MOSFET junction temp. Calculation

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abc_de

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I am working on motor control acim motor(only speed control) with switching device IRF640 with DC bus 90v.
My question if what would be Tj temp when
ID= 7amps
Rdson: .18mohm
Ta= 40degree Celsius
Rth(j-a): 62c/w
P=IxIxR
P= 7x7x.18=8.82

Tj= Ta+Rth(j-a)XP
Tj= 40+62×8.82
Tj= 586.84

Please help me to clear
 

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    abc_de

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For 7A and 0.18Ω Rds(on), the power dissipated is 49*0.18 = 8.8W which requires a heat-sink.

If you use a MOSFET with an Rds(on) of <20mΩ, which will dissipate <1W, then you shouldn't need a heat-sink.
 

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    abc_de

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crutschow said:
For 7A and 0.18Ω Rds(on), the power dissipated is 49*0.18 = 8.8W which requires a heat-sink.

If you use a MOSFET with an Rds(on) of <20mΩ, which will dissipate <1W, then you shouldn't need a heat-sink.
Click to expand...
That actually makes more sense. Spend your money on a better MOSFET rather than a large chunk of metal
 

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Remember switching losses too. What is switching frequency.?
It does seem very low power for an induction motor.
Is there current limiting anywhere?...if not, then maybe better use IGBTs

Usually induction motors are supplied straight off the mains....no drive transistors in there.
Drive transistors are usually a feature of BLDC's....or BLAC's.......or PMSMs
 

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    abc_de

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Switching frq :2khz
Continuous current of acim is : 2amp
I am assuming 7amp over current condition under any fault thats why i want to know about the Tj under 7 amp condition
--- Updated ---

crutschow said:
For 7A and 0.18Ω Rds(on), the power dissipated is 49*0.18 = 8.8W which requires a heat-sink.

If you use a MOSFET with an Rds(on) of <20mΩ, which will dissipate <1W, then you shouldn't need a heat-sink.
Click to expand...
Due to cost reason i have only mosfet option
--- Updated ---

Can someone tell me Rth(j-c) : 1°c/w what it does mean.
IRF640> PD @25°c=Tc : 125w
Actually my purpose is to keep mosfet under 65°c : Tc temp.
If i have Tc of mosfet then can we know the Tj.
# ID : 7amp
# Rds(on) : .180*k(normalisation factor of TJ)
#Vds: ID*Rds(on)
# power losses due switching : not able to cal.
Just want to know about TJ if Tc is known
 
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abc_de said:
Can someone tell me Rth(j-c) : 1°c/w what it does mean.
Click to expand...
Did you try an internet search? I bet there come a lot ofrather detailed documents.

in short: Rth(j-c) : 1°c/w
* Rth --> thermal resistance
* jc --> junction to case (junction is the semiconductor chip. case is the metal sheet that may be mounted on a heatsink)
* 1°C/W --> per watt of dissipated power the temperature rise is 1 °C

Klaus
 

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abc_de said:
Just want to know about TJ if Tc is known
Click to expand...
1) You need to be concerned about more than Tc. The thermal resistance from case-to-ambient is the biggest contributor to your junction temperature rise. You're asking to keep Tc under 65 degrees. I don't know why you care at all about the case temperature, it's the junction temperature you need to be concerned about.

2) The IRF640 is obsolete.

3) Just doing a quick search I found IRFB4227PBF. It's 200V, about 20mOhms Rds. That's 1W of power to dissipate. With a Rthj-a of 62 deg/W, that's a junction temperature of 110 degrees at 45 degree ambient. Well below the 175 Tjmax. No heatsink. Costs about US$3.

4) There are even better parts if you search for them.
 

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I just want to know if ID: 7AMPS pluse of 10usec
Ta:40°c
Then what would be estimated(not exact) temp of TC &Tj

Or another to avoid all calculation i just follow this table.
 

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Hi,

10us really is a completely new situation.

10us of a IRF640, completely ON at 7A is no problem at all.
Within 10us the heat not even has the time to travel from inside chip to outside of case.
And due to the short time this is about no energy.

Example: consider an electrical stove... that can make water to boil for coffee. Now switch it ON for 1 second and measure the difference in temperature of the water. You can hardly measure a temperature difference on the plate - even without water.
And now you know that 10us is just 1/100000 of one second...

But when you say it is just 10us ... the next question arises:
Who controls this 10us? Is it controlled by an external switch/device/bjt/Mosfet, or is it this IRF640?
In case this IRF640:
The power dissipated at fully ON is about 1W.
But during switching there is some area (of - to us - unknown time) where the IRF is partly ON.
Worst case operation point is when V_DS is 45V, then on an ohmic load you also have half of the initial current: 7A/2= 3.5A.
Multiply this 45V x 3.5A you get a peak power dissipation of about 160W!!!

This is 160 times the power when completely ON!
Obviously you want to pass this time of high power dissipation as fast as possible. This time is called the "switching time"
Switching time depends on gate drive circuit .... but we don't know about this circuit.

Pleaee - when asking a question - give all informations you have immediately. Otherwise it's a big waste of time for us.
* give your schematic
* show your PCB layout
* tell what the "load" is
* tell us how you guarantee the limit of 7A
* tell us about ambient temperature
* tell us about the complete application

Klaus
 

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barry said:
NOW you start talking about a 10us pulse?I give up.
Click to expand...
Actually i am saying if current goes upto 7amp for short duration like 10usec
--- Updated ---

KlausST said:
Hi,

10us really is a completely new situation.
Click to expand...
Actually i just consider the situation if current goes high for 10usec
KlausST said:
10us of a IRF640, completely ON at 7A is no problem at all.
Click to expand...
Thats my answer.

I just ask because i have to protect mosfet from failure. Rise in Tj become cause of failure during sudden rise in current
 
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Hi
abc_de said:
Thats my answer.
Click to expand...
Maybe, maybe not.
I guess - from reading your posts - that you are not aware that there is a discrepancy between this "idealistic mathematical" approch and the "real circuit" conditions.

You may simulate this idealistic scene without problems, but still in a real circuit your MOSFET may overheat.

Klaus
 

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May be i could not put my question here proper way. Thats ture idealistic math and real circuit never same.
My problem was simple
Under id= 7amp (lets say now continuous load)
And room temperature 40 degree Celsius whats would be junction temperature and case temperature.
I do not know where i am not able to express
 

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That’s already been answered.

You fail to say what the repetition rate of your 10 us pulse is. 20us? 2 days? That will determine your average current, and, thus, average power. As Klaus already told you, a single pulse will cause no perceptible temperature rise. Can the device handle a single 7 amp pulse?(It’s in the data sheet). If so, then don’t worry about temp rise from a single pulse.
 

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