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Power Dissipation Pd

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crutschow

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The Pd tells you the power the device can dissipate (with a proper heat sink).
It you want to switch 200A then the power dissipated is 200A times the ON resistance of the switch when it's on, plus any switching losses as it switches states.
For a switch the 75V only comes into play during the switching rise and fall times.

Incidentally, your question has no answer since you give no power value.
 

JimB

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It you want to switch 200A then the power dissipated is 200A times the ON resistance of the switch when it's on
Err..

W = I^2 x R

JimB
 
What part? What data sheet? "switch" is what?
suppose the switch I have chosen is IPT015N10N5 and here is the datasheet.
The continuous drain current requirement is 200Amps.Voltage stress across switch is 60 volts. Is the switch choosen a good one in terms of power dissipation (Ptot=375 watt, given on page 3 of datasheet ). The ambient temperature would be 30 degrees. Thankyou
 

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The Pd tells you the power the device can dissipate (with a proper heat sink).
It you want to switch 200A then the power dissipated is 200A times the ON resistance of the switch when it's on, plus any switching losses as it switches states.
For a switch the 75V only comes into play during the switching rise and fall times.

Incidentally, your question has no answer since you give no power value.
Please refer to my above comment on Mr. Ronsimpson. Thankyou
 

crutschow

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Please refer to my above comment on Mr. Ronsimpson. Thankyou
The IPT015N10N5 has an ON resistance of 1.5mΩ max when turned on with a gate-source voltage of 10V.
It's dissipation with a 200A current is the 200^2 * 1.5mΩ = 60W.
That means the device will need a good heatsink to dissipate that power, but that is a surface-mount device which is difficult to attach to a heatsink.
You should find a device in a case that can be easily mounted on a heatsink (such as a TO-220 case).

But that doesn't include any switching losses.
How often will the transistor be switched on and off?
 
The IPT015N10N5 has an ON resistance of 1.5mΩ max when turned on with a gate-source voltage of 10V.
It's dissipation with a 200A current is the 200^2 * 1.5mΩ = 60W.
That means the device will need a good heatsink to dissipate that power, but that is a surface-mount device which is difficult to attach to a heatsink.
You should find a device in a case that can be easily mounted on a heatsink (such as a TO-220 case).

But that doesn't include any switching losses.
How often will the transistor be switched on and off?
Transistor on time will around 14us and turn off time will be 6us
 

crutschow

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Transistor on time will around 14us and turn off time will be 6us
Then the power dissipation will be 14/20 * 60W = 42W plus the switching dissipation.

How will you be driving the transistor gate?
 

ronsimpson

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How will you be driving the transistor gate?
This I haven't figured out yet...
When you get tired of this, The way to drive the transistor is easy when we know input voltage, output voltage. current, power, etc.
------------------------------
If your were working on a 20A power supply I would say start out with 2A or 1A. You have chosen 200A and that has some problems.
 
When you get tired of this, The way to drive the transistor is easy when we know input voltage, output voltage. current, power, etc.
------------------------------
If your were working on a 20A power supply I would say start out with 2A or 1A. You have chosen 200A and that has some problems.
What sort of problems
 

ronsimpson

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You can't build this on a bread board.
Most people have not make a printed circuit board that will work at 200A.
You might have 40 watts of power to get out of the transistor and put into air.
Most DC to DC power supplies have a coil/inductor and you need special wire for 200A.
You will have a hard time to solder to the PCB. You will need a 200 watt soldering iron.
 

unclejed613

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here's a "f'rinstance": 200A on a 2oz pc board trace requires a trace width of 8 inches (224 mm).
 

ronsimpson

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here's a "f'rinstance": 200A on a 2oz pc board trace requires a trace width of 8 inches (224 mm).
Last time I did something like this; I used a 4 layer board and ran the same traces on all layers. The outside two layers are 2 oz and the inside two layers are 1 oz. On the "hot" traces I did not have solder mask and built up a thick layer of solder. I think the solder does not reduce the resistance much but it seems to help more as a heat sink.
This is not my board. The silver area will be covered with a thick layer of solder.
 
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rjenkinsgb

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Or to put it another way - a copper cable to carry 200A needs to have a cross-sectional area of about 70 mm^2
Roughly 8.5 x 8.5mm of copper, if it were square.

Just an insulated connecting wire is going to be as thick as your little finger.
http://docplayer.net/docs-images/43/16516288/images/page_6.jpg

That's not allowing for skin effect, which means larger cables at higher frequencies..
https://en.wikipedia.org/wiki/Skin_effect


At just 10KHz, the majority of current flow occurs within the outer 1mm of the conductor, no matter how big it is.
That "skin" region has got to have enough cross-sectional area to carry the current without excess heat production.

You are in to the realms of litz wire, which is very expensive, especially in large sizes.
https://wiretron.com/litz-wire/

Any high-current, high-frequency inductors would need to be wound with that type of wire


High current systems are a specialised field.

High frequency systems are another specialised field.

Combining the two successfully requires a lot of experience with both.

You absolutely need decent test equipment as well, such as a reasonable oscilloscope and a good multimeter - plus preferable a clip-on current probe for the scope, when working with high power switching supplies, so you can directly view current waveforms as well as voltage.



Don't give up the idea, but start small!
You _will_ destroy a lot of parts experimenting with switched-mode circuits, so use small cheap ones initially, that do not matter much and don't cause life-threatening explosions...

[And as Ron says, you need a bit bigger iron when working with large high-current cable & component joints.. The "small" one in the photo is one of our standard 50W temperature controlled bench irons..]

Irons.JPG
 
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