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MOSFETs

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magester1

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I've been looking to use MOSFETs as switches, but from what I've read, searching and in this forum I'm not sure how to.

Basicaly, what I understood is that the connection is like a normal transistor one. However, from what I read, the voltage across the FET will be less than the one at the gate, lets say I send 5v I would get an output of maybe 4v, is that correct? I'm not concidering the current.
Because what I want is to turn it full on or off with a 5v-0v digital signal (to switch 12v or 24v).

By the way, I've got two types available: IRF630 and IRF9530. Just in case that makes a difference.

Thanks for your help.
 
An IRFxxx Mosfet needs a gate voltage of 10V to fully turn on.
An IRlxxx mosfet needs a gate voltage of only 4.5V to 5.0V to fully turn on.

The IRx630 is N-channel and a +5V or +10V on the gate makes it conduct its drain to 0V.
The IRF9530 is P-channel and is off when its gate is at +10V the same as its source and is turned on when its gate is at 0V.
 
Thanks for the fast reply.
I guess it'll be easier for me to work with the p-channel one then.
Would the IRFxxx work fine with a 9v battery? Because I know that after some time they start dropping voltage.
 
Why p-channel? N channels are preferred because for two mosfets of identical construction one P and on N the N channel fet will turn on with a lower RDSon value and have a higher possible current.
Exactly what gate voltage you require to turn on a FET fully depends on the load current, a lower voltage may be alright depending on the current you need to pass. This information is available in the PDF of most mosfets.
 
Since your input is a 0V to 5V signal, you can't use the P-channel Mosfet because its gate needs to be at +12V or +24V for it to be turned off.

The spec's for an IRFxxx show its worst case performance with a 10V gate signal. all of them are guaranteed to work with a 10V gate signal. There are graphs that show a typical one but you don't know if your Mosfet is typical. If it is weaker than typical then it might not work with a 9V battery circuit when the battery has dropped to 6V or 7V.
 
I thought p chanel fets were off at 0 volts, just like N chanel fets, just the polarity was reversed to turn them on?
 
The gate to source voltage of a Mosfet must be 0V for it to turn off. But the source voltage for an N-channel Mosfet is usually 0V and the source voltage for a P-channel Mosfet is usually the positive power supply voltage.
 
Right, but that's totally unecesary for the original posters use, he just wants a simple switch. For general switching N channels fets are almost always used.
 
Actually, I said the p-channel becuase I'd rather have it fully on and partially off than the other way around.
 
You'd be mistaken however, discrete N channel fets are used almost exclusively for switching applications.
If it doesn't turn all the way on and all the way off for your application then it's not a switch. You need to determine your load current in the first place before you even know what the voltage needs to be to fully turn the FET on either N or P.
 
It's been a while since I wrote something here.
Anyway, I was busy and couldn't really try all this.
Regarding Sceadwian's last post, I can't really tell the load current because I'm not sure what I'll conect yet. But still, how would I determine the voltage if I do have the load?
I know I said about the 9530 one, but I now have the N-channel IRF540.

So, you were all telling me about the different voltages. I thought I'd make things easier and try to use an amplifier, if that is an easier solution, isn't it?
You know, to convert the 0-5v to 12v for instance. Would an op-amp work fine for this?
 
A comparator would be more useful than an opamp, comparators are meant to be switched at saturation, opamps generally aren't. or an NPN/PNP transitor pair used as a level shifter. The required gate drive depends on two things. The desired drain/source current and the drain/source voltage . If you look at the first graph of Drain/source voltage vs output current for the IRF540 datasheet you'll see a quickly rising ramp follow by a slowing curve and then it flattens out. If you look at the first one labels it's 4.5V's that's what happens when the gate is fed 4.5 volts, you get a nearly linear increase in current vs drain/source voltage until about 10 volts at 5 amps then it levels off flat where the current stays the same even if the drain source voltage rises. This is where the Mosfet is said to be in saturation, it acts as a constant current source where changes in the drain/source voltage don't change the current going through it. That means it's effectivly become a variable resistor, and where in power applications they tend to smoke because the power dissipation can be massive under the wrong circumstances.

If you notice there's a whole lot more curves there, by the list in the upper left corner of that graph each succesive line is .5 volts higher than the last up till 6 volts, then it's 1 volt for each next occurring line. For switching applications the Drain current point needs to be well bellow the bend in that graph for that gate drive voltage or the Mosfet will act as a variable resistor. So in order to get truly low series resistance you have to have relatively high gate drive relative to the drain/source voltage and dependent on the current.

There is no way to determine what you should be doing unless you know the drain/source voltage (you've said 12 volts so far this is a good place to start) but without a goal for current there is no next step. You may not even need to level adjust if your current requirements are low enough. You unfortunately can't sidelines these values, you have to pick something to shoot for, and if you need to adjust it higher later it'll require a circuit redesign.

Mosfets used to give me a real migraine because even though they are in many cases superior to BJTs and for high speed and efficient switching can be incredibly useful, you still have to know how they work and their limitations relative to your actual application.
 
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OK then.
So lets say, for learning purposes, I want to turn a LED on and off. That's about 10-20mA. Of course I realize I'm not using an appropiate FET for something like that, but this is just for me to understand how it works.
According to the graph, 4v at the gate would be enough then, wouldn't it?
And in this case that would be regardless of the Drain voltage, right?

Another thing, what about the Vgs Threshold? Does it have something to do with all this?
 
OK then.
So lets say, for learning purposes, I want to turn a LED on and off. That's about 10-20mA. Of course I realize I'm not using an appropiate FET for something like that, but this is just for me to understand how it works.
According to the graph, 4v at the gate would be enough then, wouldn't it?
And in this case that would be regardless of the Drain voltage, right?

Another thing, what about the Vgs Threshold? Does it have something to do with all this?

below vth it does not conduct... how much it conducts varies with vgs and vds. and there are pfets that work at logic levels.

if you have a fet with a vth of 3V it will likely readily drive your 20mA LED current at vgs of 4V.
 
I would recommend at least 5 volts. Just to give you an idea of what the difference between just barely meeting threshold and supply a proper gate drive at 4 volts you'll just barely be able to switch 200mas, at 5 volts at the gate you'll be able to switch 15amps. Even if you're not switching 15 amps that gives you plenty of headroom, and it's only 1 volt.

If I'm not mistaken VGS threshold is only particularly useful for people trying to operate a Mosfet in it's linear region such as as an amplifier. Also the VGS threshold is different depending on what current you state that VGS must meet. The datsheet for the 540 says the VGS threshold is around 1 volt, but that's a 250ua of DS current, which is pretty useless information for switching.
 
But what about the voltage?
You're saying that with 5v at the gate I'll be able to switch about 15amps, what happens with the Drain voltage? Do I get the 12v at Source?
Now lets say I want to use the FET for driving a motor, which is actually what I'm going to use it for. It's a small 12v one, with around 150mA without load and 500mA or a little more with load, though I'm sure it doesn't even reach 1Amp.
Acording to the above (the 5v/15amps) the FET would still switch on, wouldn't it?
 
The VGS threshold is where a Mosfet barely conducts (0.25mA) where it is almost turned off. It is a useless spec.

The curves on the datasheet for a Mosfet are for a typical one. But many are worse than typical so you should look at the printed max on-resistance spec.
 
But what about the voltage?
You're saying that with 5v at the gate I'll be able to switch about 15amps, what happens with the Drain voltage? Do I get the 12v at Source?
Using an N-channel mosfet then the source is at 0V and the load is from the positive power supply to the drain of the mosfet. With 5V on the gate then some Mosfets will conduct poorly and others will conduct fairly.
If you connect the load to the source and to 0V then with 5V on the gate the current in the load will be almost nothing.
 
Using an N-channel mosfet then the source is at 0V and the load is from the positive power supply to the drain of the mosfet. With 5V on the gate then some Mosfets will conduct poorly and others will conduct fairly.
If you connect the load to the source and to 0V then with 5V on the gate the current in the load will be almost nothing.

Exactly, so then, what gate voltage would I need to switch this 12v DC motor on and off?
 
Exactly, so then, what gate voltage would I need to switch this 12v DC motor on and off?
The voltage that the motor needs has nothing to do with it.
You want a Mosfet that completely turns on with whatever gate voltage you have. Connect the source pin of the N-channel Mosfet to 0V.
An "ordinary" Mosfet needs a gate voltage of 10V. A "logic-level" Mosfet needs a gate voltage of 4.5V to 10V to fully turn on.
 
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