MOSFETs AGAIN

hmlittle59

Hello All,

I had to step back an take a breather and look up more info, trying to under stand this part. Everyone seems to have there own spec.(FairChild,Vishay..etc) and the Voltages seem to be all over the place. I thought I was set with P-chan. enhanced and my logic was 180 deg. out. Everyone's explanation and my thought/logic was just clashing. I think I'm almost cleared of mental block so hear I go again. AS before, I currently have a BS2e directly driving a NTE R56(Normally Open) Solid State Relay($3.50)(5vdc 2sec). This works as I wanted, but I'm looking to replace the NTE part with a (5v, 3+ AMP) cheaper part.

This time around I think that I may have found the correct part. This is for ya'lls review and comments so correct me were I'm wrong. I've ordered parts to test out Vishay part (N-chan 844-IRL630SPBF) from Mouser.com. Another part that looks close is FairChild(ND6060L). Some of the symbols on the Data sheet I can figure out but others make no since.

The voltages are any were from (2.5, 4.5,5.0, 8.0,10.0, 20.0, 25.0, 30.0....100, 200, 300), I mean com on guys this was driving me crazy trying to wade thru. Why so many?

I will try and attach a simplified schematic from ExpressSCH, let me know if its viewable or should I take a picture?
I want to Pulse(BS2e 5vdc one time for 3 sec.) the Gate causing it to conduct from the Drain to the Source, this will act as a momentary push button switch shorting 5vdc to Gnd. activating the motor. Also I've seen resistors either on the Drain to load or Source to load. Taking the BS2e LOW turns off the Mosfet. A resistor in series and resistor to grd are connected to the Gate. Now is my logic correct???

thanks to all

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I'M STILL LEARNING SO MUCH...BUT STILL KNOW SO LITTLE!!!

hmlittle59

Post Edited (hmlittle59) : 4/19/2009 5:25:00 AM GMT

Phil Pilgrim (PhiPi)

You need to convert your schematic to a PDF or image file (GIF or PNG preferred), if you want to share it with the broadest audience.

-Phil

hmlittle59

thanks phil,

ExpressSCH appears not to be able to do either...GREAT!!! Let me redo it in DipTrace

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I'M STILL LEARNING SO MUCH...BUT STILL KNOW SO LITTLE!!!

hmlittle59

Phil Pilgrim (PhiPi)

Don't give up on ExpressSCH just yet. If it has a Print function, there are several free print drivers out there that you can use to create a PDF just by printing. One example is PrimoPDF. I use it and like it. I've yet to see a program that it doesn't work with.

Also, if you have a graphics program, like Corel PhotoPaint, some CAD apps will let you select and copy drawings, or portions of drawings, to the clipboard. PhotoPaint has a "New from Clipboard" function that automatically generates an image file from the clipboard that can then be saved in a multitude of graphics formats.

-Phil

Post Edited (Phil Pilgrim (PhiPi)) : 4/18/2009 3:14:51 AM GMT

Dave Hein

You can take a snapshot of the screen by pushing the "Print Screen" button.· Then bring up the MS Paint program and click on Edit/Paste.· The image can then be saved as a GIF file.

Dave

kwinn

Not sure where all those voltages come from, but the most important ratings are the absolute maximum ratings. Those should never be exceeded, and I would not operate a device at more than about half of the max voltage, current, and power of those ratings.

In other words for the IRL630 I would not run it much above 100V Vds, 5V Vgs, and 3A Id, and even at 3A continuous current it would need a good heat sink to dissipate the 3.6W of power and keep the temp reasonable.

hmlittle59

Thanks Phil,

I used the downloaded and used the program you suggested and it seems to have worked great. I've attached a simplified version of the Circuit I'm trying to get working, hope this will help to clarify any confusion.


thanks again Phil for the program.

PS. I edited this file to attach the Simplified drawing of my Question/Problem/Steep Learning Curve!!!

thanks again

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I'M STILL LEARNING SO MUCH...BUT STILL KNOW SO LITTLE!!!

hmlittle59

Post Edited (hmlittle59) : 4/19/2009 12:32:58 PM GMT

Dave Hein

hmlittle59,

Your schematic doesn't make any sense.· You show +5 going directly into the MOSFET.· This will blow the MOSFET the first time you turn it on.· I think what you're trying to say is that the one input to the motor unit is 5 volts when the control is open.· Are you driving the motor directly, or are you driving a conroller?· Do you have any idea what the current requirements are for motor controller input?· The MOSFET you selected will probably do the job, but it may be an overkill for the application.· You may be able to use a much smaller transistor, but you would need to know the current requirements.

Dave

kwinn

hmlittle59, what you want is more like the attached diagram.

OwenS

I may be wrong with this - but whenever I've looked into MOSFETs, the general advice was the less resistance on the gate terminal the better? In general I've always heard that in switching applications the lower resistance the better because the faster the FET turns fully on the less power is wasted as heat. This is all assuming the output driving the FET can take the current to charge the capacitor of course.

Of course, I could be wrong? (It must be noted most of my MOSFET research has been SMPS related)

Philldapill

OwenS, you're partially right about "the less resistance the better". The Gate of the mosfet acts as a tiny capacitor. With a high resistance between the drive signal, and the gate, the capacitor will charge slowly. This will result in the mosfet taking longer than needed to turn on - dissipating alot of heat. However, sometimes if you don't have any resistance at all between the gate and drive signal, you can get an oscillation going. This is because your traces act as inductors, and the gate acts as a capacitor, resulting in an LC circuit. When the gate oscillates, the MOSFET turns on and off at the frequency of oscillation - usally in the MHz range. Again, this results in alot of extra heat. Adding a SMALL resistor, like a 4.7 ohm, will aid in dampening the oscillation.

Don't make the resistor too small or you'll get high frequency oscillation. Don't make it too big or the mosfet will turn on too slowly. Both of which are bad news.

hmlittle59

Hello All,

The Motor and Electronic controls are in a unit already(Not my Unit), I've tried checking the current but the Electronics senses this an activates the unit. I called the company and they said it was around .5 Amp., I'm just replacing the Momentary Push Button switch with a Electronic switch and BS2e. I can see were I may need a resistor on the Drain side to protect the Mosfet. The Mosfet is only acting/replacing a Momentary switch.

Presently when the switch is pushed, 5vdc is shorted to grd inside the unit, I guess that's why they have the current so low already.

thanks

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I'M STILL LEARNING SO MUCH...BUT STILL KNOW SO LITTLE!!!

hmlittle59

Carl Hayes

ExpressSCH has an export function.· Click File > Export schematic image > Bitmap, 300 pixels per inch to create a BMP image of your schematic.· You can upload this as is, and most people can display it, but it's a big file; or you can use some program or other·(I use PhotoShop) to convert the BMP to JPG or GIF or whatever you like.


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· -- Carl, nn5i@arrl.net

Post Edited (Carl Hayes) : 4/20/2009 1:38:52 PM GMT

Dave Hein

hmlittle59 said...
The Motor and Electronic controls are in a unit already(Not my Unit), I've tried checking the current but the Electronics senses this an activates the unit. I called the company and they said it was around .5 Amp., I'm just replacing the Momentary Push Button switch with a Electronic switch and BS2e. I can see were I may need a resistor on the Drain side to protect the Mosfet. The Mosfet is only acting/replacing a Momentary switch.

You should be able to use almost any bipolar transistor or MOSFET that can handle 500 mA.· The MOSFET that you are using easily meets the requirements.· I wouldn't worry too much about the resistor values at the gate.· It won't make much difference whether you use 100 ohms or 100 Kohms for your application.
Dave

Carl Hayes

Here's what I'd try first, with that particular FET

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· -- Carl, nn5i@arrl.net

hmlittle59

Thanks Carl,

Remember I'm replacing the switch, it was just there for show ONLY, also I don't understand the diode. My understanding of the N-chan. Mosfet the Gate is Pulsed with 5vdc which will conduct(Drain to Source), causing a shorting of 5 vdc to grd(simulating pushing of a switch) to the Motor Unit itself. I stream line the schematic some more...any thoughts please.

thanks

Opps.. forgot to attach the file

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I'M STILL LEARNING SO MUCH...BUT STILL KNOW SO LITTLE!!!

hmlittle59

Post Edited (hmlittle59) : 4/21/2009 12:29:59 AM GMT

Phil Pilgrim (PhiPi)

You need a series pullup resistor on the +5V line. Otherwise, when you switch the MOSFET "on" it will create a dead short between +5V and Gnd. But more to the point, based on your schematic, why do you need the MOSFET at all, if the circutry inside the black box is doing the actual switching?

-Phil

Post Edited (Phil Pilgrim (PhiPi)) : 4/21/2009 1:47:18 AM GMT

Carl Hayes

Well, OK, leave the switch out.

Since your load is an unknown black box, the diode is there to protect the FET if the load happens to be inductive.· The load may, for example, have a relay coil inside it.· You don't say.· Perhaps you don't know; they may have sealed the thing so you can't tell.· Let's assume for the moment that it is inductive.·

Let's say the FET is conducting, and the diode was left out.· There's a current (positive-to-negative) flowing downward through the load, and downward through the FET.· The diode is not conducting because it ain't there.

Now shut off the FET.· The FET tries desperately to stop the current.· The load, being inductive, tries desperately to maintain the current.· That's what inductors do.· There is strife, and the inductive load will win, melting the FET, because there is no other path for the current to take.· Pffft.

Now put the diode back in, just as I've shown it.

Let's say the FET is conducting, and the diode is in place.· There's a current flowing downward through the load, and downward through the FET.· The diode is not conducting because it can't conduct downward the way it is positioned in the circuit.

Now shut off the FET again.· The FET tries desperately to stop the current.· The load, being inductive, tries desperately to maintain the current.· The current will continue to flow downward through the load for a little while, but the diode is there, so that same current can flow upward through the diode and doesn't get forced downward through the FET.· The FET's job is now very easy, because no one is fighting it; the diode earns a medal for heroism.· The current will circle through the diode and the load until it dies out (from the resistance of the load) or until the FET is turned on again, whichever occurs first.

That four-cent diode (I like 1N4001 for this)·will save a lot of (much more expensive) FETs in a surprisingly short time.

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· -- Carl, nn5i@arrl.net

Carl Hayes

Phil Pilgrim (PhiPi) said...
You need a series pullup resistor on the +5V line. Otherwise, when you switch the MOSFET "on" it will create a dead short between +5V and Gnd. But more to the point, based on your schematic, why do you need the MOSFET at all, if the circutry inside the black box is doing the actual switching?

-Phil

Phil, he said the company told him the current through the pushbutton was about 1/2 ampere.· To me that precludes the possibility that they're sensing logic levels.· I think they're got a relay in there.· Ergo, no pullup needed.

Alternatively, if they are sensing logic levels, and they use only a switch externally, they must have included a pullup inside the box anyway.· Agree?

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· -- Carl, nn5i@arrl.net

Post Edited (Carl Hayes) : 4/21/2009 2:35:27 AM GMT

Phil Pilgrim (PhiPi)

Carl,

'Missed the 1/2 amp part. That does imply a relay of some sort. But look at his schematic: the MOSFET shorts +5V to Gnd.

HMLittle, you need to pay closer attention to Carl's schematic and follow it assiduously!

-Phil

Philldapill

Carl, I think YOU deserve a medal for the explaination. If only my professors had explained it like that, I would have understood alot earlier. Bravo!

Carl Hayes

Phil Pilgrim (PhiPi) said...
Carl,

'Missed the 1/2 amp part. That does imply a relay of some sort. But look at his schematic: the MOSFET shorts +5V to Gnd.

HMLittle, you need to pay closer attention to Carl's schematic and follow it assiduously!

-Phil

Ja, geehrter Freund, I did look at his schematic.· Obviously it has some problems that will immediately let all of the smoke out of the FET.· That's OK, I'll just go purchase some stock in the company that makes the FET.

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· -- Carl, nn5i@arrl.net

hmlittle59

Thanks Carl/ Phil,

The Company did tell me before that there is a relay inside and their electronics handles the pushing of the button. I have this working with a (NTE R56 -D SSR) but I was looking for the BEST replacement part(cheaper) switching Mosfet/transistor that can handle twice(+) the load. The +5vdc is being shorted to Gnd within the unit when someone pushes the button and that's why I was only concerned about putting a transistor in place that could handle the +5vdc @ .5 Amp.

thanks again for your time and explanation

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I'M STILL LEARNING SO MUCH...BUT STILL KNOW SO LITTLE!!!

hmlittle59

Carl Hayes

Philldapill said...
Carl, I think YOU deserve a medal for the explaination. If only my professors had explained it like that, I would have understood alot earlier. Bravo!

Why, thank you.·

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· -- Carl, nn5i@arrl.net