I have been using the attached circuit with a resistor and led as loads and it works fine. I then tried a 6ohm 12v solenoid and it fries the Mosfet as well as the 2N4401. The MOSFET shorts out and causes the 2N4401 to draw too much current.
I have placed a freewheel diode in the drawing. Is this the correct way to add this diode? I can't add it right at the solenoid because it 7 ft away and part of a system I can't mess with.
I think I'm reaching the breakdown voltage since the MOSFET has continuity on all three pins.
Also I'm running two MOSFET's at the same time they share the 12v Source do I need to protect them an any way? Would a Schottky diode in series on the two sources help with any forward biased by the free-wheeling current?
Also I also want to know if its getting enough voltage at the gate to turn it fully on. The sheet says 4v for 5a so I think I'm good to go just want to make sure
All this is so the MOSFET lasts.
If there is anything else that doesn't look right let me know...in the end all I'm trying to do is control the 6ohm 12v solenoid with an Arduino pin.
Yes, perhaps you reach the Vgs breakdown voltage even with the D7 diode (it's a fast recovery diode?).
you can use a 30 -> 40v TVS diode in parallel with the solenoid or between the D and S of the mosfet
note: think to place D7 and TVS with very short traces and very close to the solenoid to be efficient. if not the lenght can act as an inductor and let enought voltage to reach (and destroy) the transistor before to be clamped by the diodes.
note2: you can add a serie resistor with the gate and collector of the common emitter transistor with a capacitor between the gate and source of Q4 to limit the Dv/Dt
Actually, I cannot spot the problem with your circuit. The drive voltage for your FET is clearly close to 12V, so it should have no trouble turning on fully, and presumably it it rated to carry the current your solenoids require.
There are two common blunders in understanding the action of the "freewheel" or "kickback" diode. One is that it needs to be "fast recovery". This is in fact quite irrelevant - "fast recovery" describes the ability of the diode to turn off whereas in this situation, the only requirement is that it turns on quickly and almost all diodes - including Schottky - do. There is no advantage to using a "fast recovery" or Schottky diode - all you need is an ordinary silicon power diode rated to the current rating of your solenoid.
The second misunderstanding is that there is some need or advantage to having the diode directly across the solenoid. While it seems intuitive, this is also quite wrong! When the FET switches off, the solenoid inductance acts to maintain the current flow and the diode permits this to occur (which is why I point out that the diode must be rated for the solenoid current).
If you have the diode directly across the solenoid, then you create a sudden current transient in the connecting wires between the FET and the solenoid which invites trouble. Since the two alternate paths for the solenoid current are the diode and the FET plus power supply, the diode, the FET and the decoupling capacitor on the power supply should be as close as possible and the leads to the solenoid do not matter (though they should run together). This limits the scope of switching (current) transients to the area of the above three components. The voltage transients are necessarily present no matter where the diode is placed (which is why the leads to the solenoid should run together) but are suppressed somewhat by the capacitance of the leads to the solenoid.
Incidentally, the diode you show between the bases of the two transistors, does nothing. What did you expect it to do?
fast recovery diode are in general fast switching diodes. a 1N4001 like diode is not a good choice as free wheeling diode due to its slow speed. the best choice is a fast switching diode like MBRS340 series
About the diode connections as you've described the current path when the solenoid power is turned off is in reverse polarity and this is where the free wheeling diode is working. if you draw the current path in this case it start by one pin of the solenoid to the diode and to its other pin.
so if the diode is near the solenoid this reverse current is quickly dumped but if you have a long lenght you have a parasitic inductor created by the wires thru the diode. this kind of parasitic inductance can generate oscillations/high voltage/EMI.
this is why the diode must be as close as possible of a coil like a relay.
For the turning on phase the Di/Dt is limited by the solenoid inductance itself. (current inrush delayed)
I'm using the diode + zener technique in my designs.
your design is like a SSR so:
page 2: fast recovery preffered if solenoid is switched quickly before the standard diode recovers and page 4 diode + zener technique:
The second misunderstanding is that there is some need or advantage to having the diode directly across the solenoid. While it seems intuitive, this is also quite wrong!
Disagree!
I have seen many times (over 15) where a diode placed several feet form the inductive load (in switching equipment) caused interference on communications equipment.
The impedance between the load and the diode was enough that a spike was still at the load causing noise spikes on receive inputs.
Was obvious on a scope trace and disappeared when the diode was moved to the load. All communications problems disappeared after moving the diode.
Hi, just looking at the circuit, you are high side switching and I wonder if your problems are due to the need for the transistor in the gate circuit to do the level and logic conversion. The spike from the coil collapse could be turning the transistor back on.
I would be replacing the transistors with opto-couplers, they will also keep horrible switching and inductive spikes from getting to the Arduino.
Or you look up high side switching MOSFET IC's they are made for automotive applications such as operating fuel injectors.
4.7K with near 1nF of gate capacitance is really fast I think. And the linear (resistive mode) portion of the conduction will be very short (see Id f(Vgas) 10A at Vgs = 4.5v
Since the dc resistance of the solenoid coil is 6Ω (2A @ 12V), I would select protection diodes from the 1N5400 Series (3A rated) and not the 1N4000 series. Also, I would suggest using a higher reverse voltage rating, like the 1N5406, 1N5407 or 1N5408.
If the protection diodes are placed directly across the solenoid coils, the flyback current would not be forced to complete its path through your ground connections.
I will try the same circuit with the diode I added and make sure it can handle the current in the circuit. Also it seemed when I turned off power to the MOSFET if I turned it back on it no longer would shut off result in continuity on all three pins.
Also the reason I added a diode from pin 7 base to pin 6 base is to be able to turn on both transistors if pin 7 is activated but only pin 6 if activated. If this isn't the correct way or there is a better way let me know know.
I see someone touched on a opto coupler. Would this really be the way to go? I can also switch out the transistor for a opto just not sure which one would work best.
What are people's thoughts on a diode on the source lines to prevent forward bias of the diode inside the MOSFET? still not sure what they are trying to say...if one is on and the other is off it can forward bias the other diode in the MOSFET causing possible damage and the solution is a diode on the source lines to prevent this?
dark1990:
Also the reason I added a diode from pin 7 base to pin 6 base is to be able to turn on both transistors if pin 7 is activated but only pin 6 if activated.
But it won't.
You need another resistor - all three could be 1k.
The diode inside the MOSFET is happy as it is . Leave it alone.
I should be interested if anyone can illustrate any fault in my explanation of the "kickback" diode.
without diode to get a spark 