As Kubeek said the energy dissipated in the MOSFET is not affected by whether it's from an avalanche mode or normal operation. In all cases it's just the current through the MOSFET times the voltage across the MOSFET while it is carrying the current.

 

As Kubeek said the energy dissipated in the MOSFET is not affected by whether it's from an avalanche mode or normal operation. In all cases it's just the current through the MOSFET times the voltage across the MOSFET while it is carrying the current.

Ok, and the mosfet will absorb energy the same way whether it is on or off?
I am interested about how it absorbs energy while the mosfet is off. i.e.while it is stopping current from flowing through

 

You seem to be having difficulty with Watts Law. Power = Current times Voltage. When the current is zero, the power is zero. When the transistor is off, the current is zero. When the current is zero the power is zero. Under these conditions, the only way the mosfet will absorb energy is if you place something warm near it or put it in sunlight.

 

So does everyone agree with rude #12??
a Mosfet cannot absorb energy while it is off.

i.e if a mosfet is used to interrupt a circuit, the mosfet absorbs no energy during the period that is off.

is this correct.

ps: i have no difficult understanding ohm's law stop insisting that i do.

 

ps: i have no difficult understanding ohm's law stop insisting that i do.

It seems to me that this is not true.

a Mosfet cannot absorb energy while it is off.i.e if a mosfet is used to interrupt a circuit, the mosfet absorbs no energy during the period that is off.

Like any other piece of matter, a mosfet can absorb energy - for example if you heat it up with soldering iron, it will absorb thermal energy.

I guess you mean electrical energy coming from the circuit it is in. And we´re back to P=U*I. If the mosfet is off, i.e. no current is flowing, than the mosfet doesn´t absorb any electrical energy from the circuit (apart from a tiny leakage current of maybe a few μA).

However when you exceed it´s voltage rating there will be some current flow, hence it will absorb some energy, even though the driving circuit drives the gate to the normally would-be off state.

 

I was talking about the electrical energy coming from my circuit
yes those tiny leakage currents

i previously designed a circuit or rather to be more accurate Ron H helped me design a circuit
i just wanted to know if the mosfet could absorb the electrical energy when the mosfet turns off to interrupt a circuit.

So based on what you just said the energy the mosfet absorbs is the leakage current*Voltage the mosfet is operating at

 

Yes, but this holds only when the mosfet is truly off.
Looking at your circuit, I see a mosfet switching a motor, which will have large inductance. When the motor is on, there will be some current flowing through the mosfet. Now when you turn the mosfet off, the inductance of the motor tries to maintain this current, which causes the voltage on the drain rise very high until it reaches the breakdown voltage of something to complete the circuit, which is usually the mosfet. This is why you always need to use a diode connected across the motor or any other inductive load to provide an alternative path for the current and not to damage the switching device.

Like this: http://www.softwareforeducation.com/wikileki/index.php/File:MOSFET-Switch.gif
The diode needs to be rated for the current the motor is normally drawing.

 

The energy it would need to absorb is basically equal to the energy stored in the motor´s inductance - E=1/2*L*I^2

 

is it true that some power mosfets are actually designed to operate in that breakdown voltage state. i know some mosfets will get damaged if the voltage exceeds breakdown and current is allowed to keep flowing through. hence the need for a diode in parallel with the motor to provide alternative path.

What of those power mosfets which wont be damaged by the current flowing through
Would i diode still need to be connected in parallel with the motor

ps: you said it only holds true when the mosfet is truly off, what if the mosfet isnt truly off? Based on the circuit designed i can't get the mosfet to truly turn off. thats why i am asking

 

That circuit seems to be able to turn the mosfet off. Anyway if it is not off then you need to measure the voltage and current to see how much power is lost on it.

 

The energy it would need to absorb is basically equal to the energy stored in the motor´s inductance - E=1/2*L*I^2

I don't think so. That's the total energy that needs to go somewhere, but doesn't a significant fraction of it flow to ground when it flows through the broken-down MOSFET? The amount of energy the MOSFET has to handle is once again the reverse voltage across itself times the leakage current. That cannot be more than the energy of the collapsing field, but it can be less.

Maybe you're right - the MOSFET is probably the biggest obstacle by far, so I guess it does have to absorb most of the energy. Should've thought more before writing.

 

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

Folks, I designed that circuit and posted it in another thread.
That MOSFET never avalanches (the voltages are way too low), and in fact, never even gets warm, because it is subjected to less than half an Amp for less than 5 microseconds each time the switch is activated, and Rds(on)=4 milliohms. That's 700 microwatts for 5 microseconds, with an extremely low duty cycle.

Newbie, why are you suddenly concerned about energy absorption and avalanche in this transistor? Neither one is even a remote issue.

 

Really, every microwatt is caused by current flow times voltage. All you have to do is calculate the current flow times the voltage at that moment. You can choose a steady state flow at a useful rate, you can do the math for a microsecond of switching time, you can figure out the leakage current at some particular temperature and multiply that by the applied voltage, you can calculate the current during "avalanche" conditions and multiply that times the applied voltage. There is no magical exception to this rule where a sudden change causes the component to sympathetically generate heat all by itself.

ps, I don't think that was rude.

This is rude

 

Really, every microwatt is caused by current flow times voltage. All you have to do is calculate the current flow times the voltage at that moment. You can choose a steady state flow at a useful rate, you can do the math for a microsecond of switching time, you can figure out the leakage current at some particular temperature and multiply that by the applied voltage, you can calculate the current during "avalanche" conditions and multiply that times the applied voltage. There is no magical exception to this rule where a sudden change causes the component to sympathetically generate heat all by itself.

ps, I don't think that was rude.

This is rude

feel free to exit the thread. #12
at everyone else thank you.

@ronH i was just wondering how the mosfet absorbs energy during a fault.
I did install a heatsink but i noticed i didn't even need it so i was wondering why that is.

just wanted a quick explanation as to how mosfets absorb energy.

 

I'm new to Mosfets too, but what I understand about them goes like this. There are many types of Mosfets. They vary in the way they look, but the look is mostly a product of the power that the Mosfet can handle. TO92 for very low power, TO220 for more power. One of the most important aspects of a Mosfet is the ON Resistance, or RDS. A very low RDS is desirable, as it produces less of a voltage drop, and less heat. The best way to get the most power through a Mosfet is by using pulses, or PWM. Not all applications could utilize pulses, but when they can, the Mosfet can pass much power without the use of a heat sink. The best, a most costly Mosfets have a very low RDS. The best price I have seen on Mosfets is on EBay.
Will someone correct me if I am wrong. I am in a learning mode and I like to get my facts In order, it makes further learning much easier. I'm new to the forum,

Joe

 

Massimj, I don't think the PWM part is correct, when talking about heat. I to am very new to this. From what I understand the most heat is made in a mosfet during the linear mode of conduction. If you are using PWM it will be going through the twice for each pulse. Where if used in the on or saturated mode it never goes into linear mode, except when being turned on or off.

The other thing with mosfets is to make sure you use at least the gate voltage that is shown in the data sheet in the Vgs. Using too low a Vgs puts it in the linear mode. Using to high a Vgs is hard on the gate structure, for little gain in amperage output.

 

Thanks, I do understand. One point I am not clear on is about saturated, or Linear mode. I was understanding that the Mosfet was either full on, or full off, and could operate at very fast switching rates. So I did a Google search and see that Mosfets are very popular for audio amplifiers. DIYAudio was a good source for the info. Thanks for clarification.

Joe

Massimj, I don't think the PWM part is correct, when talking about heat. I to am very new to this. From what I understand the most heat is made in a mosfet during the linear mode of conduction. If you are using PWM it will be going through the twice for each pulse. Where if used in the on or saturated mode it never goes into linear mode, except when being turned on or off.

.

 

I was understanding that the Mosfet was either full on, or full off, and could operate at very fast switching rates. So I did a Google search and see that Mosfets are very popular for audio amplifiers.

These two things fit together. The MOSFETs can be operated as fast switches - not in their linear mode - so fast that the switching is not detected in the audio frequency range. MOSFETs do have a "linear", less-than-saturated range but it is relatively narrow compared to a BJT, for instance.

 

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The other thing with mosfets is to make sure you use at least the gate voltage that is shown in the data sheet in the Vgs. Using too low a Vgs puts it in the linear mode. Using to high a Vgs is hard on the gate structure, for little gain in amperage output.

Using the datasheet Vgs(threshold) value as a switching voltage is a common application error. You want the voltage much higher than the Vgs (th) voltage for proper switching. Normally you want a gate-source voltage equal or greater than the Vgs used for specifying the ON resistance (typically 10V for a normal MOSFET and 3 or 5V for a logic-level type). Of course you must always keep the voltage below the maximum Vgs rating.