How can i calculate the power dissipation on a mosfet and also is there any way i can calculate the Rds(on) of a mosfet?

 

 

Know how to to that in LTspice, but not Pspice.
Can Pspice do waveform arithmetic on its plots?

 

Can Pspice do waveform arithmetic on its plots?

Yes, it can do it. Just like any spice program.

How can i calculate the power dissipation on a mosfet and also is there any way i can calculate the Rds(on) of a mosfet?

Is this some kind of a homework question?
But first of all your MOSFET (BS250) is a P-channel MOSFET and you got it connected wrongly on your diagram.
As it is now your MOSFET behaves like an ordinary diode connected between the drain and the source.
Lika I show here


Because you have forgotten that every single MOSFET will have a built-in body diode.
This diode is even shown in the datasheet
https://www.vishay.com/docs/70209/70209.pdf

So, this body diode will conduct all the time and the Rcoil voltage will be around (12V - 0.6V) = 11.4V and the current is 11.4V/360Ω = 31.7mA.
And the power dissipation on a mosfet is P = 0.6V *31.7mA = 19mW

Now if you properly connect your MOSFET (swap drain with the source)


The "body diode" no longer conduct any current because as you can see the "body diode" is in reverse biased.
And the MOSFET can now work properly as a switch in this application.

And in datascheet we can find Rds(on)_max = 14Ω if Vgs = 10V hence the P = (12V/360Ω)^2 *14Ω = 15.6mW. But typical power dissipation in the MOSFET will be lower than this becouse Rds(on) will be typical in the range of 4Ω.

Also what is the purpose for a R2 resistor?

 

r

Yes, it can do it. Just like any spice program.


Is this some kind of a homework question?
But first of all your MOSFET (BS250) is a P-channel MOSFET and you got it connected wrongly on your diagram.
As it is now your MOSFET behaves like an ordinary diode connected between the drain and the source.
Lika I show here
View attachment 147504

Because you have forgotten that every single MOSFET will have a built-in body diode.
This diode is even shown in the datasheet
https://www.vishay.com/docs/70209/70209.pdf

So, this body diode will conduct all the time and the Rcoil voltage will be around (12V - 0.6V) = 11.4V and the current is 11.4V/360Ω = 31.7mA.
And the power dissipation on a mosfet is P = 0.6V *31.7mA = 19mW

Now if you properly connect your MOSFET (swap drain with the source)
View attachment 147506

The "body diode" no longer conduct any current because as you can see the "body diode" is in reverse biased.
And the MOSFET can now work properly as a switch in this application.

And in datascheet we can find Rds(on)_max = 14Ω if Vgs = 10V hence the P = (12V/360Ω)^2 *14Ω = 15.6mW. But typical power dissipation in the MOSFET will be lower than this becouse Rds(on) will be typical in the range of 4Ω.

Also what is the purpose for a R2 resistor?

R2 resistor is if the nand gate gets destroyed we will be sure that 0 volts will go to the mosfet.

 

R2 resistor is if the nand gate gets destroyed we will be sure that 0 volts will go to the mosfet

But 0V at the gate of a P-channel MOSFET will turn it ON.

 

But 0V at the gate of a P-channel MOSFET will turn it ON.

The Resistor R2=100k is used to make sure that the voltage source-gate will be 0V in case CD4011 gets destroyed or removed.(That's what the book says)

 

On pspice im finding the PD=219,89mV(voltage drain -source)*IL=32.72mA=7.26mW and my calculation were Pd=14.41mw.
Can the difference be so big or am i doing something wrong?

 

Which book? And maybe in the book, they are using N-channel MOSFET instead of a P-channel one?

As you can see here


To "open" (drain current (Id) will start to flow) the P-channel MOSFET the voltage at the gate need to by at least Vgs(th) lower then the voltage at the source of a MOSFET.

On pspice im finding the PD=219,89mV(voltage drain -source)*IL=32.72mA=7.26mW and my calculation were Pd=14.41mw.
Can the difference be so big or am i doing something wrong?

Have you corrected the circuit diagram?

 

Which book? And maybe in the book, they are using N-channel MOSFET instead of a P-channel one?

As you can see here
View attachment 147514

To "open" (drain current (Id) will start to flow) the P-channel MOSFET the voltage at the gate need to by at least Vgs(th) lower then the voltage at the source of a MOSFET.
The circuit is supposed to be like that this is our homework im not doing something wrong.



Have you corrected the circuit diagram?

The circuit is supposed to be like that,that's our homework(i think that the mosfet is rotated so we can have posotive voltage to drain and negative to source)

 

But this circuit will only work as it should only when the source terminal is connected to the positive terminal of a voltage source and the drain terminal is connected to the Rcoil side.
Because now with the drain connected to the positive side of a battery the "body diode" will conduct the current all the time and the relay will be energized all the time also, independent of a gate voltage.

As for the power dissipation, because we do not know the exact value for Rds(on) we cannot get the exact resoult for Ptot in the transistor. We can only find the worst case power dissipation.