5V Relay Module 4-Channel
This is a 5V 4-Channel Relay interface board.
Be able to control various appliances, and other equipments with large current.
It can be controlled directly by Micro-controller (Arduino , 8051, AVR, PIC, DSP, ARM, ARM, MSP430, TTL logic) .
5V 4-Channel Relay interface board, and each one needs 50-60mA Driver Current.
Equiped with high-current relay, AC250V 10A ; DC30V 10A.
Indication LED’s for Relay output status.
I am concerned that it says "each one needs 50-60mA Driver Current". I understand that the Arduino can only output up to 40mA on each pin. Does that mean I can't connect this board directly to the Arduino using the output pins and power and ground to the Arduino or does it mean the circuitry on the relay board handles that? I just want to make sure I don't ruin either board. Also can someone please explain the "Optocoupler High Trigger" function?
Also when it says AC250V 10A - DC30V 10A does that mean I can connect (4) AC120V 10A or (4) DC30V 10A appliances at the same time or is that total voltage and amperage? Why does it say 250V and not 120V or 125V AC?
I am concerned that it says "each one needs 50-60mA Driver Current".
That's the relay coil current and you should be concerned for the reasons you stated.
Does that mean I can't connect this board directly to the Arduino...
You can use it with your Arduino as long as you drive the relay coils directly from a power supply, not from the Arduino I/O pins. I have a similar board. There is an optocoupler driving each relay. You should find a jumper which, when removed, allows you to power the relays from a separate supply so the Arduino only has to drive the optocouplers.
Also can someone please explain the "Optocoupler High Trigger"
This probably is saying that you need to drive the optocoupler pin high to activate the relay which would be backwards from my relay board.
Why does it say 250V and not 120V or 125V AC?
Because this is a maximum voltage rating. You also can use it to control a lower voltage.
I think they've shot themselves in the foot on this one. I highly doubt that it requires so much current to drive them. Looking at the image it appears that they have opto issolators driving them. I'd be surprised if they need any more than 10ma to switch those relays.
Nevertheless, I'd be inclined to steer well clear. Maybe the picture isn't exactly what you'll get.
The relay board and its description seems OK to me as it's about as good as it gets when languages get translated. It appears to be a common relay board - this is from another supplier:
I am concerned that it says "each one needs 50-60mA Driver Current
This is the current required to energize the relay. The opto-isolator's transistor will see this current. The Arduino control current will be 15-20mA through each control input (IN1-IN4).
Recommended connections for full isolation:
Remove the JVCC-VCC jumper and provide a separate 5V power source to JVCC.
Connect the GND from this source to the GND terminal. Do not connect to Arduino GND.
Connect Arduino +5V to the VCC terminal.
IN1-IN4 is connected to desired Arduino outputs.
Also can someone please explain the "Optocoupler High Trigger" function?
That description cannot be found on the link you provided. Perhaps it came from "People who viewed this item also viewed" or "Recommended Products" section of the page.
From the link I've provided with identical board the control is described as follows:
"Input IN1 / IN2 / IN3 / IN4 signal wire low level is effective"
This would mean logic LOW on the Arduino output would energize the relay. Logic HIGH will turn it off.
Also when it says AC250V 10A - DC30V 10A does that mean I can connect (4) AC120V 10A or (4) DC30V 10A appliances at the same time or is that total voltage and amperage? Why does it say 250V and not 120V or 125V AC?
This is the maximum contact ratings of each relay. If the load is AC, then the maximum load the contacts can switch is 250VAC 10A (2500VA). If the load is DC, then the maximum load the contacts can switch is 30VDC 10A (300VA).
These ratings are for each relay. One relay could be switching a DC load while the others are switching AC loads. For AC, the total voltage and amperage is 250VAC, 40A, 10,000VA. This represents an equal load of 250VAC, 10A on each relay.
dlloyd:
This is the current required to energize the relay. The opto-isolator's transistor will see this current. The Arduino control current will be 15-20mA through each control input (IN1-IN4).
I do believe you've hit the nail on the head. But until I saw this explanation, I had jumped to the same conclusion as the OP.
Ok so I am still a little confused. I am very new to this. Are you saying that I need to provide a separate 5v power supply for the relay board and connect that power and ground directly to the relay board but don't connect the ground from the relay board to the ground of the Arduino? So I would hook up the digital output pins from the Arduino directly to the IN1-IN4 pins on the relay board and the 5v of the Arduino to the VCC of the relay board along with the 5V from the separate power supply? Where would I connect the ground of the Arduino to?
If this diagram is correct, Am I correct in assuming that I can have up to 4 devices that don't draw more than 250V 10A all connected to the relay board and I can power them on and off with the Arduino without damaging anything? Sorry for my limited understanding of this concept. I really am trying to understand. It seems that the Internet just confuses me because so many people are saying so many different things.
Thanks again!
Ok so I am still a little confused. I am very new to this. Are you saying that I need to provide a separate 5v power supply for the relay board and connect that power and ground directly to the relay board but don't connect the ground from the relay board to the ground of the Arduino? So I would hook up the digital output pins from the Arduino directly to the IN1-IN4 pins on the relay board and the 5v of the Arduino to the VCC of the relay board along with the 5V from the separate power supply?
The Arduino 5V connects to VCC.
The 5V from the separate power supply connects to JVCC. (jumper removed).
Where would I connect the ground of the Arduino to?
Not used.
IN1-IN4 when LOW, will sink current to ground, illuminating the relay board's LED and the opto-isolator's IRLED.
If this diagram is correct, Am I correct in assuming that I can have up to 4 devices that don't draw more than 250V 10A all connected to the relay board and I can power them on and off with the Arduino without damaging anything?
Yes. Those are the maximum ratings for each relay. Of course, life expectancy of the contacts (switching cycles) will increase if your load is less than this.
The Arduino 5V connects to VCC.
The 5V from the separate power supply connects to JVCC. (jumper removed).
What is the JVCC (jumper removed) ? Is that a separate pin on the relay board?
Also if I connect a 5v wall wart to the Arduino's dc port is that considered a "separate" power supply? If it isn't I'm not sure how to get a 5V external power supply. I have a 6v battery. Would that be too much?
What is the JVCC (jumper removed) ? Is that a separate pin on the relay board?
Why don't you look on your relay board and see if there is a jumper with a label similar to "JVCC". The photo in your link appears to show such a jumper at the right side.
Also if I connect a 5v wall wart to the Arduino's dc port is that considered a "separate" power supply?
You may be able to power your relays from the same power supply that is feeding your Arduino.
You will have to explain what you mean by the term 'dc port' since there are several different ways to power the Arduino but none of the connections are labeled 'dc port' as far as I know.
Have you considered powering your Arduino via the USB lead and using your wall wart for the relays?
Sorry I meant to say the port that is on the Arduino UNO that you can connect a wall wort to. I see the jumper you are referring to. So I just need to remove the jumper and connect a wire from the external power supply to the JVCC pin and the other wire from the external power supply to the ground on the relay board? The other way I was going to do it was to use the Arduino connected to a 5vdc wall wort for the relays. I was just wandering if that is considered an "external" power supply and wouldn't damage the Arduino.
Sorry I meant to say the port that is on the Arduino UNO that you can connect a wall wort to.
That is called PWRIN (Power In) and your Arduino will not work properly (if at all) if you connect your 5 volt wall wart here. It requires 7-12 V.
So I just need to remove the jumper and connect a wire from the external power supply to the JVCC pin ...
Yes
... and the other wire from the external power supply to the ground on the relay board?
No. The other side is is connected to VCC and if you connect it to GND you will be shorting whatever is feeding the regular VCC input.
The other way I was going to do it was to use the Arduino connected to a 5vdc wall wort for the relays.
As i said before the Arduino will not function properly (if at all) if you connect your 5vdc wall wort to "the port that is on the Arduino UNO that you can connect a wall wort to."
If it is a regulated 5v supply and if it's current rating is high enough you could power both the Arduino and the relays from it. You would have to connect it to a +5V pin on the Arduino.
Have you considered answering the question I asked you in my previous post?
Would this board work without the micro-controller simply as a voltage sense, 12VDC max, or lower if need? If I were to tie the sense voltage to the IN1, IN2, ... pins and a separate power supply for the board and the relays, as described, then the relay(s) theoretically would close. Correct?
What the board would be doing is activating 2 hours meters to only run when a specific subsection of test equipment is operating.
I am using a similar board with two relays connected to an Arduino UNO. 5 volt and ground from the Arduino board power the relay board, as has been noted in other posts. The control for each relay goes to digital pins(output) on the UNO board. Works fine with two relays. I see no reason not to be able to use 4 relays.
Design your system so the normal operation of the external devices is for the relays to be not-transfered. It takes about 1 second for the UNO to energize a relay after you power up the board.
Just to give you some idea of the power required by the relay circuit board. I have a 2 relay board and the following are the 5 volt current measurements:
no relays energized= no current
1 relay energized = 52-54 ma.
2 relay energized = 86-87 ma.
The current varies from time-to time. I don't have a good reason for that.
That's total overkill for driving relays, fine f you really need the isolation.
Simple transistor, or high current shift register are all that is needed, as I used on these relay shields, pics on page 2 http://forum.arduino.cc/index.php?topic=276450.15
dlloyd:
... This is the current required to energize the relay. The opto-isolator's transistor will see this current. The Arduino control current will be 15-20mA through each control input (IN1-IN4).
Not true Although I fully agree with the fact that the board requires separate power supply (Arduino power supply would not be able to energize all relays), the optoisolator driving current will be much smaller than 15-20 mA.
Each relay control input, connected to Arduino control output, is active low, that means it must be connected to ground to activate the relay. However, Arduino output voltage at low level is not 0V, Atmel ATmega328P datasheet shows Vol voltage (voltage on output pin set to "0") at sink current below 10mA being approx 0.6 V.
The control current runs from Vcc voltage through 1kohm limiting resistor, through optoisolator LED and through red indicator LED to control input (which is ~ 0.6 V in active low state). Voltage drop on 817 optoisolator LED is approx. 1.2V, voltage drop on red SMD indicator LED (in series with optoisolator) is around 1.8 V. When connected to +5.0 V power supply, voltage on 1kohm limiting resitor will be around 1.4 V (5.0 - 1.2 - 1.8 - 0.6), resulting in 1.5 mA sink current from each control input.
Although I fully agree with the fact that the board requires separate power supply (Arduino power supply would not be able to energize all relays), the optoisolator driving current will be much smaller than 15-20 mA.