The LEDs are high current 1.4A at 3V each (max) and aI am regulating their intensity using the Arduino.
Well, everything was working fine with 1 LED. Power supply was regulated to 3V instead on 12, but other than that is the same.
Then I tried 2 LEDs with 6V and it was working fine as well.
When I went on to use the 4 LEDs the Mosfet became hot so I swiched it off and checked if it was still working with 2LEDs. But it isn't, it is not regulating as I have 6V constant voltage no matter what the output in the gate is. I am measuring the voltage across the LEDs (drop). And I have 0.7V between gate and drain. I assume the Mosfet is damaged, but I don't understand WHY, when it is supposed to be working fine up to 50V/14A D-S and up to 10V in the gate.
I will have to get new mosfets as a replacement so before I do that I would like to understand what did I do wrong to damage the mosfet, as maybe I need a different one or just to wire something differently? Maybe I missunderstood one of its parameters?
I think it's because you need to apply the voltage between gate and source. Gate higher voltage than source.
Also, with N-channel mosfets, the drain is the positive side (to be connected to the supply rail), while the source is the negative side (lower voltage side).
Thanks for your resply
So if voltage at the gate is 12V I need 13V to open the gate?
Then I don't understand how would it work when I had a 6V circuit with 2LEDs.
However, then I would need 4 separate circuits one for each LED so that the 5 V of the arduino is always higher than the source voltage?
Most welcome. Try this one. Basically similar to what you had, but just flipped around. Applying 5 to 6 Volt DC (gate to source) should do the trick. Gate voltage higher than source voltage by 5 to 6 Volt.
efv_t:
Could you shortly explain why and if that would work with the same kind of mosfets? Cheers
Yeah.......they talk about an intrinsic diode (inherent to the mosfet) which is usually not shown, and not everybody knows about it maybe. It's there as a result of fabricating this kind of device. Look for the picture in the link .....click here.....
Once you see that intrinsic diode in the diagram, then you'll know why conduction can occur in the direction from source to drain if Vsource is higher than Vdrain.
efv_t:
Thanks for your resply
So if voltage at the gate is 12V I need 13V to open the gate?
Then I don't understand how would it work when I had a 6V circuit with 2LEDs.
However, then I would need 4 separate circuits one for each LED so that the 5 V of the arduino is always higher than the source voltage?
Thanks again!
You have source and drain swapped. Source should be ground, load on the drain.
Most obviously you have blown the MOSFET with overcurrent/overpower by having no
current limiting resistor on the LED string (a no-no) - you may have cooked the LEDs too.
You probably drew the max current the supply could give. Remember LEDs are current
driven, the voltage rating is merely a guide, never drive an LED with constant voltage.
That MOSFET has far more on-resistance than is normal, pick something more like 0.01 ohm on
resistance. On resistance is everything, ignore the current rating (if you are anywhere need the
max current you'll be water-cooling the device).
The golden rule is check twice, buy once. You will remember this!
efv_t:
Thanks for your resply
So if voltage at the gate is 12V I need 13V to open the gate?
Yeah..... MarkT gave excellent recommendations already (previous post).
For an N-channel mosfet, if the SOURCE is at 12V, then you can turn on the device if the voltage at the GATE is adequately higher than the voltage at the source. Usually, the gate voltage being 5V 'higher' than source voltage should do the trick.
EDIT update:
Also.... note that the voltage source (that powers the LEDs) has a '-' terminal that is not meant to be connected to arduino 0V (aka arduino GND). We can assume that arduino '0V' pin is ground. So make sure that the 0V (GND) arduino pin is not electrically connected to the '-' terminal of the battery/voltage source.
I'm not entirely sure what you did to blow out the MOSFET; that fet is absolutely sufficient for switching a 1.4A load. 100mOhm @ 1.4A = 140mV drop, for ~200mW, which is fine for that package.
Only way I can think of for you to have trashed the fet is if the gate was grounded while the source was still seeing 12V, and thus the gate to source voltage spec was exceeded?
Others have already corrected you w/r/t wiring - that fet should work fine if wired as a low side switch (source to ground, drain to negative side of load, and gate to arduino pin with a 10k resistor from the gate to the source to keep the mosfet off when the arduino is disconnected or in reset state. .
The MOSFET is backwards, its just a forward biased diode in series with LEDs with no current limiting resistor,
as I understand it, so even a couple of amps will rapidly heat the device. 12V isn't normally a fatal gate-source
voltage for a logic-level FET (there are exceptions).
Thanks a lot for all your answers, you're legends.
I have a benchtop power supply with both Voltage and Current regulation, but maybe I need a resistor anyway.
As I gather I need to:
1- place a current limiting resistor
2- have current flowing Source-Drain instead of Drain-Source and then 5V will be enough to regulate the gate.
I'll try that, only one concern about the modified diagram by Southpark
So make sure that the 0V (GND) Arduino pin is not electrically connected to the '-' terminal of the battery/voltage source.
But in the picture attached the current flowing out from the Mosfet's Drain would be short circuited to Arduino GND ? If that is not the case, I don't understand why it wouldn't go that way.
Another alternative I thought is to work with the 4LED's in parallel, which means that voltage in the Mosfet D-S or S-D would still be 3V but almost 6A. Then 5V from the Arduino would be high enough in theory. My concern is... 6A is a lot of current. Is that crazy? Do I need to take extra precautions? And, would that work? D-S or S-D?
efv_t:
I have a benchtop power supply with both Voltage and Current regulation, but maybe I need a resistor anyway.
As I gather I need to:
1- place a current limiting resistor
2- have current flowing Source-Drain instead of Drain-Source and then 5V will be enough to regulate the gate.
Have the current flowing from Drain to Source (for an N-channel mosfet). The words 'drain' and 'source' can be confusing, I know this. It is often an issue when 'source' is interpreted as the part connected to the positive side of the voltage source..... but that's not the correct meaning in the case of Mosfets. If I could change history, I would ask them to change the wording.
efv_t:
But in the picture attached the current flowing out from the Mosfet's Drain would be short circuited to Arduino GND ? If that is not the case, I don't understand why it wouldn't go that way.
In your original picture (in post #2), yep..... the '-' terminal and the arduino GND were connected together. That's no problem.
In the modified circuit diagram that I loaded up, the '-' terminal and arduino GND are not connected together.
efv_t:
Another alternative I thought is to work with the 4LED's in parallel, which means that voltage in the Mosfet D-S or S-D would still be 3V but almost 6A. Then 5V from the Arduino would be high enough in theory. My concern is... 6A is a lot of current. Is that crazy? Do I need to take extra precautions? And, would that work? D-S or S-D?
One important thing is..... need to know the usual safe operating current for one LED. How much current does it need to use to become illuminated at your desired intensity.
maximum current i 1.4A. The pwm pin in the Arduino regulates according to desired (according to a sensor reading in a more complicated circuit, but I'm working with that disconnected at the moment and just setting the LED brightness manually 0-255)
Alright, I was a bit confused this morning. The mosfet symbols are really confusing!
I already had current Drain-Source. So I guess the symbol was wrong... SORRY for wasting your time, I didn't know better!
I have two options now and I drew them down with the appropriate symbol from the Mosfet datasheet. Options are parallel and series. Can I ask if my diagrams are correct?
And maybe opinions in what is better? In series I will still have Vds bigger that Vgs, so it still wouldn't work?
efv_t:
Alright, I was a bit confused this morning. The mosfet symbols are really confusing!
I already had current Drain-Source. So I guess the symbol was wrong... SORRY for wasting your time, I didn't know better!
I have two options now and I drew them down with the appropriate symbol from the Mosfet datasheet. Options are parallel and series. Can I ask if my diagrams are correct?
And maybe opinions in what is better? In series I will still have Vds bigger that Vgs, so it still wouldn't work?
Your new diagram is fine. When the arduino pin is high, about 5V will be applied to the MOSFET gate, the MOSFET will be 'on', and the drain will be close to GND potential, turning the LEDs on.
The only thing missing is that it's better to put a series resistor between the Arduino pin and the MOSFET gate, to limit initial inrush current when the MOSFET switches on, until the gate capacitance has charged.
N.B. SouthPark's corrected schematic in reply #8 is also wrong. The 12V supply is connected in the wrong place.
It's -ve terminal should be connected to the MOSFET source and to Arduino GND, and it's +ve terminal should supply 12V to the LEDs. (All of which you've done correctly in your latest diagram, shown below.)
Just go with your latest diagram, but add that series gate resistor. 270Ω is a good value, and will limit initial current from the Arduino pin to about 19mA.
In your latest diagram..... will need to consider what kind of voltage supplies we can buy that will provide the required amount of power.
If you run the LEDs in parallel, then that gives reliability.... since if 1 LED goes out, then not all of them will go out. So that's ok.
You could get 5V DC supplies that provide a fair bit of current. You could also get 12V DC supplies, maybe more common. It'll be a balance..... smaller supply voltage will mean smaller current limiting resistors.
If you choose a 12V DC supply, and you want 1.3 Amp through each LED, and the forward voltage is 3V for the LED, then the series resistor for each LED can be a 5.6 Ohm resistor, with say a 10 Watt rating.
OldSteve:
N.B. SouthPark's corrected schematic in reply #8 is also wrong. The 12V supply is connected in the wrong place. It's -ve terminal should be connected to the MOSFET source and to Arduino GND, and it's +ve terminal should supply 12V to the LEDs. (All of which you've done correctly in your latest diagram, shown below.)
Thanks Steve. But.....when the MOSFET in my diagram is switched on, it will complete the circuit (the loop), and the battery/voltage source will drive the LEDS.
Normally, I would just put the voltage source in the usual spot.... such as move it around the loop so that the '-' terminal is at the same potential as the arduino GND, and then move the resistor and LEDs to their usual spots.
