It isn't a 'slider'. It's a "Read More" link .
All I did was click on it. (to reveal the specs)
It is the frame which you use the slider to read in the original page. Opened out especially for you!
So would you recommend I use 3V?
No, I would recommend finding the minimum voltage that gives you enough force in the stroke. Only you can do this because only you have your mechanism. Use a bench supply or a variable buck converter to find this.
Forget trying to find a battery until you find this voltage.
Ok, I found how to open it.
Grumpy_Mike:
No, I would recommend finding the minimum voltage that gives you enough force in the stroke. Only you can do this because only you have your mechanism. Use a bench supply or a variable buck converter to find this.
Sorry for the delay but I ordered a variable buck converter and it arrived today. The minimum voltage I found was 1.75V. So should I just find a battery that is 1.75V and can output 2.5 Amps (5(1.75V / 3.5 Ohms))?
samlbaker:
I am using the other side of the solenoid (side without the spring) for push functionality rather than pull.I do not need very much strength as I am using them for a ratchet mechanism and they just need to reach the 3mm length in order to push in a very weak pawl.
Time is much more important for my application as they may need to stay powered up to a minute.
So would you recommend I use 3V?
Sam
This is where you get to experiment!!!! You do not know how much current is required to move the project to the end position, not how much current is necessary to "hold" it there for the required time.
Connect your solenoid project up and use 3 volts and see if it works and measure the current being drawn. Next add small value resistors in series with the 3 volts and try again. At some point it won't work properly and then you know how much current (power) is necessary for your project.
If 3 volts never works, use 5 volts and repeat, etc.
Paul
samlbaker:
The minimum voltage I found was 1.75V. So should I just find a battery that is 1.75V and can output 2.5 Amps (5(1.75V / 3.5 Ohms))?
Basically yes, but the voltage of a battery depends on the chemistry used in its construction, and remember that 1.75 is the minimum that will work.
The voltage on most batteries declines as its charge gets exhausted so this 1.75 is what should be available at the end of battery life.
I ordered a variable buck converter and it arrived today
What is wrong with using that with say three AA batteries and the buck converter to get the voltage you need and then use the voltage from the three AAs to connect directly to the 5V pin to power the 5V Arduino.
samlbaker:
So should I just find a battery that is 1.75V and can output 2.5 Amps (5(1.75V / 3.5 Ohms))?
Sorry no;
Current = Volts/Resistance
Current = 1.75 / 3.5 = 0.5 A is the current you will need to supply.
You will not be able to find a 1.75V battery and as @Grumpy_Mike in Post #26 you will need to take into account that the battery. which will have a slowly dropping output voltage.
You will be better to try as @Paul_KD7HB, in Post #25, has suggested and use some series resistors to limit the current from a higher and more readily available battery voltage.
Tom... 
What I see is the problem with using resistors to limit the current is you are just burning off the extra power. When designing a battery powered device you want to maximise the usage of your charge which is why I would recommend using a buck converter.
TomGeorge:
Current = 1.75 / 3.5 = 0.5 A is the current you will need to supply.
Yes but since I am powering 5 solenoids, wouldn't the total current draw increase 5 times as well? I tested it with my variable buck converter and found when all 5 are powered they draw a little under 2.5A.
Grumpy_Mike:
What is wrong with using that with say three AA batteries and the buck converter to get the voltage you need and then use the voltage from the three AAs to connect directly to the 5V pin to power the 5V Arduino.
To clarify I am only using the buck converter to test the minimum voltage of the solenoids. I ordered a cheap one with an LCD so that I can change the voltage and see the current draw.
Although I most likely will end up using a different buck converter as Mike said since I can get constant voltage, less power loss, and increase the maximum current output. < Correct me if I am wrong about that. If I go that route and say I want to provide 2V output from the converter, what type of battery do you recommend I use?
Also, something to add, when powering a single solenoid 1.75V is sufficient, but as soon as I add another 2.5V is needed, but then it doesn't need an increase when adding the other 3 solenoids. Just something I found interesting.
something to add, when powering a single solenoid 1.75V is sufficient, but as soon as I add another 2.5V is needed, but then it doesn't need an increase when adding the other 3 solenoids.
It might’ve something to do with the way the buck converter handles an inductive load. Try adding more large capacitors across the output of the buck converter.
samlbaker:
Also, something to add, when powering a single solenoid 1.75V is sufficient, but as soon as I add another 2.5V is needed, but then it doesn't need an increase when adding the other 3 solenoids. Just something I found interesting.
I think I found the culprit. When connecting more than one solenoid I was using a breadboard. I think that since the resistance of the solenoids is already so low, the resistance of the breadboard was enough to decrease the current, causing my need to increase the voltage.
2.5A/0.5 =0.5A (EXACTLY 5 TIMES (1.75/3.5)
Each solenoid draws 0.5A so of course 5 solenoids
draw 2.5A. It has nothing to do with the resistance
of the breadboard.
FYI,
5*1.75=8.75=~9Vdc
If you wire them in series they all draw the same
current and you can use a 12V battery with a 2.5A
dc to dc converter adjusted for 8.75 to 9.0V.
Of course this only works if you don't need to
power them individually (not all on at some time)