from what i have been reading power Mosfets absorb energy during avalanche operation.
However the sources i am using are slightly confusing
Need help breaking it down.
How does a power mosfet absorb energy?

 

Nothing can absorb energy without heating up.

 

i know that
heatsink can dissipate the heat
what i would like to know is the operation of a mosfet when its absorbing energy.

 

When you turn on the bjt, the Vce will be around 0.2V, but when you turn on the Mosfet there is different, whatever it is a N Channel or P channel, because the mosfet has different Impedance inside between DS, so the Vds is not regular as bjt, the Vds depends on the Impedance of Rds and the current flow through DS when the mosfet is a N channel and the current flow through SD when the mosfet is a P channel.

The Vds will be like this Vds = I * Rds, and the Watts = Vds * I, that's what the mosfet absorb energy and transfering to the heat.

Different MOSFET has different Rds, so the Vds is different, you can check some of the Rds of mosfets.

 

Ok say it is an N channel Mosfet the IRF150 specifically
The current flow from drain to source is between 0.1 and 4.1
Rds of IRF150 is 0.055ohm

so could you please help me explain how the mosfet absorbs energy in a circuit if the Vds at t=0 is 0.1 and at t=1ms increases to 4.1
Then a few microseconds later drops back down to 0.1

How would the mosfet absorb energy in that situation

 

Ok say it is an N channel Mosfet the IRF150 specifically
The current flow from drain to source is between 0.1 and 4.1
Rds of IRF150 is 0.055ohm

so could you please help me explain how the mosfet absorbs energy in a circuit if the Vds at t=0 is 0.1 and at t=1ms increases to 4.1
Then a few microseconds later drops back down to 0.1

How would the mosfet absorb energy in that situation

According to the given data from you to calculate the following values.

Vds = (t=0) → 0.1V → (t=1ms) → 4.1V → (x uS) → 0.1V

(t=0,0.1V) ==> a low Volts on Vgs and a current(0.1V/55mΩ=1.818A) flowing through Rds
(t=1ms,4.1V) ==> a high Vlots on Vgs and a current(4.1V/55mΩ=74.545A) flowing through Rds
(t=x uS,0.1V) ==> a low Volts on Vgs and a current(0.1V/55mΩ=1.818A) flowing through Rds

So there are something wrong about the given data of current and voltages.

 

Sorry i dont understand what is wrong please explain
can anyone else clarify

 

Sorry i dont understand what is wrong please explain
can anyone else clarify

See the V and I data that what you given and the values of current that what I calculated.

 

I only gave you current data 0.1A and 4.1A and the RDS on which is 0.055
you did a few calculations
which i dont understand
What i would really like is for you to please explain to me how a mosfet actually absorbs energy

 

It's about Watts Law: P = IE The current through the mosfet times the voltage across it creates heat. The mosfet absorbs energy and turns it into heat the same way a resistor does, or even a sausage connected to an AC power line. There is nothing magical about amps times volts in a mosfet.

 

charming!!....:/
could you please explain what the whole avalanche operation in a mosfet is about
does that have anything to do with how it absorbs energy?

 

charming!!....:/
could you please explain what the whole avalanche operation in a mosfet is about
does that have anything to do with how it absorbs energy?

Please google Ohm's Law.

Ohm's Law:
E = I * R or V = I * R
I = E / R or I = V / R
R = E / I or R = V / I

W = E * I or W = V * I

Ohm's Law Calculator.

 

Jesus people i know Ohm's law that is not what i am asking at all
Avalanche operation does that have anything to do with the mosfet absorbing energy
yes or no
I know the power dissipated is current through the mosfet* voltage across it
Power dissipated is energy absorbed

Are you saying the mosfet avalanche operation mode has nothing to do with the Mosfet ability to absorb energy
if so please just say yes

based on the attached schematic could you please tell me how much energy the mosfet absorbs during the fault condition

 

Jesus people i know Ohm's law that is not what i am asking at all
Avalanche operation does that have anything to do with the mosfet absorbing energy
yes or no
I know the power dissipated is current through the mosfet* voltage across it
Power dissipated is energy absorbed

Are you saying the mosfet avalanche operation mode has nothing to do with the Mosfet ability to absorb energy
if so please just say yes

based on the attached schematic could you please tell me how much energy the mosfet absorbs during the fault condition

What do you mean "fault condition", does the mosfet open or shorted?

If you want to reduce the heat, the one is to choose some other mosfets has less Rds as bolow, also you need adding the heatsink:

FDP5800,Nch,80V/60A,4.6mΩ.
FDP8440,Nch,40V/80A,2.4mΩ.

 

the mosfet is to turn off when the current in the 2.2k ohm increases past 9.09mA
Its at 9.09mA when the switch in series with the 47ohm resistor is open

my question is not about how to reduce the heat
i already attached my mosfet to a heatsink

i really just wanted to understand the process by which a mosfet absorbs energy

same way a capacitor can absorb energy by charging
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html

 

I found this App Note.

Does it help?

 

I read that already
Its quite short
have u had a look at it?
from what i understand
avalanche is when max drain to source voltage when mosfet is off has been exceeded.

what i am unsure of is how a mosfet would for example absorb energy during a fault condition. As per my attached circuit

 

I've no experience with what you're asking. Just trying to give your thread a boost. Sorry.

As a suggestion, you could consider calling/emailing an application engineer of the manufacturer of the part in question.

 

I still don´t quite see what you´re asking. The energy lost on the mosfet simply gets converted into heat, and it doesn´t matter whether this is through normal operation or avalanche breakdown. The mosfet has some thermal capacitance, so the energy is stored there and the temperature of the mosfet rises. If you ask it to store this energy too often the temperature will rise high, possibly with hot spots in the structure, ultimately causing the mosfet to fail.

 

You focus on the absorbs energy, but in your circuit that you need is adding a rectifier diode in parallel with the motor, and you have to reverse the polarity, and the negtive is conecting to D of mosfet, the positive connecting to the another pin of motor.

where do you want to protect when you using the rc integrator circuit? (Vgs?)