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I have an Arduino Uno. I need to use an electric linear actuator to control the movement of a joint in which I have installed a rotary encoder to know in what angles it's moving.

The actuator is a 12 Volt DC and has only two wires to control it, no libraries or anything. I have a 'double relay' which I think is useful to make the actuator run in both directions, but I'm not sure if I'm going to fry it or not. I also need to connect the arduino and the actuator to batteries or an outlet and I really don't know how NOT to fry them and this is my real problem right now since I don't know what kind of power source to use.

The encoder is to prevent the actuator to move between some angles and to measure the rotation speed, and the movement of the actuator is supposed to be controlled by two buttons, one to go forward while is being pressed, and the other to move backwards.

To picture the design a little better, imagine a leg in which the actuator is in the thigh and it pushes the calf in order to move the leg. The encoder is located inside the knee joint.

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  • \$\begingroup\$ You bought an actuator without a detailed datasheet? \$\endgroup\$ – Spehro Pefhany Nov 24 '17 at 22:40
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    \$\begingroup\$ Any serial numbers or just go back to the supplier - or was it off the back of a lorry , no receipt or guarantee... \$\endgroup\$ – Solar Mike Nov 24 '17 at 22:45
  • \$\begingroup\$ start with a 1.5V battery across the wires. see if you get movement. if you get some, reverse battery and see what happens. increase to 2 batteries and so on ... use AA batteries at start so there is not a huge amount of current available \$\endgroup\$ – jsotola Nov 24 '17 at 23:00
  • \$\begingroup\$ It's a 12V linear electric actuator, it runs 10cm with 1200Newton at 8mm per second. That's all I know. \$\endgroup\$ – norman123123 Nov 25 '17 at 14:37
  • \$\begingroup\$ I've done some research and I found a couple actuators that have the same specs as mine and they work on 6 amps aprox (I think) and say that they have Current: 1 [Amp Draw] @ no load. \$\endgroup\$ – norman123123 Nov 25 '17 at 15:33
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If you know for sure that the actuator is intended to be operated from 12 volts DC, simply apply 12 volts DC and measure the resulting current.

The actuator will attempt to draw its rated current - if this happens to be more than your power supply can produce, the supply may reduce the voltage, blow a fuse or otherwise attempt to protect itself.

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Without details like what actuator you bought or a datasheet it is a little difficult to give you exact information and guarantee its correctness. However we can do a bit of educated guessing.

If there are only 2 wires going to the actuator, note not 2 control wires but 2 wires total then we have to assume thats for power and ground. Power them one direction and the actuator moves power them another and it moves the opposite direction. We can determine the maximum current the actuator can consume by measuring the resistance between these 2 leads. 12volts/resisitance(in ohms) will give you the maximum current draw. So if you measure 1 ohm then the current draw is 1 amp for 12 volts.

This is way to much for the arduino to handle. If I were setting this circuit up I would use whats called an H bridge. However using relays might be a bit easier as a beginner. You will need 4 relays that way you can turn either wire positive or negative. Depending on the relay you probably will not be able to trigger it with the arduino directly... Hook one side of the coil to your 12 volt power supply and wire a logic level n channel mosfet up to the the other. Put a resistor between the gate of the mosfet and your microcontroller pin. That will allow you to turn the relay on and off with your arduino

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  • \$\begingroup\$ It's a 12V linear electric actuator, it runs 10cm with 1200Newton at 8mm per second. That's all I know. I've done some research and I found a couple actuators that have the same specs as mine and they work on 6 amps aprox (I think) and say that they have Current: 1 [Amp Draw] @ no load. I'm not sure if that makes it more complicated to control or not or how to supply that amperage. \$\endgroup\$ – norman123123 Nov 25 '17 at 15:35

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