Thank you.You don’t show the connections to the coil. They’re important!
At a minimum, you need to have a diode across the relay coil connections. The anode should be on the negative or ground connection of the coil.
Do you have such a diode? Post all the connections to the relay, it’s coil and the Mega!
Bill, the Mega 2560 is an Arduino board with greater capabilities than an Arduino Uno. For example, it has much more flash memory, 54 GPIO pins, double the number of analog pins, more PWM pins and 4 hardware serial ports.I am presuming that The MEGA 2560 is some sort of microcontroller.
Thank you.You need a diode across the relay coil as I said.
Also, unless RL2 is a low current, 5V relay, you need a transistor or MOSFET to drive that relay.
RL1 will do nothing if it’s really wired as in that diagram. And I’m not sure that R1 and C1 do anything either. Why do you have two relays?
Thank you.The real question is what is the TS intending to have happen? Diode suppression of the inductive spike when a coil switches off is only one method, and it has the very serious consequence of causing problems if the diode is backward, or if the diode failures in a short circuit mode.A resistor, capacitor, or pilot lamp will also work, with the pilot lamp being the most useful choice. And no, the second circuit will not function because of no power source to drive that second relay. And that second relay does not need a coil diode because the control contacts can handle the spike very well. But, once again, what is the TS intending to have happen? I am presuming that The MEGA 2560 is some sort of microcontroller.
The recent schematic is improved.Thank you.
1. Sorry about the drawing missed a power for RL1;
2. added diode1 and diode2 to the coil of the relays;
3. The RL2 is a 5 V low current relay, I added a transistor there;
4. R1 and C1 works as arc extinguishing, I heard the contact will generate arc?
5. The purpose to used two relays is try to isolate the electronic interference from RL1 which pass large current.
Thanks
View attachment 180005
OK, and thanks for the clarification. I figured that it had to be some kind of microcontroller.Bill, the Mega 2560 is an Arduino board with greater capabilities than an Arduino Uno. For example, it has much more flash memory, 54 GPIO pins, double the number of analog pins, more PWM pins and 4 hardware serial ports.
thank you.OK, and thanks for the clarification. I figured that it had to be some kind of microcontroller.
For total immunity to relay and load noise you need an isolated opto-device and a separate transistor to control the relay. If a FET device with the internal diodes is used then even a coil shunt device would not be needed.
Thank you.The TS just stated that he was attempting to drive a high current 24V motor. We can assume that the motor does not need
to be reversed. We don’t know what kind of motor nor specifically how much current it needs? @LAOADAM , how much current are you switching?
I’ve driven multiple servos requiring a total of 8A with a Mega. I’ve also driven 5A of LEDs with a Uno. So it isn’t difficult.
Short answer is YES, BUT you will still be smart to have an opto-isolator because of the size of that relay. I had not realized that you were controlling such a big motor. And the mosfet you use will need to have adequate current and voltage ratings for the relay coil. The big deal with the motor is that the short term starting current will be quite high, possibly more than the 30 amps, but only for a few milliseconds. So it will be hard on relay contacts, meaning that you need a relay rated for starting that size of DC motor. Those are certainly available, but not everywhere. You may need to visit an auto parts store to find one locally. The circuit shown in post # 12 should work quite well if the ratings for that mosfet are adequate for the relay you use.
The general rule is a breaker is sized to 1.5x the normal running peak current, this is to prevent nuisance tripping.Thank you.
The motor is 24VDC Mower motor, the original circuit breaker is 30 A, that's why I think the largest current can be 30 A.
Unless they are products rated as being intended for motor starting applications.The general rule is a breaker is sized to 1.5x the normal running peak current, this is to prevent nuisance tripping.
Max.
In almost all instances I provide motor overload protection separately from the short circuit protection. Thus the code is satisfied while still allowing the motor to be adequately protected. There are two different goals and thus two different systems. My observation has been that "When one size fits all, it does not fit any of them very well." That is almost always correct, and yet never acknowledged by those who set the rules that we are forced to follow.The NFPA70 rule is that overloads for a motor circuit shall be between 115% to 140% FLA, depending on the motor rating, most installations I have been involved in has used the 140% - 150% rule.
Max.