power supply temperatures

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[bobsmith1492]

Components in an electrical circuit are not perfectly efficient, so they heat up. They also have a maximum temperature. The hotter the ambient temperature, the hotter the components will get. When capacitors, in particular, run at or above their specs, they degrade more rapidly. Transistors also will run into problems like electromigration where the wires inside wear out.

However there is not a hard line where if you run for a certain amount of time at a certain temperature it will fail. If you plot expected life span vs. temperature, there is a curve; up to the rated temperature or so, the expected life is very long. Over that, the life expectancy starts to drop exponentially.

That said, 100C is not very hot for a transistor. 150C junction temperature ratings are common, but you don't want to be that hot. I've designed circuits to run in 120C ambient temperatures (in transmission fluid); we tried to keep the component internal temperatures below 130 or 140.

Electrolytic capacitors are typically the weak link in power supplies. More expensive capacitors can run at higher temperatures but you can save money on the supply but using cheaper, 85C-rated parts.

 

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[bobsmith1492]

What I know about transformers is the winding insulation is the main limiting factor. That, and the core will saturate at lower currents which causes, naturally, more heat.

As far as the heatsink temp, it depends completely on the design. That's really hot and I wouldn't run anything at those temps but there's no technical reason why it couldn't work.

Of course, the 25C rating means the air is 25C, not the heat sink. The temperature difference between the air and the heatsink is caused by the amount of power being lost in the supply.

You can determine the output power capability of a supply by the following:

Pout = Pin - Pwaste -> (efficiency = Pout/Pin)

You want to limit Pwaste so that the supply doesn't overheat.
Pcomponent = (Tcomponent-Tambient) x Theta(component-ambient)

Where Pcomponent is the amount of power a transistor or transformer can dissipate, Tcomponent is the max temperature allowed, and Theta(component-ambient) is the thermal resistance from the part to the air. Theta is the standard symbol for thermal resistance.

So, you set Pwaste = Pcomponent:

Pout = Pin - (Tcomponent - Tambient) x Theta(component-ambient).

From these equations you can see several things that affect the supply output power capability:
1. Efficiency of the supply: higher efficiency means less heat wasted in the supply so it heats up less and parts are not stressed as much
2. Max temp ratings of components: higher temp ratings mean they can dissipate more power, allowing the supply to provide more power
3. Thermal resistance: better cooling and lower ambient temperatures mean the supply can waste more power while providing more output power.

 

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[William Gaatjes]

On cheaper supplies I noticed the transistors share the same heatsink. Is it common in industry to cut costs and have transistors with different case potentials attached to the same heatsink with thin mica as insulation? Or is this an unacceptable risk and all transistors attached to a heatsink are at the same case potential?

Unless you have a really under the counter power supply, it is the case that power supplies are designed and tested. There are minimum regulations and tests that any power supply must pass succesfully before it can be legally sold. When all transistors/ rectifierdiodes are mounted on a single heatsink, the insulation must be able to withstand the voltage. With mica it can be a few thousand Volts at a millimeter thick or 1/25 inch before the breakdown voltage is reached. The breakdown voltage is also known as dielectric strength. So no, it is not an unacceptable risk. Just do not toy with it if you feel like doing that. Live power is life threatening !

http://www.allaboutcircuits.com/vol_1/chpt_12/8.html

Insulator breakdown voltage


The atoms in insulating materials have very tightly-bound electrons, resisting free electron flow very well. However, insulators cannot resist indefinite amounts of voltage. With enough voltage applied, any insulating material will eventually succumb to the electrical "pressure" and electron flow will occur. However, unlike the situation with conductors where current is in a linear proportion to applied voltage (given a fixed resistance), current through an insulator is quite nonlinear: for voltages below a certain threshold level, virtually no electrons will flow, but if the voltage exceeds that threshold, there will be a rush of current.

Once current is forced through an insulating material, breakdown of that material's molecular structure has occurred. After breakdown, the material may or may not behave as an insulator any more, the molecular structure having been altered by the breach. There is usually a localized "puncture" of the insulating medium where the electrons flowed during breakdown.

Thickness of an insulating material plays a role in determining its breakdown voltage, otherwise known as dielectric strength. Specific dielectric strength is sometimes listed in terms of volts per mil (1/1000 of an inch), or kilovolts per inch (the two units are equivalent), but in practice it has been found that the relationship between breakdown voltage and thickness is not exactly linear. An insulator three times as thick has a dielectric strength slightly less than 3 times as much. However, for rough estimation use, volt-per-thickness ratings are fine.





Material* Dielectric strength (kV/inch)
===========================================
Vacuum ------------------- 20
Air ---------------------- 20 to 75
Porcelain ---------------- 40 to 200
Paraffin Wax ------------- 200 to 300
Transformer Oil ---------- 400
Bakelite ----------------- 300 to 550
Rubber ------------------- 450 to 700
Shellac ------------------ 900
Paper -------------------- 1250
Teflon ------------------- 1500
Glass -------------------- 2000 to 3000
Mica --------------------- 5000


* = Materials listed are specially prepared for electrical use.


REVIEW:
With a high enough applied voltage, electrons can be freed from the atoms of insulating materials, resulting in current through that material.
The minimum voltage required to "violate" an insulator by forcing current through it is called the breakdown voltage, or dielectric strength.
The thicker a piece of insulating material, the higher the breakdown voltage, all other factors being equal.
Specific dielectric strength is typically rated in one of two equivalent units: volts per mil, or kilovolts per inch.

EDIT:

I have not looked at it fully, but if you (or anybody reading this) are interested in electronics, this site seems like a good place to start...

http://www.allaboutcircuits.com/vol_1/index.html

 

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[bobsmith1492]

Quote:
Originally Posted by bobsmith1492 View Post
What I know about transformers is the winding insulation is the main limiting factor.
How is the insulation affected by temperature?

If it gets too hot the insulation breaks down - "melts," shorting out the windings.

Quote:
Quote:
Originally Posted by bobsmith1492 View Post
That, and the core will saturate at lower currents which causes, naturally, more heat.
I don't understand what you are trying to say. Can you explain this more clearly?

A transformer is just a coil that is used to create a magnetic field. There is typically a second winding that picks up that field and converts it back to electricity. There is also usually a core, made of a material like iron or a ferrite that channels the magnetic field like a wire does electricity. However it can only channel so much magnetic flux before it saturates and pinches off the flow of magnetic flux (is that redundant??). Once you hit that point, you can't push much more energy through the transformer; it is saturated. That point reduces in typical materials, like ferrites, as the temperature goes up.

OK. Do you know the answer to the question: Is it common in industry to cut costs and have transistors with different case potentials attached to the same heatsink with thin mica as insulation?

This happens all the time in lots of different electronics, not just computer PSUs. Open up a car amp and you'll see whole rows of them clamped onto the edge of the case.

 

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[bobsmith1492]

Look at inductance vs. current on page 2: http://www.coilcraft.com/pdfs/mlc.pdf That shows saturation, where the inductance rolls off with increased current.

Look at page 29 to see saturation flux density roll off with temperature in different types of core materials: http://www.digipower.it/magnetics/Part2.pdf

 

[William Gaatjes]

So if the insulation fails, the device will short? I have also seen transistors bolted onto the metal heatsink. I believe I saw a car amp with bolted on transistors.

That is because you can buy transistors that have a electrically isolated casing (but have a higher thermal resistance). And you can buy transistors with a metal casing (and afcourse have a lower thermal resistance). It all depends on how much power the transistors need to dissipate. If the amount is low, get a version with an electrically isolated casing. If the amount of power dissipation is too high for an isolated version get one with a metal casing. If the amount of power dissipation is still to high in combination with mica plates or other insulators while using non isolated version, do not use mica plates but use a thermal paste and connect the transistor directly to the heatsink. If the last case is the case, and the transistor caries a high voltage you must use a seperate heatsink. If it i still not enough, use active cooling. In the end it all comes down to preventing the silicon in the transistors from melting and becoming a conductor like a copper wire. Because then your power supply (or amplifier)starts behaving differently.


It is a trade off of costs of components , the best design for a certain price.

As i mentioned before, there are rules and regulation to prevent dangerous situations. Since electronics is omnipresent, people take electronics for granted but it is still a field of engineering.


An isolated transistor :
http://uk.farnell.com/toshiba/2sk3569/mosfet-n-600v-to-220sis/dp/1300779
datasheet :
http://www.farnell.com/datasheets/55583.pdf
thermal resistance, channel to case : 2,78 c/w.

A not isolated transistor :
http://uk.farnell.com/fuji-electric/2sk3682-01/mosfet-n-to-220ab/dp/1208660
datasheet :
http://www.farnell.com/datasheets/39857.pdf
thermal resistance, channel to case : 0,463 c/w.


When you look at the thermal resistance number, you will see that the non isolated is better at conducting heat away. These transistors have similar dimensions. Both have a case called TO-220.

EDIT :
Thermal resistance means the efficiency of heat transfer. The higher the number, the lower the amount of heat energy you can transfer in a given time.


look at this transistor :
It is as big as a mars bar.

http://uk.farnell.com/fuji-electric/1mbi200s-120/igbt-module-1200v-200a/dp/1208669

This baby can dissipate in excess of 1000 watts of heat when properly cooled.

http://www.farnell.com/datasheets/39743.pdf

The reason why this transistor can dissipate more is because it's thermal resistance is so much lower , because it has a bigger surface to transfer heat.
thermal resistance : 0,096 c/w. (for the transistor only)
thermal resistance : 0,260 c/w.(reverse current diode)

thermal resistance, contact surface : 0,0125 c/w. (with thermal paste).
You have to add all these numbers for specific scenario's.