Go Down

### Topic: MOSFET (Read 2385 times)previous topic - next topic

#### vams

##### Sep 27, 2012, 12:37 am
Hi everybody,

Here's the datasheet of the mofset i've bought :

http://www.irf.com/product-info/datasheets/data/irf9540n.pdf

If i get how it works, this little guy has a transconductance of 5 S, and a threshold voltage of -3 V (min -2 , max -4...i guess it's 3 then xD ).
Does it mean if I put a 5 V voltage supply attached to the gate and the source,  the mofset will generate a source-drain current of ((-5 -(-3)) V * 5 S ) = 10 Amps
Clearly I'm misunderstanding something...

#### Grumpy_Mike

#1
##### Sep 27, 2012, 01:01 am
Quote
Does it mean if I put a 5 V voltage supply attached to the gate and the source

No.
The gate threshold is when the FET just starts to conduct. To be fully on you have to have a 10V gate signal.

#### vams

#2
##### Sep 27, 2012, 01:19 am
Mmm, and may you tell me how i can find/calculate this spec in the datasheet? Cause the only thing i found was that value.

#### retrolefty

#3
##### Sep 27, 2012, 01:48 amLast Edit: Sep 27, 2012, 02:10 am by retrolefty Reason: 1

Mmm, and may you tell me how i can find/calculate this spec in the datasheet? Cause the only thing i found was that value.

They are usually shown in graph form on most mosfet datasheets. On yours look at the graph named figure 1. The black heavy lines represent different gate/source voltages choices (from 4.5 to 15) and what current they will allow to flow Vs the source voltage applied. Figure 2 is the same but showing the effect of operating at a much higher device temperature. When using a mosfet in a simple switching application (on/off Vs linear amplification) you need to drive the gate voltage high enough to fully saturate the device on so that it will dissapate the smallest amount of heat as possible, and that value is normally 10vdc for standard mosfets and 2-4 volts for 'logic level' mosfets.

Lefty

#### DVDdoug

#4
##### Sep 27, 2012, 01:52 amLast Edit: Sep 27, 2012, 01:57 am by DVDdoug Reason: 1
Look at figure 3. (Gate-to-source voltage vs. drain-to-source current.)

Quote
Does it mean if I put a 5 V voltage supply attached to the gate and the source,  the mofset will generate a source-drain current of ((-5 -(-3)) V * 5 S ) = 10 Amps
It doesn't generate any current, but it can control the current.  FET devices require very little gate current, so they have "essentially" infinite current gain.  (But, they really are not "current gain" devices, since the "output" current is a function of gate voltage.)

#### dhenry

#5
##### Sep 27, 2012, 02:42 am
The gs spec makes no sense for a switching application (it makes sense for a linear application).

If you look at the Vgs(th) spec, you will see conditions.

Generally, you want to drive a mosfet so that its d-s becomes "resistive" (also called "linear region", confusingly so). One of the charts will show Ids - Vds curve, by Vgs. Vertical mosfets show signs of being resistive at Vgs > 3.5 - 4v, at low Ids levels. As Ids levels go up, the required Vgs to drive the mosfet into resistive goes up as well.

If I were you, I would go back and look at the current you need to switch and see if the Vgs requirement exceeds your mcu drive levels. This assumes that you aren't switching the load too fast - in that case, gate charge + switching frequency come into play.

#### Grumpy_Mike

#6
##### Sep 27, 2012, 09:07 am
Quote
Mmm, and may you tell me how i can find/calculate this spec in the datasheet?

Look at the gate voltage quoted for the on resistance? That will tell you.

#### vams

#7
##### Sep 27, 2012, 05:47 pmLast Edit: Sep 27, 2012, 05:50 pm by vams Reason: 1
Thank you very much, i was misunderstanding a couple of thinghs.
I've studied a little more the mosfets and i think i understand how to use them in some application now.
To be sure of that, could you confirm me that the following is a viable way of setting up a digital switch (for digital i mean it's controlled by two different input values) that must control a known load?
Im assuming the mosfet to be n-channel (V_th>0)

1) Find the current needed by the LOAD, I_L (for example by current = power/voltage or things like that depending on the load, i am thinking of a lamp right now)
2) Find the minimum gate voltage needed to trigger correctly the switch (allowing him not to saturate before the needed current is drawn).
In order to do this, I check the I_ds vs V_ds graph in the datasheet, and I find the first line who get to the right I_L before saturating (becoming flat, not dipendent by V_ds), and I pick the corresponding gate voltage value, V_gs.
3) I build a circuit like this one : http://tronixstuff.files.wordpress.com/2010/04/examp3p1.jpg.
Where of course, instead of the npn transistor i have my beloved MOSFET,the load is the relay, and in place of the digital 2 pin I have any controlled voltage supply who can furnish the V_gs value i found at point 2)
4) I add all the stuff not related to the transistor usage that can improve the switch, like a pull to ground resistance for the gate or other things I don't know.

OK guys, will i blow up my house, or i will switch on the lamp?

#### Grumpy_Mike

#8
##### Sep 27, 2012, 11:48 pm
Step 2 is wrong. You need enough voltage to turn the FET on fully, having too much voltage here is not a problem, but too little and the FET only partly switches on and so will get very hot.
Make the resistor in the gate 100R not 1K.

#### winner10920

#9
##### Sep 28, 2012, 02:34 pm
I don't think 1k would be a problem on the gate, depending on your switch rate and the gate charge, for low frequency switch rates it is probably fine
however I hope you realize this is a pchannel fet and so that circuit will not work as you expect, that is a simple fix although makes driving it a little more complex from the mcu

Go Up

Please enter a valid email to subscribe