Reverse Voltage Issue

[raschemmel]

I must add diodes in parallel to every E.Magnet connected to that relay module (Yes, in the  
          overall setup I have more than one) and that will guarantee that any reverse voltage coming  
          back to the relay and from there to the Arduino is eliminated.
  

The diodes suppress back-emf. What guarantees that it won't get to the arduino is the fact that the magnet is running off a separate supply and there is no GND return from the magnet supply to the arduino supply because the GNDs are not common.


BTW/FYI,
Someone commented that the relay board might have a 5V regulator on board. I don't see one. Do you ?



Also, there's no JD-VCC jumper either.

[Miro80]

raschemmel, a small question:

I forgot to mention that the Arduino is always connected to a PC through the USB jack - I need it for the serial monitor.

If I'll power it with a 5V P.S, won't the Mega ignore the external P.S and will draw it's power from the PC instead ?

[dlloyd]

A. I do NOT need to connect the grounds of the 12V P.S which drives the loads connected to the relay module with the 5V P.S which drives the relay module itself.

True.

B. I must add diodes in parallel to every E.Magnet connected to that relay module (Yes, in the overall setup I have more than one) and that will guarantee that any reverse voltage coming back to the relay and from there to the Arduino is eliminated.

No, just a single diode connected in parallel across the coil is required. The 1N5819 is rated for 1A and 40V reverse voltage.

Additional question regarding the free-wheel Diode to use, can you please advise regarding the 1N5819, will it be sufficient ?

Yes

According to the specs of the E. Magnets I use, they are all 12V and consumes between 100mA-430mA so the 1N5819 should handle the EMF isn't it ?

Yes ... make sure that it's reverse connected. If the coil is low-side switched, the cathode of the diode goes to the switched side, anode to the +'ve side of the coil.

[raschemmel]

I forgot to mention that the Arduino is always connected to a PC through the USB jack - I need it for the serial monitor.

If I'll power it with a 5V P.S, won't the Mega ignore the external P.S and will draw it's power from the PC instead ?

NO.
Why ?
Look at the schematic

What is USB_5V - 5v_SUPPLY_5V ?

If the USB supplies 5V and the power supply supplies 5V, the difference of potential between the two will be very small, (millivolts).


It probably will not be zero but will be small enough that you will not be able to measure any effect to the 5V power on the 5V pin.

Check it yourself.

If there is a voltage adjust pot on the 5V frame supply set it to whatever the USB 5V is.

There are other options like a dropping resistor but let's not get into that yet.

[Wawa]

Someone commented that the relay board might have a 5V regulator on board. I don't see one. Do you ?

In the top right hand corner of the board is room for three caps and a 5-pin buck chip.
Those parts are missing on this picture.
Many, boards like that on ebay do have these parts.
But ebay sellers are known for posting different pictures.
That's why I asked for a picture from OP.

5volt on the 5volt pin of the Mega, and connected to the PC, will backfeed 5volt into the PC.
The Mega has a "disconnect mosfet", but that works from >=7.3volt on the DC socket.
Leo..

[raschemmel]

5volt on the 5volt pin of the Mega, and connected to the PC, will backfeed 5volt into the PC.

The PC USB  voltage is 5V . 5V - 5V = 0 V. (no difference of potential)

The USB is fused at 500 mA (0.05 A)

If the difference between the two(USB=5.00v, Ext. P.S. = 5.025V)   is 0.025 V,

then P = I * V = 0.5 * 0.025 =0.0125 (12.5 mW)

Are you going to worry about 12/5 mW ?

It is very easy to connect a DMM in current mode between the USB 5V and the external P.S. 5V and measure the current between the two. I am going to do that simple test at home but I think it will be very small.

[Wawa]

All true, but you assume that the PC is always on.

I didn't say it was wrong. I only stated the facts.
Leo..

[raschemmel]

I can't recall turning off the PC with the USB cable still connected but anything's possible.

Test Results

A-UNO USB Voltage: 5.26 V
B-External P.S. voltage: 5.16 V
[EDIT]  (just realized meter was on mA scale so units is mA, not A)

Current between  A and B = 0.270 mA - 0.300 mA ( for 0.1 V difference)

Current < 1 mA

[Miro80]

What would you say about this:
I understand that the whole issue of me feeding the Mega through the DC Jack with 12V is that it causes the regulator to heat, basically it's the only disadvantage....
So what if I can find a P.S of 7-9 Volts and still feed the Mega through the "regulated" input.
The regulator shouldn't heat that much and I won't worry about any backfeeds to the PC ?

[raschemmel]

So what if I can find a P.S of 7-9 Volts and still feed the Mega through the "regulated" input.
The regulator shouldn't heat that much and I won't worry about any backfeeds to the PC ?

That would be better, yes. I want to emphasize that "worry" is not something we do here.
We TEST and VERIFY. WE DO NOT WORRY. This is ELECTRONICS, not ELECTRONICS+PSYCHOLOGY.

Forget about "worry" and put your DMM in current mode and connect it BETWEEN the 5V pin of the arudino with the USB cable connected and the 5V  pin of the 5v supply (with all  GNDS connected together) and measure the current (on the mA scale) and post a photo of the meter showing what scale it is on and the value shown in the display.
Measure the voltage of the USB 5V and the extednal stadalone 5V P.S. before connecting them together..

Then ask your question again with the TEST results in your post.

We only deal with facts here. If you worry that the earth is going to be hit by a giant asteroid there's nothing we can do about that but if you measure the current between the two 5V supplies and it is very small and you still worry when we tell you there is nothing to worry about then what can we do ?

would it be "better" to use the 7 to 9V supply ?  Yes.
Would anything bad happen if you powered the Mega from the 5V pin of the 5V frame supply ?
Let's wait to see your test results.

What Wawa said about the PC being ON when doing the above is correct. If you did use the frame supply connected to 5V pin with USB cable connected you would need to make sure you DID NOT turn OFF the PC. If working with such conditions makes you nervous then maybe the 7V supply is a better solution for you. The USB has a resetable polyfuse that opens if the current exceeds 500 mA but I measured < 1 mA backfeed current when I connected a 5.16 walwart to the 5V pin of an UNO that had a USB 5V voltage of 5.26 V (5.26-5.16= 0.100 V difference).
Do you see the comparison there ? 9 < 1 ma <=> 500 mA)

If you want to learn electronics you need to learn how to verify things "empirically". What that means is that when engineers read datasheets or product info that claims some specification, they design tests to verify the truth of the product manufacturer's claims, and they generate a specification document that contains product specs vs actual specs obtained by testing. They conclude the product claims are either the same as the test results or the product is "underrated" (claims less than it delivers) or the product is "overrated" (claims more than it delivers). At the end of the day, they know what's what. That's what you need to learn. We can tell you anything but you are responsible for what you do, not us. It is your responsibility to verify what we tell you if you are capable of doing so.

At some point you should reach a point where anyone can tell you anything and you can figure out what's what.

The phrase "Knowledge is power" comes to mind.
"Knowledge about power is also good to have".