mosfet junction to ambient temperature

[cdez]

hello,

on IRFB7540 (and many others) datasheet it indicates junction to ambient temperature = 62C/W


what does this mean ?

does it mean that if I touch the mosfet package I would feel 62 * P(dissipated) degrees ? 8-O

let's say I dissipate only 3 watts so I would have 62*3 = 186 degrees more than Tambient ?
I am sure it's more complicated than that but "62 degrees per watt" seems pretty explicit to me :-?

And do I need a heatsink if I let 25amps & 12V through this mosfet (IRFB7540) drived by high duty cycle 20khz pwm ?


thanks

 

[drium]

Hi cdez!

All Mosfets have a internal resistance between Source and Dren. direct proportional to gate tension. if the current flow in the source-dren. are ,for example, 2A and the RDSon(internal resistance between source and dren. when tension between gate to source tension are > VGTthd ) are 0.1ohm the P(dissipated) are W= I^2*R... in this case W= 4*0.1= 0.4W.

 

[D.A.(Tony)Stewart]

hello,

on IRFB7540 (and many others) datasheet it indicates junction to ambient temperature = 62C/W what does this mean ?
does it mean that if I touch the mosfet package I would feel 62 * P(dissipated) degrees ? 8-O
let's say I dissipate only 3 watts so I would have 62*3 = 186 degrees more than Tambient ?
And do I need a heatsink if I let 25amps & 12V through this mosfet (IRFB7540) drived by high duty cycle 20khz pwm ?
thanks

Each physical interface has a thermal resistance which adds in series. For J=Jcn, C=Case, S=Sink(heat) A= Ambient

For J->C->-S->A serial paths
RΘJC + RΘCS+ RΘSA = RΘJA

Consider a CPU sized heatsink but no forced air ( big ) if you intend to use max power dissipation
RΘSA ~ 1°C/W

0.95 °C/W Max + (TO-220) 0.50 Min + 1°C/W = 2.5 °C/W which is almost best-case passive cooling, while No heatsink 65 °C/W is worst case assuming free air flow. If it was enclosed, with no air flow (Insulated) RΘJA) would be even worse.

Design Approach
A good conservative design considers worst case ambient, max junction temp of 85'C and chooses heat sink according to worst case power dissipation which is determined by duty cycle * pulsed I²RdsOn

Thus if Ambient is 40°C Max then choose a heatsink according to 45°C Temp rise and then design heatsink requirements for °C/W based on worst case W.

Forced air at 2m/s can reduce heatsink RΘJA /5 to maybe /10
For low power copper surface area RΘsA may be computed for surface mount types.

 

[FvM]

And do I need a heatsink if I let 25amps & 12V through this mosfet (IRFB7540) drived by high duty cycle 20khz pwm ?

Simple answer, probably yes.

You should specify the duty cycle of the 25 A transistor current. If it's 100%, you get about 2.5 W conduction losses. Switching losses add up, very much depending on how you drive the switch and also on circuit inductance that might slow down switching respectively cause large overvoltages over the nominal 12V.

 

[cdez]

thanks for your answers

The duty cycle will mostly be around 90%.
Does the duty cycle changes the losses noticeably ? like +- 1 watt or more ?

I have another question:
How can the TO220 would allow 25 amps knowing that the pins have around only 0.5mm² section, 25 amp requires something 4mm² to 6mm² section ( around 10 awg )

 

[D.A.(Tony)Stewart]

thanks for your answers

The duty cycle will mostly be around 90%.
Does the duty cycle changes the losses noticeably ? like +- 1 watt or more ?

I have another question:
How can the TO220 would allow 25 amps knowing that the pins have around only 0.5mm² section, 25 amp requires something 4mm² to 6mm² section ( around 10 awg )

You will need 3mm flat braid wire on board to connectors to reduce wire resistance < 1 millohm.

RdsOn=5.1millohm @10Vgs means W=90%*625*5.1m=2.9W

then add Gate power 10% ( Rule of thumb for PWM) to get 3.3W

So for a 45deg C max rise on junction, for RΘJC + RΘCS+ RΘSA = RΘJA
RΘJA= 45 °C/3.3 W = 13.6 °C/W thus your heat sink needs to be RΘSA=10 °C/W

Got it?

 

[FvM]

90% means you have nearly full conduction losses Pv = 0.9*Id²*Rdson. Switching losses add on, but shouldn't be high with reasonable gate drive current.

The relative thin TO-220 pins can carry high currents because they are short and the generated heat is mainly conducted. The total pin resistance per pin is in a 100 µOhm range and only contributes a smaller part of total conduction losses.

 

[cdez]

You will need 3mm flat braid wire on board to connectors to reduce wire resistance < 1 millohm.

RdsOn=5.1millohm @10Vgs means W=90%*625*5.1m=2.9W

then add Gate power 10% ( Rule of thumb for PWM) to get 3.3W

So for a 45deg C max rise on junction, for RΘJC + RΘCS+ RΘSA = RΘJA
RΘJA= 45 °C/3.3 W = 13.6 °C/W thus your heat sink needs to be RΘSA=10 °C/W

Got it?

Yes thanks.
I have bought ~15mm*20mm to 220 heatsink from china ( poor quality I guess ), I hope they have at least 30 to 40 °C/W so I would have around 90°C rise on junction, it's quiet hot :/

 

[D.A.(Tony)Stewart]

add on a 35mm fan