HI Jmac,

I calculated the values for the following mosfet: PSMN4R0

Junction temp(max)= 175 [C]
ambient junction temp =60 [K/W]
V in= 20[V]
i in=22[A]
Pmax=440[W]
let duty cycle be 'x%'
Pmax at 'x' duty cycle= 440*(x/100)
now temp rise = 440* (x/100) *60[junc-amb temp rise per watt]
= 264x
this should be not higher than the junction temp i.e. 175+20(25 is just a guess)
now
264x=200
x=0.75%

now when we use this dutycyle reversly
pmax= 20 *22 *(0.75/100)
= 3.3 w
NOW TEMP RISE
=3.3 * 60 =180 K
which is quite acceptable temp with the heat sink

can i approach the power dissipation problem as follows:

thanks for all the help

 

What is your load? In the drawing you don't show any load other than the parasitic inductance.

To calculate power dissipation you need to cmultiply the current (24A) by the voltage drop across the mosfet (VDS). You need to know your load to calculate the voltage across the MOSFET when the circut is sinking 24A.

No, your mosfet certainly won't be able to dissipate 480W! You determine what a mosfet can handle by making sure its junction temperature stays below the specified limit - as I described in a previous post. Lets take a common TO-220 mosfet, the IRF640 with thermal resistance=40degC/W. At 480W, the temperature rise is Trise=(480W)(40C/W)=1920C and the mosfet is a puff of smoke

So you really need to properly calculate the power dissipation by figuring out VDS, multiplying it by the current (24A), applying the duty cycle, then multiplying it by the thermal resistance of your mosfet.

About the load...the total load is just the MOSFET and the sense resistor..

 

About the load...the total load is just the MOSFET and the sense resistor..

Then why do you need a switched current sink? Are you trying to charge a battery?

 

More like discharge a battery, or make a convoluted space heater.

 

Your calculations are wrong ...

P = (20V)(22A)(50% duty cycle) = (20)(22)(.5)=220W

Trise = (220W)(60C/W) = 13200 C

Your mosfet is toast

------------------------------------------------------------

Lets work the calculations in reverse to see what is the maximum power you can dissipate across the mosfet...

Trise + Tambient = MaxJunctionTemp

(P x 60C/W) + 25C = 175C

P = (175-25)/60

P = 2.5W

So the maximum average power you can dissipate in this mosfet is only 2.5W. This is 5W ant 50% duty cycle, etc...

------------------------------------------------------------------------------------------

BTW, as others said, your circuit makes no sense. Why are you making a current sink that has no load? All you're doing is drawing a constant current from your voltage source. Why?

 

I want to build an electronic load....I would like to put wave form input and see the respose of the power supply circuits on changing load...somewhat like this...i think the concept works virtually...all i need to do is choose the correct component ...thanks

 

OK, sounds reasonable. I've made something similar for testing the step response of switching power supplies. Except that I used a power resistor for the load, and just turned the MOSFET on/off with a signal generator.

Just bolt your mosfet to a large heatsink (LARGE!) and blow a fan on it.

 

Running the mosfet in a linear region is not a very good idea, compare the thermal mass of the transistor die with the thermal mass of a large heatsinkable resistor. With such power I suspect the life of the transistor will be very hard and short.

 

OK, sounds reasonable. I've made something similar for testing the step response of switching power supplies. Except that I used a power resistor for the load, and just turned the MOSFET on/off with a signal generator.

Just bolt your mosfet to a large heatsink (LARGE!) and blow a fan on it.

one short question ...m not sure if the mosfet is operating in saturated or linear on the above ciruit...till now i assumed i was operating mosfet on saturated mode...the opamp rail to rail is -15 and +15

 

No, the transistor will be mostly in linear mode because the opamp tries to maintain the set current through the sense resistor.

 

No, the transistor will be mostly in linear mode because the opamp tries to maintain the set current through the sense resistor.

Does this somehow mean that the opamp will be changing the resistance of mosfet constantly...??

 

Basically the opamp drives the mosfet with such voltage to make the current through the sense resistor equivalent to the voltage on the positive input. So if the input or the mosfet power supply keeps changing, then the mosfet gate voltage will change as well.
The mosfet will go into saturation only when the control voltage is so high that the mosfet supply cannot output any more current.
Also during the transient in the control voltage the mosfet might be saturated for a short time until the loop stabilizes due to parasitic inductance and capacitance.

 

Think about it this way ...

You have a 20V power supply. You have a current sense resistor with a 195mV drop across it. Where is the rest of the 19.825V dropped?

Across the MOSFET of course! This puts it in its linear region, dissipating lots of power. LOTS!

 

The MOSFET saturation region

is pretty much the opposite of the BJT saturation region.

This issue comes up here fairly frequently.
Due to context, we probably all know what is meant by "saturation" and "linear" in this thread. Still, definitions are important.

 

Basically the opamp drives the mosfet with such voltage to make the current through the sense resistor equivalent to the voltage on the positive input. So if the input or the mosfet power supply keeps changing, then the mosfet gate voltage will change as well.
The mosfet will go into saturation only when the control voltage is so high that the mosfet supply cannot output any more current.
Also during the transient in the control voltage the mosfet might be saturated for a short time until the loop stabilizes due to parasitic inductance and capacitance.

Bravo Kubeek....Every line was a important note to me...every line...thanks a lot..

 

One more quesiton:
Form wikipedia:
1. Triode mode or linear region (also known as the ohmic mode)
When VGS > Vth and VDS < ( VGS – Vth )

2. Saturation or active mode
When VGS > Vth and VDS > ( VGS – Vth )

As JMac said...when the load voltage is 20[V] and suppose the gate voltage is 15[V], does that mean that the mosfet is in saturation..according to wiki...and again ...according to JMac...how could be the voltage drop be about 19.835[V], cause 22[A]*Rdson=22[A]*20[mohm]=440[mV]...

I am a little bit confused ..could you please calrify..a bit...

 

One more quesiton:
Form wikipedia:
1. Triode mode or linear region (also known as the ohmic mode)
When VGS > Vth and VDS < ( VGS – Vth )

2. Saturation or active mode
When VGS > Vth and VDS > ( VGS – Vth )

As JMac said...when the load voltage is 20[V] and suppose the gate voltage is 15[V], does that mean that the mosfet is in saturation..according to wiki...and again ...according to JMac...how could be the voltage drop be about 19.835[V], cause 22[A]*Rdson=22[A]*20[mohm]=440[mV]...

I am a little bit confused ..could you please calrify..a bit...

Your question should be "How can Rds(on) be 20mΩ when Vds=19.835V and Id=22A?".
The answer is, Rds(on) is the slope (dV/dI) of the V-I curve in the ohmic region. Rds(on) is only relevant if the transistor is operating in the ohmic region, i.e., Vds<≈1V. Your transistor is operating in the saturation region. It has ≈20V across it.
There are only 3 elements between +20V and ground: The inductor, which has no resistance, the transistor, and the sense resistor. How do you think that 20V is distributed? As Mac said, the transistor has≈19.9V across it, and the resistor has ≈200mV across it.

See the attachment.
Keep in mind that this is a simulation. Every MOSFET is different, even those that have the same part number. Your operating point will be the same as that shown on the attachment, except the Vgs may be different.

 

Your question should be "How can Rds(on) be 20mΩ when Vds=19.835V and Id=22A?".
The answer is, Rds(on) is the slope (dV/dI) of the V-I curve in the ohmic region. Rds(on) is only relevant if the transistor is operating in the ohmic region, i.e., Vds<≈1V. Your transistor is operating in the saturation region. It has ≈20V across it.
There are only 3 elements between +20V and ground: The inductor, which has no resistance, the transistor, and the sense resistor. How do you think that 20V is distributed? As Mac said, the transistor has≈19.9V across it, and the resistor has ≈200mV across it.

See the attachment.
Keep in mind that this is a simulation. Every MOSFET is different, even those that have the same part number. Your operating point will be the same as that shown on the attachment, except the Vgs may be different.

Hi Ron,
Thanks for the help. I think I was following totally different way of understanding MOSFET now..but now I gathered some conclusion...would you please see if they are correct....

1. MOSFET holds its RDSon value only in the linear region...where is acts as constant resistor allowing ohms law....I proportional to V.
2. When outside linear region, eg. Vds=15[V] and Vgs=10[V]
Then, Vds>Vgs
Mosfet is saturated
(what I was thinking before was, when the MOSFET is saturated, it has its RDSon)
(Which is not true...I think now)..
At saturation MOSFE has higher resistance.......

In my case...
Vload=20[V]
Id=22[A]
Vrsense=200[mV]
Vmosfet=Vds=19.8[V]
Vgs=10[V]
Vds>Vgs=>saturated
Resistance of mosfet = Vds/Id=19.8/22=0.9[Ohms]=900[m Ohms]

 

nepdeep,

I'll let you continue your general mosfet discussion with Ron. But for your specific circuit...

You have an op-amp driving the gate of your mosfet in a closed loop. The op-amp will drive the gate of the mosfet to WHATEVER voltage is required to make the inverting and non-inverting inputs of the op-amp equal.

You are therefore setting a small voltage across your sense resistor in order to set your load current. If you have the small voltage across your sense resistor, and a 20V power supply on the drain of the mosfet, obviously the rest of the voltage (20V - Vsense) is dropped across the mosfet. This is certainly not in the saturated region, its operating linearly, like a voltage controlled variable resistor.

 

nepdeep,

I'll let you continue your general mosfet discussion with Ron. But for your specific circuit...

You have an op-amp driving the gate of your mosfet in a closed loop. The op-amp will drive the gate of the mosfet to WHATEVER voltage is required to make the inverting and non-inverting inputs of the op-amp equal.

You are therefore setting a small voltage across your sense resistor in order to set your load current. If you have the small voltage across your sense resistor, and a 20V power supply on the drain of the mosfet, obviously the rest of the voltage (20V - Vsense) is dropped across the mosfet. This is certainly not in the saturated region, its operating linearly, like a voltage controlled variable resistor.

Yes I got about the opamp concept....it drives its output ....Votage at gate so that it can equalize the inverting and non-inverting input..isnot it...okey voltage drop is kind of constant at sense resistor then....even if the load voltage changes...only thing i want to know now is....what is the resistance of mosfet on saturation VDS>>VGS...or am i not getting term saturation...i am studying about it everywhere....sometimes i seem to understand...but now....
Frankly I am totally confused now....could you please comment on my conclusions a bit...is the information provided by wiki...not correct the...that when Vds>Vgs and Vgs>vth it operates on saturated region...my case is similar...isnot it...y