I was under the impression that common grounds were necessary so this seems like it solves that problem.

Thanks for your help

You have to check the voltage regulators on the Arduino boards (put your finger on it). They might get too hot with 12V. The current needed by an Arduino board can be 1uA to 1A (that is a range of a million times). Which Arduino boards are you using ?

What kind of wall wart are you using ? A regulated 12V 1A good quality wall wart is okay. An unregulated cheap wall wart might output 18V and get too hot above 300mA.

You can power all three with a single wall wart. You have to connect the grounds of course. Do you have a transistor or mosfet to operate the linear actuator ?

You can power all three with a single wall wart.

Technically , that's true but it is bad practice. Your logic supply should be separate from the motor supply although they can share a common ground. Motor inrush current pulls down the regulator input voltage so if you are absolutely set on running them off the same supply you need to put a honking big electroytic cap (at least 1000uF) across the 5V bus) and probably another one across the Vin with a cap rated for the voltage you are using (12V ==> cap voltage rating > 15Vdc) . If you don't have 1000uF then you can probably get away with 470uF but I wouldn't go less than that. You should probably use a 3 ohm (3W) dropping resistor on the input to the arduino Vin to drop the 12V down to 9V so the onboard regulator doesn't get hot dissipating the extra 12-7=5V that it needs to drop since the regulator ideal input voltage is 5V + 2V headroom = 7V and that is 5V LESS than 12V. Without the 3 ohm (3W) dropping resistor your regulator is going to get hot.

I'm using this wall wart: http://www.cui.com/product/resource/epsa-12w.pdf It says it's 5% line/load regulation so that seems like a good thing.

It's an Arduino Uno with a L298N motor shield drawing 36 mA logic current. The grounds are connected. I'm using a motor controller board for the linear actuator.

I've got plenty of 1000 uF caps kicking around. I'm not sure where you're saying to connect them, though. One is Vin pin to ground, right? I don't know what the 5V bus is.

12V is within the recommend voltage limit for the Arduino Uno, and I'm intending to power through the jack. Is it really going to be a problem?

Thanks

I think the wall wart is okay.

The 12V is within the limit, but there is not a hard limit. It depends on the current that is used. The voltage regulator on the Arduino board converts 12V to 5V, thereby wasting 7V (12-5) to heat. So if you use a number of leds or other things from the 5V, that voltage regulator could get hot.

Which linear actuator is it ? I'm curious if it has a large start current. The 1000uF can be useful in that case, but perhaps the 1A wall wart is more than enough.

The 5V 'bus', is the 5V pin of the Arduino. That is a 5V output, and together with the GND it can be used to power a few small things.

Peter is saying basically the same thing I said . The point of my suggestion was that it is better to dissipate heat in a 3W dropping resistor than to dissipate it in your regulator. If you drop the voltage with a resistor the resistor may get warm but it is a power resistor and can easily handle it and your 5V regulator will stay cool.

Okay, I'll grab the resistor.

http://www.robotshop.com/media/files/pdf/l12-datasheet.pdf I'm using the L12 210 gear ratio 100 mm actuator. Doesn't say start-up current, but the maximum current is under 150 mA. It can certainly start with the wall wart now, but the (single now, if I haven't mentioned it yet) Arduino isn't connected yet.

If the actuator only draws 150 mA and the logic circuit only draws 50mA then total current is about 200mA
If you add another 50mA for good measure then the dropping resistor becomes :

Rdrop = (Vin-Vload)/I load =(12-7)/0.25=20 ohms

and the power rating becomes P= I x E = 0.25A x 5V = 1.25 W (use 2W for safe measure)

Dropping resistor for a 5V drop @ 250mA = 20 ohm / 2 W

My original calculations were based on a higher anticipated current.

Wait, I was powering the linear actuator in parallel with the Arduino since it's 12V. I'm only putting 50 mA into the Arduino itself. I think I need a bigger resistor. Right?

My mistake. I forgot they were in parallel. (oops!)
Thanks for catching that.

The formula is :
Rdrop=(Vin-Vload)/Iload
R = (12V-7V)/0.050A=5V/0.05A=100 ohms.

Yes. Your correct. You need 100 ohms.
Power rating:
P = I x E
PRdropping=IRdropping x Edropped=0.050A x 5V=0.25W = 1/W Watt
You might get away with using a 1/4W 100 ohm but just in case the arduino draws more than 50mA I would use 100 ohm , 1/2W.