Mod edit,
This topic, from this question down to reply #6 is the same as the follow on topic from reply #10. @awais110565 and @kusspuss100 are the same person.
End of mod edit.
I have controlled two actuators with 1 L298N driver control , howeever i am confused how should i control 4 actuators with 2 L298N driver.. I can do programming , problem is wriring
Please need guide
Most linear actuators require more current than the L298N can handle. Post a link to the product page for the actuators, so forum members can see whether your project is even possible.
Please post a diagram of how you have the 2 actuators wired up to the one L298N.
If 2 are working off one L298N, then 4 should be able to work off two.
Yes sure will upload tomorrow
i am trying to extend and retract actuators at same time but one is moving other actuator is not moving
please guide
const int ENA= 12; // Assigning pins to channel A in L298N motor control driver (Actuator 1)
const int IN1= 11;
const int IN2= 10;
const int ENB=9;
const int IN3=8;
const int IN4=7;
void setup()
{
// Assigning output Pins in L298N driver of two channels ENA and ENB
pinMode(ENA,OUTPUT);
pinMode(IN1,OUTPUT);
pinMode(IN2,OUTPUT);
pinMode(ENB,OUTPUT);
pinMode(IN3,OUTPUT);
pinMode(IN4,OUTPUT);
digitalWrite(ENA,HIGH);
digitalWrite(ENB,HIGH);
}
void loop()
{
// The following code in void loop will run untill external voltage stopped( both actuators extend and retracts within specific time )
// Actuator 1 will extend with in specific time
// For actutaors to extend or retarct we should assign High or Low in digitalwrite(pin,high or low)
digitalWrite(IN1,HIGH);
digitalWrite(IN2,LOW);
delay(18000);
//Actuator 2 will extend
digitalWrite(IN3,HIGH);
digitalWrite(IN4,LOW);
delay(18000);
// stopping linear actuator 1
digitalWrite(IN1,LOW);
digitalWrite(IN2,LOW);
delay(19000);
////stopping linear actuator 2
digitalWrite(IN3,LOW);
digitalWrite(IN4,LOW);
delay(19000);
// Actuator 1 will retract
digitalWrite(IN1,LOW);
digitalWrite(IN2,HIGH);
delay(18000);
// Actuator 2 will retract
digitalWrite(IN3,LOW);
digitalWrite(IN4,HIGH);
delay(18000);
//stopping linear actuator 1
digitalWrite(IN1,HIGH);
digitalWrite(IN2,HIGH);
delay(19000);
//stopping linear actuator 2
digitalWrite(IN3,HIGH);
digitalWrite(IN4,HIGH);
delay(19000);
}
thanks in advance
Have you checked that this outdated driver is sufficient for your actuators?
Please give links to the actuator data sheets.
Which actuator power supply do you use?
i am not sure how to find this that this driver support two actuators or not , however i have tried with one actutsors its working perfectlt but when two actutaors they are not moving parallel .for supply iused adapter and give power at 12V , however the maximum supply of adapter is 24V pics below
Please learn to use code tags when posting code on the forum; it's described in How to get the best out of this forum.
Next apply to your post.
thank you for guide about forum , done
The motor pics suggest 24V operation. What's the actuator resistance?
the link of datasheet is
(https://f.hubspotusercontent40.net/hubfs/7717445/PDFs/Actuator%20datasheets/PA-14P%20datasheet.pdf)
actually i am not sure actuator resistance , there is not such tghing written on actuator
Don't worry, the data sheet rates the operating current of 0.5 to 2.5A @ 24V.
So your PSU and driver should support up to 2x2.5A @ 24V.
so can i connect two actuators with one L298N driver
That first 'delay(18000)' means that the second actuator isn't operated until 18 seconds after the first actuator.
i see aha , please can you suggest how to do this parallel
Isn't it a fine time to start learning how to read data sheets?
In the L298N data sheet you'll find 2.5A output current (each channel) for 10ms or 80% ON time (PWM controlled)
or 2A permanent.
Now the advanced stuff:
The driver typically consumes 5V @ 2.5A, or 2.55V@ 1A,
Ohms law states R=U/I so that the resistance of your actuator should be 24V/2.5A or about 10 Ohm. You can verify this calculation by measuring the actuator resistance with your DMM.
As I=U/R and the driver reduced voltage this results in (24V-5V)/10 or about 2A peak current. This means that you are at but don't exceed the driver limit when powering the actuators by 24V. A reduction to 20V will reduce the power (heat) again, at the cost of linear actuator speed. In any case a sufficiently big heat sink should be attached.
A reduction to 12V, as you state, reduced by some V by the driver itself, may prevent the actuators from moving at all. Swap the actuators and watch which one or which driver channel won't work as expected. Also measure the output voltage which should reside in the expected area of Vcc minus a few driver volts. If nothing helps then increase the motor supply voltage, but only after mounting the required heat sinks.
Don't worry if you killed the driver module already or in your following experiments. Simply get a modern driver module with MOSFET transistors and consequently almost no heat generated in the driver chip.
start both at the same time.
i already killed two drivers with experiment , i will supply 12V constant to see what will happen ,
