MOSFET IRLZ44N to drive a industrial relay

Hey,

I am experiencing some problems controlling an industrial relay with my Arduino, although I have done a lot of research. When I power the system, the 68 Ohm resistor gets so hot i can barely hold a finger at it. First I thought I was sending way to much power trough it, so I brought a larger one that can handle 2 watts, but its still getting hot. In my calculations it should be about 1.1w going trough.

The circuit is shown in the attached photo.
Thanks in advance.

WHY on earth do you have the 68 ohm resistor between MOSFET and GND? It makes absolutely no sense.

Also, your contactor is connected between GND and MOSFET switching to ground - it will never receive power.

Right now, the MOSFET is dumping power into the resistor, the contactor gets no voltage at all.

Rethink your schematic - it's all wrong.

// Per.

I have used this schematic before just with less current, working totally fine.
Something is obviously wrong, but the relay/contactor is receiving power. Only when the relay is off the resistor is getting hot.

The 68 ohm resistor is between MOSFET and 10VDC.

Zapro: WHY on earth do you have the 68 ohm resistor between MOSFET and GND?

I read it that the 68R is across the relay, which is to all intents and purposes a short of 10V to ground through the closed mosfet.

Did you mean the 68R to be a flyback diode? You do need one, and it would go where that resistor is.

This is my understanding of OP’s circuit.

20180124_150315.jpg

Sumitsubo: I have used this schematic before just with less current, working totally fine. Something is obviously wrong, but the relay/contactor is receiving power. Only when the relay is off the resistor is getting hot.

The 68 ohm resistor is between MOSFET and 10VDC.

No, it is not working fine.

First off, you are trying to power the contactor trough the Arduino's diode on the input - a big NO-NO. Power it directly.

Second, your contactor is connected between GND and GND - it will NEVER receive power.

Thirsd, your 68 ohm resistor IS connected from VCC to GND trough the MOSFET - it WILL be hot - There is NO reason to have this 68 ohm resistor there.

Please post a picture of how you ACTUALLY connected it - your schematic is NOT how you did it, if your contactor receives power at any point in time.

// Per.

Here is some photos of the project. Link to dropbox (photos too large to post here)

https://www.dropbox.com/sh/zjjemywwr3gtgza/AACa2n0zkja0mnn6g7GjxXlxa?dl=0

Sumitsubo: Here is some photos

Why don't you post a schematic of the circuit? Is my interpretation as in #4 correct?

If so, what's the 68R across the relay for. Lose it and stick a reverse biased diode in as flyback protection.

Your circuit should look like this. No resistor across the relay.

Such a resistor could possibly make sense because it might cause faster switchoff than a diode. In that case, however, it should be done by someone who remotely knows what he is doing and it should be appropriately dimensioned.

ElCaron: Such a resistor could possibly make sense because it might cause faster switchoff than a diode. In that case, however, it should be done by someone who remotely knows what he is doing and it should be appropriately dimensioned.

Faster switchoff? What on earth are you talking about man ... There is no way in hell to get a mechanical relay to switch off any faster, other than to remove the power to it.

// Per.

Zapro: Faster switchoff? What on earth are you talking about man ... There is no way in hell to get a mechanical relay to switch off any faster, other than to remove the power to it.

Of course you need to remove the power. But the the collapsing field of the inductor will keep the current going. The energy stored in the inductor needs to be dissipated somewhere. Initially, right after the power is cut off, the current will be the same as before, but going round in the loop with the flyback diode. The diode dissipates energy according to forward voltage*current. A resistor in place of the diode can drop potentially drop more voltage, thus dissipate more energy, thus collapse the magnetic field of the inductor faster, potentially leading to a faster release of the relay. The obvious disadvantage is that if not combined with a diode, the resistor will dissipate energy during the whole switch on time of the relay, so the resistor needs to be able to take that energy and the supply needs to be able to supply the additional current. It is certainly not advised in everyday situations. Also, the bigger the resistor is, the higher the voltage created by the inductor will be, so you need to be careful with the breakdown voltage of the mosfet. Here is an interesting stackoverflow article about different methods to deal with the collapsing field energy.

Now, don't ask me when it would be sane not to add a diode to the resistor, but I think I have heard of such applications, they are certainly possible and in principle do the job. It is certainly a very special case. Or maybe I am not remembering correcly, and there was always an additional diode that was just left out in some discussion about the field breakdown because it made a minor contribution.

Zapro: Faster switchoff? What on earth are you talking about man ... There is no way in hell to get a mechanical relay to switch off any faster, other than to remove the power to it.

Yes, a resistor (or nothing at all) makes a relay turn off faster than a kickback diode alone. The value has to be choosen wisely, so the now higher kickback voltage stays below the safe drain voltage of the fet. IIRC about twice the supply voltage is generated if the resistor is the same value as the resistance of the coil. A diode in series with that resistor (cathode/ring towards the supply) stops the resistor using power (heating up) when the relay is active. Leo..

What is this industrial relay? How much voltage/current does the coil require?

The original attachment has the 68R in SERIES with the relay, actually it is acting as a current limiting resistor because the relay is hooked up kind of wrong. With the mosfet OFF, the relay coil pulls all it's current through the 68R. With the mosfet conducting, it drops the voltage to the relay coil, and the 68R carries the current through the mosfet AND relay.

Best I can say is to get the relay coil in series between Vss and mosfet drain, with the source directly to ground. Lose the 68R, and go with a diode in PARALLEL with the coil. The gate resistors are fine.

Thanks to all who have contributed with their knowledge on this thread!

The relay-coil is rated at 12VDC 500mW, it requires 8VDC at minimum to turn active.

I have a diode module attached to the relay (picture as attachment).

Is the rewired circuit and the diode attachment, okay now?

No!!

Do it like post 8.

Allan

Yes, circuit is ok now. That relay seems to have a buildin snubber circuit (varistor), and an indicator LED. No external diode across the relay coil needed. Leo..

Hi,
You need to learn to draw schematics of your circuits, not use fritzy pictures.
A schematic like this is easier to read and analyse.
Also your relay needs to be connected A1 to positive and A2 to negative as in this schematic.
Mega_Relay.jpg

gl.jpg
Tom… :slight_smile:

TomGeorge: Hi, You need to learn to draw schematics of your circuits, not use fritzy pictures. A schematic like this is easier to read and analyse. Also your relay needs to be connected A1 to positive and A2 to negative as in this schematic.

Hi Tom.

Totally agree with you, this schematic is better than mine on Fritzy. Thanks! I did not get the Fritzy schematic mode to turn out so well... What program do you use?

//Mats

Hi,
I use a simple CAD called ExpressPCB.
It is a simple package, freeware and does not come with any advertisments or conditions.

google ExpressPCB download

Tom… :slight_smile: