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Author Topic: Burned FET  (Read 1224 times)
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By "turning LED reflectorsr" do you mean you're only turning the LED light on/off or are you rotating the reflector with a motor?

With LED reflector i mean only LED light. In our country we say reflector meaning light in case for mounting and rotary light for the one with motor smiley
To be exact, i am using that kind of LED light - LED reflector, without any motor -->

There's a difference between "turning a reflector" and "turning on a reflector" (or "turning a reflector on"), even in your country - isn't there?
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Now i putted another FET (IRL2505) in the same circuit and after more than hour it's almost completely cold. Without heatsink on it, everything is in completely sealed plastic box.


You should not have a problem with the IRL2505.  The dissipation would be around 100mw.
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Dirt Biker

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I saw that maximal Gate-to-Source voltage for IRL2505 is 16V while for IRL540 is 10V. Since i am powering the whole circuit with a little bit more than 12V, this could be problem?

No, the problem was you were dissipating about 800mW with the IRL540.  Since it was in a sealed plastic box, this allowed it to get hot.  Too hot.  Once it get's hot enough the Rds goes up rapidly, increasing the dissipation and making it even hotter.   Thermal runaway ensues and the device burns up.  This can happen really quickly.
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Dirt Biker

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There is huge argument about MOSFET thermal runaway:-

BJTs suffer from a property known as "thermal runaway." Thermal runaway happens because the conductivity of a BJT increases with temperature. Because transistors tend to heat up in proportion to current flowing through them this means that the conductivity and temperature of BJTs can increase exponentially. This can damage the BJT and makes designing circuits for BJTs more difficult. MOSFETs do not suffer from thermal runaway.

thermal runaway, ehow.com

Power MOSFETs typically increase their on-resistance with temperature. Under some circumstances, power dissipated in this resistance causes more heating of the junction, which further increases the junction temperature, in a positive feedback loop.

thermal runaway, wiki

Here we need an expert to show us a light.  

@OP, FET is normally referring as JFET. Your title "Burned FET" means "Burned Power JFET" to me and make me click it rightway. smiley-cool




« Last Edit: May 05, 2013, 01:06:35 pm by sonnyyu » Logged

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@sonnyyu,

I agree.  It is difficult to cause a power MOSFET to go into thermal runaway, under normal conditions, that is.  But operating a TO220 MOSFET at near a watt where it is not a major source of resistance in the circuit and providing no cooling at all is one way this can be achieved.

Looking at the normalized Rds(on) vs. temperature curve for an IRL540 and taking it piece by piece shows us that at 850mw, the temperature of the 540 will want to rise by 52 degrees in free air at an ambient of 25 degrees.  At 77 degrees (25 + 52) the Rds will rise by a factor 30%, increasing the dissipation to 1.1w, causing the temperature to rise to 93, causing the dissipation to go to 1.4W, causing the temp to rise to 111, causing....The curve actually just gets steeper and steeper.

And all that assumes free air with a fixed ambient temperature of 25.  Sealed in a plastic box, this will happen much quicker as the ambient temperature will rise, being heated by the MOSFET.

The self-stabilizing effect becomes prominent when the MOSFET itself is a major contributor to the circuit's overall resistance.  Then a 1.5 factor increase in Rds will reduce the overall current dramatically and actually lower the dissipation of the device.

The fact that the load in this case is the major contributor to the circuits resistance makes the matter worse since the IRL540 going from .08 ohms to .12 ohms will have little effect on the drain-source current, but a large effect on the power dissipated by the device.

Actually, the load here is quite likely going to be a constant current load too.
« Last Edit: May 05, 2013, 02:46:15 pm by DirtBiker » Logged

Dirt Biker

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Thermal runaway for BJT's is a big problem when they are paralleled (hence the emitter resistors in high current power linear supplies) because in getting hotter one will prevail in low resistance (due to manufacturing tolerances) and start to take more current so getting hotter so taking more current and so on.

mosfets don't suffer in this way so are easier to parallel
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I calculated power dissipation too late, after everything was burned smiley-grin

There's a difference between "turning a reflector" and "turning on a reflector" (or "turning a reflector on"), even in your country - isn't there?

You're right, that's happen when you're translating literally smiley-wink
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