# Picking Capacitor Size and datasheet clarity

Hey guys,

We are doing a project with 5 solenoids rated at 5V. The datasheet for the solenoid isn't in English and googling the part number results in attaining the same data sheet. From what we understand, for fast movement we require 700mA.. The datasheet is here: http://www.robotshop.com/content/PDF/datasheet-rob-11015.pdf

And there will be 2 fans that are 5V with 200mA.

So, 700mA*5 + 400mA = 3.9A

The arduino can only provide 650mA, thus we would like to add a capacitor to raise the 650mA to aprox. 3.9A. The capacitor will not be part of the circuit unless all of these conditions are true using a solid state relay with the arduino if that would be a problem.

Thank you for your help guys We appreciate it

The arduino can only provide 650mA, thus we would like to add a capacitor to raise the 650mA to aprox. 3.9A

That is not going to work, the capacitor will be too big and take a long time to charge. You need a proper external supply of 4 A or more.

Okay, that sounds good.

There is this 5V 4A AC DC power supply I can connect to the arduino which will give off 4A in the 5V output pin. Would having this too many amps in the circuit? Or too many amps going into the solenoids creating too much wattage? Or would it work out?

Thank you for the quick response Mike

The problem is, you cannot run the 4 amps through the arduino. Not even just from the input jack to the 5V output pin.

You need to supply that current to the hungry devices somehow bypassing the arduino. Which is inconvenient.

Note also, that device you linked to has a 2.5 mm jack and I think the arduino is 2.1 mm

There is this 5V 4A AC DC power supply I can connect to the arduino which will give off 4A in the 5V output pin. Would having this too many amps in the circuit? Or too many amps going into the solenoids creating too much wattage? Or would it work out?

The power supply does not "push out" 4 Amps. It puts-out a constant 5V, and the current draw depends on the resistance/impedance of whatever is connected to the power supply. (Resistance means resistance to current-flow.)

Most of the stuff we work with is considered "constant voltage". For example, the electrical outlets in your house put-out a constant 120 or 220V (approximately, depending on where you live). If nothing is connected, the voltage is still present, but no current flows. A toaster "pulls" more current, consumes more power, and generates more heat than a small lamp, If you try to get too much current out of the wall (or from a battery or power supply), something will "give". The voltage will drop, something will "fry", or you'll blow a fuse or something.

Car batteries are rated 12V at several hundred amps. But if you touch a car battery, only a few microamps flow through your body, because of your body's high resistance and you don't die. However, the spark plug voltage or household voltage can be dangerous!

The relationship between voltage, current, and resistance is determined by [u]Ohm's Law[/u]. (Current + Voltage/Resistance, or Amps = Volts/Ohms).

Power (Watts) is calculated as Voltage x Current. With algebra and Ohm's Law, you can get P = Voltage squared/Resistance or P = Current squared x resistance.

You need to supply that current to the hungry devices somehow bypassing the arduino. Which is inconvenient.

It might be but that is the only way to do it. You need transistors or FETs to convert the small signal from the arduino into the large signal to drive the fan or solinoide. You need to cut the end off the power supply and feed it into the 5V pin. Then you can take the power straight off that 5V pin without the current passing through the arduino.

akashroy: Hey guys,

We are doing a project with 5 solenoids rated at 5V. The datasheet for the solenoid isn't in English and googling the part number results in attaining the same data sheet. From what we understand, for fast movement we require 700mA.. The datasheet is here: http://www.robotshop.com/content/PDF/datasheet-rob-11015.pdf

And there will be 2 fans that are 5V with 200mA.

So, 700mA*5 + 400mA = 3.9A

The arduino can only provide 650mA, thus we would like to add a capacitor to raise the 650mA to aprox. 3.9A. The capacitor will not be part of the circuit unless all of these conditions are true using a solid state relay with the arduino if that would be a problem.

Thank you for your help guys We appreciate it

Not trying to be sarcastic, but you need to invest a little time at the library and get a basic electronics book and understand the consequences of your actions. It's far more costly in terms of time and money to experiment without some more basic knowledge.