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MOSFET which can on fully at 5V

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ok means at 4A if Rds it 30mohm P= 0.48W

But what if I connect 30V across drain and source , will this calculation remain same?
If you connect 30V across drain and source, you will get either 0 Amps (if the gate is OFF) or 1000 Amps (if the gate is ON), which will instantly destroy any reasonable MOSFET. But this is not a condition you should ever see in your induction heater. I assume your circuit has a load in series with the MOSFET. That load will prevent the voltage across the drain and source from ever getting that high, except when the MOSFET gate is OFF. So while you can see 30V across the MOSFET and you can have 4 Amps flow through the MOSFET, you should never see both of these things at the same time.

So if your load ensures that the current will never be higher than 4 Amps, the MOSFET will never need to dissipate more than half a Watt.
 

Search Logic Level MOSFET

Vgs has many options with Vgs (th) thresholds that have reduced from 4V to 2.5 to <1 V

Specify RdsON Vds and logic level.

Do you know the "Sharade" game? Where one asks the other to find the answer by giving only one syllable at a time or use images instead of complete descriptions.

This is similar to how specifications are given before learning how to design.

Always define every critical parameter or define all the inputs, outputs and function.

I am guessing, the function is high voltage single-sided synchronous resonant switch to induction heater coil.

This then is followed detailed specifications for power, input and output voltage, frequency, thermal resistance, output capacitance, input capacitance, switch resistance
maximum junction temperature rise.

Often major choices are for ferrous or aluminum pots that affects switching frequency and device voltage.

Sorry I meant YOU to search for RdsON and Vgs, Vds until you learn to give everything else to finish the game.
halfbridge.JPG
 
If you connect 30V across drain and source, you will get either 0 Amps (if the gate is OFF) or 1000 Amps (if the gate is ON), which will instantly destroy any reasonable MOSFET. But this is not a condition you should ever see in your induction heater. I assume your circuit has a load in series with the MOSFET. That load will prevent the voltage across the drain and source from ever getting that high, except when the MOSFET gate is OFF. So while you can see 30V across the MOSFET and you can have 4 Amps flow through the MOSFET, you should never see both of these things at the same time.

So if your load ensures that the current will never be higher than 4 Amps, the MOSFET will never need to dissipate more than half a Watt.

According to me 1000A will never flow , if my gate voltage is 4.5V, because CHARACTERISTICS curve showing 20A saturating current at 4Vgs
 

Hi,

You have to know/calculate power loss.
Then in every device's datasheet you will find the thermal resistance values.
Multiply both values and get the temperature rise.

Klaus
 

Hi,

You have to know/calculate power loss.
Then in every device's datasheet you will find the thermal resistance values.
Multiply both values and get the temperature rise.

Klaus

Means if rds on is 30m ohm and current is 4A
Then power dissipation 0.48W

Maximum Junction-to-Ambient 90 °C/W

temperature rise 43.2 °C

Is that right ?
 

Hi,

Yes.
(With the same conditions like the datasheet says for 90°C/W)

Klaus
 

A rise in temperature is a rise in temperature (above the ambient temperature), it is not the temperature that is reached.
Note that the on-resistance of a Mosfet increases when its temperature increases which makes it hotter.
 

THermal rise is always above ambient and thermal resistance in 'C is linear like Ohm's Law.

Slew rate depends on mass and air flow, so very high air speed results in fast slew rate of temperature. Heat sinks range in thermal resistance and depends on surface area and thermal insulation of surface. A CPU heat sink can be as low as 0.1'C/W where a TO-220 small tab mount might be 8'C/W

Free air convection is assumed and not an enclosed box.
Using a heat sink from case to ambient then uses Rjc (device) + Rca ( sink)
WIth no heatsink Rjc+Rca=Rja where j= jcn, c=case,a=ambient.

Use it like Ohm's Law.
 

A rise in temperature is a rise in temperature (above the ambient temperature), it is not the temperature that is reached.
Note that the on-resistance of a Mosfet increases when its temperature increases which makes it hotter.

Means if temperature rises than on-resistance increases and when on-resistance increases temperature increase again. How to handle it?
Is this process continues ?
 

Means if temperature rises than on-resistance increases and when on-resistance increases temperature increase again. How to handle it?
Is this process continues ?
Why don't you look in the datasheet for your Mosfet that has a graph that shows the amount of on-resistance rise when the temperature rises? The on resistance rises about 1.5 times when it gets VERY hot. Then its heating also increases about 1.5 times. So you handle it by cooling it with an adequate heatsink and/or by limiting the current.
 

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