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[SOLVED] POwer dissipation in datasheets

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baby_1

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what is power dissipation meaning?

becase in all mosfet and transistor datasheet the manufacture write

Maximum Characteric

PD=....

for example for irf3205
PD=200w

but id=110 and vds=55 so with calculate p=V*I , p i more than PD?

what is pd in datasheets and how can know the maximum power for devices in their datasheets?
 

Re: Power dissipation in datasheets

Is this the data you're referring to? http://www.irf.com/product-info/datasheets/data/irf3205.pdf
You have to read all the notes and conditions that apply to the data. The charts help to illustrate the relationships between the various factors.

In the example you cite, you give VDSS as 55V, but when you look on page 2 you see that this is a breakdown voltage measured with ID at only 250µA, not 110A.
 
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Yes of course,

according datasheet:

PD @TC = 25°C Power Dissipation 200 W
 

Re: Power dissipation in datasheets

The description of the TO-220 case at the top of the first page mentions that the package is good for 50W. The 200W figure is a maximum with adequate heat sinking and/or cooling. In other words, it's a 50W package by itself but in an air conditioned room with a big heat sink and fan on it, you can get it to 200W.
 
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so you assume that i use a heat sink for mosfet
and this component should drive a motor with 24 volt and 50 amper current subsequntly p=v*i , p=50*24=1200

that is so much more than power dissipation
 

Re: Power dissipation in datasheets

Most of your power is dissipated in the motor, not the mosfet. Try I²R using your load current and RDS(on) of 8mΩ (20W) or using data from the typical output characteristics chart, for 50A, VDS < 1V (P = I * V = 50W.

Someone who understands this far better than I will probably give you a really good answer later. Be patient.:smile:
 
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A heatsink is not perfect so the Mosfet will heat up which increases its on-resistance to about 12mΩ. Then with a current of 50A its heating will be 30W and a pretty big heatsink (and maybe a fan) is needed.
 

Thanks about all useful information
but when you see the page2 of datasheet ,its

"RDS(on) Static Drain-to-Source On-Resistance ––– ––– 8.0 mΩ VGS = 10V, ID = 62A"

their calculate rds and examine this mosfet at VGS = 10V, ID = 62A
p=V*I=62*10=620w

main problem is this power is more and can be dangerus for this mosfet with max pd=200w
 

Thanks about all useful information
but when you see the page2 of datasheet ,its

"RDS(on) Static Drain-to-Source On-Resistance ––– ––– 8.0 mΩ VGS = 10V, ID = 62A"

their calculate rds and examine this mosfet at VGS = 10V, ID = 62A
p=V*I=62*10=620w

main problem is this power is more and can be dangerus for this mosfet with max pd=200w
The Mosfet is not dissipating 620W! Ohm's Law calculates its 0.008 ohms and 62A to dissipate "only" (62A squared x 0.008 ohms=) 30.75W.
Its load is dissipating most of the power depending on its supply voltage. The 10V gate of the Mosfet dissipates nothing.

The spec says that the Mosfet has its chip at only 25 degrees C somehow (liquid nitrogen?) because its on-resistance increases as it heats which makes it heat even more. It can dissipate 200W only if its case is held to only 25 degrees C somehow which is almost impossible (maybe with liquid nitrogen?).
 
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    baby_1

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A term, that should be introduced to the discussion is switching operation of a MOSFET. You try to operate the MOSFET either on or off, like a mechanical contactor. But you have to care, that you don't leave the linear operation mode with a few milliohm Rds,on, e.g. by applying a short. Then the power disspitation can get very high and actually destroy the device within a few milliseconds.
 

according your explaintion for bjt for calculate power dissipation we can use P=VI and for mosfet p=R*I*I
?
 

P=V*I applies to all devices. But you have to put in the on-state voltage Vds,on of the device, not the supply respectively load voltage. Vds,on = Id*Rds,on in linear (ohmic) mode.
 
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