If im not wrong, this is what u meant right? Using a transistor to provide a 12V power supply to the Vgs of the Pchannel and Nchannel.
http://www.armory.com/~rstevew/Publi...d/h-bridge.htm

 

Here is how to calculate the drain current of a MOSFET with a resistive load: When the Vds is 5 volts, the current in the load is zero, so put a point on the charactisitic graph at zero amp and 5 V. When the Vds is zero, the current in the load is 15 amps, so put a point on the graph at 15 amps and zero volts. Draw a line between those points. Where the Vgs line crosses the load line, read the drain current and Vds. I get about 3 volts Vds and 10 amps current. That represents a big power loss, you would do better with another MOSFET.

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I dont understand how come the drain current is 0A when Vds is 5V. Is this done practically or am i suppose to assume that when Vds is 5V, Id=0? i also dont get the Vds=0,Id=15A part .

How come there is a big power loss? Totally loss :cry:

 

He's assuming a 5 volt power supply... if Vds were 5v, then there would be zero volts across a resistive load so the current would have to be zero. I don't know where he's getting the 15A from... but you would want to subsitute the current value if Vds were zero, based on the resistance of your load. for instance, say you had a 2 ohm load; if Vds were zero, then the full 5v would be across the load, so I=V/R=5v/2ohms = 2.5 amps. Since it is a resistive load you can draw a straight line from the zero Vds to the zero Id points and where that "load line" intersects the curve for the value of Vgs you are using, that is what your current will be (your operating point, or Q-point, as they call it)

P=Vds*Id. at 3 volts and 10 amps, that's 30 watts of power dissipation. using a better MOSFET you can easily get a value of Vds more like less than one volt for a drain current of 10 amps, which will cut your power dissipation immensely. The more power you are dissipating in the transistor, the less power is going to your motor.

If you can wade your way through everything we've just said, congratulations... I just realized how big of a crash course you just received in MOSFETs. keep asking questions if you need help and I'll do what I can to straighten it out for you... I know this is a lot of info to absorb when you're starting from zero.

 

Enlighten me, where do you get the 3Vds from? I don't have my glasses on but I think that I see a value of about 1.25V-1.50 @ 10A :?:

 

yeah looks more like a little under 2v now that i actually look at it. still, about 20 watts is a lot of power when it can be reduced to less than half that by just switching mosfets.

 

Be aware that the term "saturation" only means turning a transistor into a low resistance "ON" state, like a switch, only when referring to BJTs.

Due to the total anarchy in the electrical engineering world, much like how we never decided once and for all which way current "flows", "saturation" means the EXACT opposite for MOSFETs. It means the magnitude of current through the drain-source is being controlled by gate voltage. If the gate voltage is high enough that it leaves the MOSFET in a low impedance state from drain-source for that circuit, it is now OUT of the saturation region.

Stupid engineers.

 

THX THX.
Supposing my Vgs is 5V. Looking at the Vgs Versus Id curve, my Id would be around 15A. Is this how i plot my load line for my Y-AXIS? using the 0V(Vds) and 15A(Id). For motors, the resistance depends on the load, therefore i dont think i can use the I=V/r

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curr_109.jpg (53.0 KB, 265 views)

 

I forgot to ask one more thing. Since the load line would determine the power dissipation and Vds, then What is the Rds(on) for? I thought that as long as the MOSFET was ON, we would able to use the Rds(on) which is equal to 0.077ohm. Assuming we have a current,Id of 5A, the Vds=RdsxId=0.077x5=0.385V. However, looking from the load line, the Vds is apporximately 2V.Is there a relation between the load line and the equation i just calculated here?

 

Is reading the Rds vs. Id curve important?

 

Rds(on) is measured at specific values of Id and Vgs. For instance, on the IRF540N, Rds(on) is measured at Id=16A and Vgs=10v.

Rds(on) gives a good linear approximation of the equivalent resistance of the MOSFET when it is in the OHMIC region (not saturation)

so as long as you are switching it with a high enough gate voltage to keep it from going into saturation at high current, then Rds(on) is a good approximation of the equivalent resistance of it as a switch.

 

ok i understand. thank you very much. But i still dont get the LOAD LINE part.
When i use the formula Vds=Rds x Id, i will be able to obtain the voltage drop across the Drain-Source. By obtaining the Vds, i would be able to calculate the power dissipation of the mosfet.
However, it seems that each of you(experts) calculate using power dissipation using the load line. Is there any difference or am i getting this all wrong?

 

The power dissipation is always going to be Vds*Id. the load line is used to find the actual values of Vds and Id when using it with a specific power supply voltage and load resistance. (the operating point) from there you can of course calculate power dissipation.

 

I've understand what you have explained previously. But finding the Vds using (Vds=Rds x Id) is really bothering me. The Vds calculated here provide such a small value. However, using the load line, the Vds IS awfully BIG. This part is killing me.

Thx for any explanation

 

why don't you show us some calculations.

bear in mind, that Rds depends on Vgs. the higher Vgs is, the lower Rds becomes. Rds is often measured around 10v for Vgs, whereas if you are operating it at Vgs=5v then Rds is going to be higher than the rated value.