Relay board blowing 3.3v pro mini. What am I doing wrong.

I am using 3.3v pro minis to power an opto relay boards. Usually when I connect the battery the pro mini powers up ok, but then the relay turns on there is a click and the pro mini regulator blows. If I replace that pro mini the new one usually works ok. Sometimes I have to replace with a third one.I am wiring everything up the same way every time. When I get a pro mini that works, it will always work from that moment on. I am confused by what is happening.

So my relay board is 5v and 12v coils. I have removed the leds on IN lines because I am using on 3.3v board. 12v battery is connected to JD-Vcc and ground for relay coils. Same 12v battery is connected to raw and ground of pro mini.Pro mini VCC to VCC of relay board. IN1 to pin 2, and IN2 to pin 3. The relays activate when pins are set to LOW.

Does anyone know what I'm doing wrong, or why some boards seem to work great?

Thanks.

Well, the first thing you are doing wrong is posting before reading How to use this forum

Beyond that, with the limited information that you provided [having not read the above]:

  • Your description of it blowing out when the Relay clicks, suggests there is no protection diode across the relay coil. Check out the following: Relays | Electronics Club
  • I've had the problem of clone Pro Minis not handling 12V [where proper Pro Minis handle 12V just fine].
  • Do you really expect us to know what Relay Board you're talking about? Think about that :wink:

Most relay modules operate from 5V, how do you power it from a 3.3V mini?

Hi Sorry,

These are the boards I am using 5V, 12V or 24V 1-2-4-8-16 Channel Relay Module Arduino PI ARM AVR DSP PIC | eBay.

The coil is 12v, and the 5v side works ok on 3.3v. I remove the led and bridge its contacts to prevent voltage drop across led. 3.3v is enough of a high signal for the relay board.

The regulator blowing makes me think its drawing to many ma's, but I thought the control side of these boards didn't draw much current.

Which relay board. There are many on that page.
I guess it's not the 16-channel one, because that doesn't have a JD-VCC jumper (or opto isolation).

A diagram and/or picture is a much quicker way to get things sorted.
Leo..

Seedler:
Hi Sorry,

These are the boards I am using 5V, 12V or 24V 1-2-4-8-16 Channel Relay Module Arduino PI ARM AVR DSP PIC | eBay.

The coil is 12v, and the 5v side works ok on 3.3v. I remove the led and bridge its contacts to prevent voltage drop across led. 3.3v is enough of a high signal for the relay board.

The regulator blowing makes me think its drawing to many ma's, but I thought the control side of these boards didn't draw much current.

If you removed the LED and bridged across the pads, you now have only the voltage dropping resistor across the 3.3 volts. What does that accomplish other than to load you 3.3 volt signal.

Paul

Paul_KD7HB:
If you removed the LED and bridged across the pads, you now have only the voltage dropping resistor across the 3.3 volts. What does that accomplish other than to load you 3.3 volt signal.


I think he means bridging over the green LED, thus removing a 2 V or so drop from the input circuit. You then only have the 1.2 V drop in the optocoupler.

You normally power opto VCC from 5volt (marked Arduino 5volt).
Then combined Vf of the LEDs prevents the opto from working even if you have only 3.3volt on the IN pin.
But if you power a 3.3volt Arduino from 12volt, then you don't have a 5volt source.
Leo..

I can't say for sure what is happening by your description but try this:

1.Get a 7805 regulator to step your 12v down to 5v

  1. Connect your Pro mini through the regulator

Make sure you are using a 3.3v relay like this:

[Make sure you are using a 3.3v relay like this 2 Channel 3.3V/5V 10A Relay Module for Arduino RPi ESP8266 ESP32 5060606740156 | eBay](http://Make sure you are using a 3.3v relay like this 2 Channel 3.3V/5V 10A Relay Module for Arduino RPi ESP8266 ESP32 5060606740156 | eBay)

Also, if you can change over to a slightly more robust chip like a NANO with 5v output capability, try that too. I get "strange things" happening on all arduinos at times when I try to power relays directly from output pins, so I almost always (now) go from output pin, through a 1k resistor to a transistor (like a tip120) to power the load, or, a tip to switch a coil relay for larger loads to supply path to ground and then 5v to the relay directly. I have had shotty luck many many times with those 5 in 1 relay boards whether using opto isolated mode or not - I'm pretty much done with those things and prefer
separate relays these days. Anyway...

You probably want to put decent sized ceramic cap on your load (like motor or solenoid valve etc) to help with switch release spiking. Those relays probably already have protective diodes on them already. If you use separate relays, always put a diode across the input pins (forward biased) for spike protection.

Paul__B:

I think he means bridging over the green LED, thus removing a 2 V or so drop from the input circuit. You then only have the 1.2 V drop in the optocoupler.

Thanks. I remember reading somewhere to do this if using 3.3v on 5v logic. Thanks for the diagram. So the 5v side (3.3v in my case) is only powering green led and led in opto-isolator. That wouldn't be enough current to blow the regulator, would it? How much can these regulators supply?

UsernameD:
I can't say for sure what is happening by your description but try this:

1.Get a 7805 regulator to step your 12v down to 5v

  1. Connect your Pro mini through the regulator

Would I not have to use a 3.3v regulator on 3.3v pro mini? I use the 3.3v pro mini because it uses a lot less power than the 5v version.
The 3.3v relay modules look good, but I need ones with 12v coils.

Is there an alternative regulator I could use with low power and high current?

I'm still confused whats happening though considering the Arduino pins are only driving the led in opto-isolator.

Thanks for your help

We still don't know which relay board you're using (asked in post#4).

If you have a board with 12volt coils, then it should NOT have a VCC jumper.
If it does, and you have the jumper in the wrong place, then the full 12volt is connected to VCC of the Nano.

Opto LED current (sink) is only ~2mA per relay.
Leo..

Seedler:
The 3.3v relay modules look good, but I need ones with 12v coils.

Why does the coil have to be 12V? Can't you use a 5V relay instead? May even use less power overall (quite interesting that you're concerned about the few mA of power use difference between two Arduinos, but not with the much greater power use of the relays).

Seedler:
Does anyone know what I'm doing wrong, or why some boards seem to work great?

Clearly, making a major wiring mistake outside of what you think you are doing. :astonished:

I think you appear to be saying that you are powering the 3.3 V Pro Mini from 12 V using its on-board regulator. That would be a bad idea as the regulator is not adequately heatsinked for 12 V operation, and the situation is even worse than for a 5 V version. Now whether that would in itself "blow" the regulator is a curious question, most regulators are supposed to enter thermal shutdown if overloaded, but - may not be adequately robust and of course, cheap Chinese (eBay) parts are questionable.

As Wawa points out, powering the input side of the relay board - "Vcc" - from 5 V is an excellent way to operate it with a 3.3 V Pro Mini without disabling the green LED as long as the 3.3 V supply is derived from that 5 V supply so it cannot be lost unless the 5 V supply is lost also. Arguably, using a (buck) regulated 5 V supply to feed Vcc on the relay board and "Vin" on the Pro Mini should be quite reliable (presuming the on-board regulator has a dropout voltage less then 1.7).

Wawa:
We still don't know which relay board you're using (asked in post#4).

If you have a board with 12volt coils, then it should NOT have a VCC jumper.
If it does, and you have the jumper in the wrong place, then the full 12volt is connected to VCC of the Nano.

He has explained that it does have that jumper - which is as you would expect since the 12 V version with the jumper works just fine, given that you have to use 12 V logic on the input. He also explains that he has removed the jumper, so there is no problem there.

Now...

Seedler:
So the 5v side (3.3v in my case) is only powering green led and led in opto-isolator. That wouldn't be enough current to blow the regulator, would it? How much can these regulators supply?

With the green LED disabled so you can (ugh!) feed it from 3.3 V, the opto having a 1.2 V drop and a 1k resistor ("102") - but check that! - each opto will draw 2.4 mA. That certainly will not damage the regulator. I'm punting for a wiring error. Or ...

Seedler:
Would I not have to use a 3.3v regulator on 3.3v pro mini?

For the Pro Mini itself, if you insist on running it on 3 V, yes. With the regulator removed, you can run it at 5 V for almost as much efficiency since the current draw is affected more by the lower clock speed than the operating voltage.

Seedler:
I use the 3.3v pro mini because it uses a lot less power than the 5v version.
The 3.3v relay modules look good, but I need ones with 12v coils.

Why?

Different relays will require much the same power to actuate, so it is the case that operating suitably specified relays from a linear regulator is much less efficient than directly from the primary supply voltage.

Seedler:
Is there an alternative regulator I could use with low power and high current?

Tricky. An adequately rated switchmode regulator dropping 12 to either 3.3 or 5 V as explained above should resolve the situation.

Seedler:
I'm still confused what's happening though considering the Arduino pins are only driving the led in opto-isolator.

Short of a recurrent wiring blunder, the problem will be overheating of the on-board regulator.

wvmarle:
Why does the coil have to be 12V? Can't you use a 5V relay instead? May even use less power overall

Well, according to the datasheet, all voltage versions of the relay consume the same power at nominal voltage.

So using the lower voltage relay loses power in the regulator and control transistor - especially if you use a linear regulator.

Ok I found an mcp1702 regulator lying around. TO-92 package, so it was a struggle getting it wired in place of the pro mini sot23-5 regulator. Anyhow I managed to do it and everything now works. So I guess the problem was the cheap ebay clone regulators.

Pity they don't do the MCP1702 in a sot23-5 package. Are there any other packages that may fit easier?

Is there a simpler way to connect TO-92 package to pro mini?

Paul__B thanks. 99% of the time only the Arduino is running, 3.3v pro mini uses a lot less power. Plus I have other 3.3v components connected. It seemed simplier to use 3.3v.

Thanks.

Seedler:
Ok I found an mcp1702 regulator lying around. TO-92 package, so it was a struggle getting it wired in place of the pro mini sot23-5 regulator. Anyhow I managed to do it and everything now works. So I guess the problem was the cheap eBay clone regulators.

OK, so you are going to feed this with 5 V, are you not? Not 12 V and certainly not from a car battery on a charger as that would easily exceed the voltage rating. 12 V alone will likely cause it to overheat unless you fit a decent heat sink. It is however, "guaranteed" (with caveats) to shut down instead of burning out (unless you reverse the supply voltage in which case it will be toast like the others).

Seedler:
Is there a simpler way to connect TO-92 package to pro mini?

Yes: don't solder it to the board.

Take a piece of perfboard, find the appropriate capacitors for this regulator (see data sheet), solder it all up on protoboard, and connect the output to the Vcc pin of the Pro Mini.

Paul__B:
OK, so you are going to feed this with 5 V, are you not? Not 12 V and certainly not from a car battery on a charger as that would easily exceed the voltage rating. 12 V alone will likely cause it to overheat unless you fit a decent heat sink. It is however, "guaranteed" (with caveats) to shut down instead of burning out (unless you reverse the supply voltage in which case it will be toast like the others).

My project uses 11.1v lipos, are you saying that's too much? I don't want to have to go from 11.1 to 5 to 3.3v. The regulator says it is rated upto 13.2v input.

Would something like this be ok? 5x MP2307 Mini DC-DC Buck Converter Step Down Module like LM2596, UK SELLER FAST | eBay

wvmarle:
Yes: don't solder it to the board.

Take a piece of perfboard, find the appropriate capacitors for this regulator (see data sheet), solder it all up on protoboard, and connect the output to the Vcc pin of the Pro Mini.

Doah thanks :slight_smile:

A 3.7V LiPo goes to about 4.2V when charged, so your 3x cell pack reaches 12.6V when charged. That should be OK for the regulator. You do lose about 70% of your input power in the form of heat in that regulator, just 30% is used to power the Arduino. Very wasteful, you want your battery's energy to power the Arduino, not to turn its enclosure into a sauna.

But there's a simpler solution. Place those cells in parallel, giving you a 3.7V output (nominal; real output will be 3.2-4.2V). Then you don't need regulators AT ALL. Find a relay with a suitable coil for that voltage, and connect the Arduino to the battery through the Vcc pin. The processor can handle it for sure; check the other components (mostly tantalum capacitors though I expect those will be rated 6.3V or higher).

For big improvements of your battery life look for latching relays, those take a short pulse to switch, then hold their position. Regular relays need to have the coil powered at all times, easily drawing 200-300 mA.