ventilation of power supply?

[mealman]

I am not new to power but I am totally new to heat dissipation and ventilation.

I recently purchased KL- 7212($149/pcs) with 860W, 72VDC/12A.

I am planning to put it into a steel utility case and drill some holes for ventilation.

My question is, is a fan and heat sink for ventilation necessary for this type of power supply? Because I want to keep the design as simple as possible.

Thanks in advance

[pstret]

The problem with switcher? power supplies is to keep the electrolytic caps as cool as possible as they are the most likely cause of failures due to overheating. When ventilating a cased supply, use decent sized slots immediately above the hottest devices (usually power FETs, diodes and transformer). I recommend a minimum slot size of 5mm x 40mm and the more the better. If there is a safety issue with this size of slot (Probe test finger) then double skin the case 5mm apart in the hot areas and use staggered slots. Small holes or slots are not effective as air flow can actually be zero (check out Reynolds numbers). When you have cased the supply, check temps of hot bits on full load with thermocouples. Aim for max rise of 30C above ambient. Hope this helps. Paul

[apache405]

i would use a fan. a heat sink is nice, but that is typically handled by the frame of the PSU (power supply unit); i.e. the rectifier is connected to the chassis. before you put the PSU in the enclosure take some temperature readings with the unit running under normal load. then put it in the enclosure and repeat the process. the differnce in temperature will allow you to compute the amount of heat the PSU is making, as well as the air flow needed to remove the excess heat. one note, if the temp gets mush above 120deg F (48.9 deg C), shut the PSU down to prevent damage to the circuit card and components. hope this helps...

[mealman]

thx guys, it definitely helps

[bobs bots]

I'm not familiar with the power supply you mentioned, but power supplies are typically 90% efficient, so you will need to dissipate 86watts at full power.
You can actually calculate the air flow necessary to remove this amount of heat, (as the Cp is known and dT should be ~ 30C) (I'm too lazy to look it up at the moment)
Switchmode supplies are smaller and lighter (but not necessarily more efficient ) than unregulated linear supplies. Additionally linear supplies dissipate additional heat when they are regulating. At your power level you will almost certainly need a fan, typically an 80mm fan would be appropriate, easiest way is to use a hole saw and cover the hole with a matching finger guard for the fan. The fan should BLOW onto the hot components of the power supply (as turbulence increases the heat transfer coefficient). The air exit holes should be roughly the same total area as the fan hole,(~4000sqmm) but no hole smaller than say 10mm (i.e. around 50 holes at 10mm, 12 holes at 20mm). Alternatively, if you make the box as two U shapes, then you can leave a large gap between the halves. eg two gaps 10mmx200mm = 4000sqmm.
Useful facts and rules of thumb:
(a) Design temperatures for most equipment assume a 40C rise above a 40 ambient, so component temperatures ~ 80C. The air passing through the power supply will be heated by about 30C. Very few components have higher operating temperature, e.g. ceramic resistors (200c), some rectifier diodes(125C) and Thermistors(150C)
(b) Electrolytic capacitors (as mentioned in an earlier post) are usually the most unreliable component, and will last only 20,000hrs (just over two years) when operated at the design temperature (90C) and voltage. Lowering the capacitor temperature 10C will double the life. Increasing voltage 10% over the rating will halve the life.
(b2) Electrolytic capacitors generate internal heating due to ripple current and "ESR" (Equivalent Series Resistance), and at end of the 20,000hr time mentioned above, the ESR is deemed to have doubled. (which also increases the self heating!). This point is often overlooked by power supply designers, which means that if your power supply has marginal regulation and the capacitors run hot at day one, it can only get worse! You need to design your supply assuming it has the ESR of 20,000hr old capacitors.
(c) Thermal shutdown should operate at a heat-sink temperature of 105C, as the junction temp will be about 125C. Irreversible changes to the semiconductor begin to occur at junction temperatures of 150C.
(d) The surface area of heat-sink required in still air is typically ten times larger than that with fan assisted ventilation.
If you have some idea about the size of enclosure you envisage, then let me know and I can advise further.
Cheers, Bob T

[bobs bots]

Ahh ,
Here's some numbers, courtesy of Google, see http://www.nidec.com/aircooling/fantech.htm
about 0.6watts per cfm per deg, so with an average 80mm fan at 40cfm, and 86watts, the exit air will be ~ 5C hotter than inlet.
This suggests
(a) you can be sloppy about where you put the fan or
(b) you can get away with a 60mm fan at 20cfm and ~10c hotter exit air

Oh and another rule of thumb
(a) You get about 2m/s for most "typical" fans irrespective of area (up to double for screaming high flow variants) (and down to half for "low noise " variants). This assumes the air is blowing through a tube of the same area as the fan. Once inside your box the airflow will slow according to the cross sectional area.
(b) Most heat-sink manufacturers rate their fan cooled heat-sinks for 2m/s (This means you pretty much have to bolt the fan onto a heat-sink of the same size to get the rating)
(c) To get a real world heatsink performance you need to scale according to the actual air speed , so as an example, assume 20cfm from a 60mm fan, area ~2500sqmm, blows into a box 100mm x 250m cross section (area=25000sqmm) , so air flow drops from 2m/s to 0.2m/s. It should be noted that this is approximately the same as the performance of the heatsink in free air if it were outside the box, so the fan has not increased the R theta of the heatsink compared to that in free air, what it does do is to remove the heat generated, so it doesn't stew in its own juice.

Cheers, BobT

[mealman]

Thank heaps BobT!!!

[signature]

Bob,
Most supplies are designed to disipate the thermal energy they are producing. If your design is to protect the supply I would design a carbon steel box using perforated sheets. This will protect it and allow air to circulate.

Good luck,
Dan O'

[bobs bots]

Dan has raised a good point,
If you can find some perforated sheet you can make a box out of this and rely on passive ventilation. I've found a photo of the KL7212 now, It is built on a shallow U base made from aluminium? So one could use a sheet of perforated material for just the top and sides , and screw it to the flanges of the 7212.
(It should be noted at this point, that here in Australia, we can't just stroll down to the nearest ACE hardware and pick up a 2ft square sheet of perf. metal)

For best results the supply should be mounted vertically, with the transformer at the top, and the capacitors at the bottom. The transformer will be dissipating around 70W at full load and 35W at idle, so will create a chimney effect and draw cool air over the capacitors. The bridge rectifier will be dissipating around 24W at full load, and the capacitors dissipating around 6W total.

I've also discovered an error in my earlier post, turns out your average electrolytic capacitors are rated for 2000 hr life at full ripple current at rated temperature (I was mislead by some other high performance capacitors I was dealing with earlier). The KL7212 uses 2x 10,000uf @ 80V capacitors , I don't have the part number , but typically this value capacitor has ripple currents around 6A at 85c for 200Hrs, and starts with 30milliohm ESR (each). So If you want to drive this power supply at full rated output for 24Hrs/ day at 40C ambient, you will definitely need some supplementary cooling to keep the caps well below 85C.
On the other hand if you intend using it for hobby CNC, say 20Hrs/week, in say 20C ambient, and your loads are less than full rated current, (and I would expect this to be the case), then you will get a long useful life out of the unit, even with quite modest natural ventilation.
BobT

[bobs bots]

============== KL7212 Power supply lifetime calculation =========
OK for the curious , I will do a worked example of a lifetime calculation.
(a) assume capacitor is the life limiting component, (2000hrs at 85C at 6A each with 30milliohms)
(b) transformer should be OK as they will usually run at 125C for decades
(c) assume the current drawn is full rated for 1minute and 25% of rated for 9 minutes in every 10minute period. (Typical for CNC applications)
(d) assume self heating in the capacitor due to rated current and ESR produces a 40C increase in still air.
(c) assume a 20C ambient, as the power supply is not inside a larger sealed enclosure, and assume at sea level.
(d) assume you mount the supply vertical, with modest natural ventilation with 12 x 20mm holes in the bottom and top

===Step one , work out cap temp rise with our duty cycle====
The self heating effect is due to ohmic heating due to the ESR (Equivalent Series Resistance) times the current squared.
For continous current of 6Amps per capacitor, (which have a 30millohm ESR) the power is 6^2 * 0.03 = 1.08W (which we assume produces an internal temperature rise of 40C in the cap).
At continuous 25% rated current (1.5 A per cap) the power is 1.5^2 * 0.03 = .26W , which will raise cap temperature by only 10C.
For our duty cycle we need to add the 10% of heating at full current to 90% of heating at 1/4 current. or 0.1* 1.08W + 0.9*0.26W = 0.35W , raising cap temperature by 13C.

=== Step 2 work out cap temp ====
We need to work out the ambient temperature in the vicinity of the cap, which will always be higher than that in the room. I'm just going to pluck a number out of the air and say the bridge rectifier heats up the local air by 10C, so local ambient is 20C + 10C =30C, now add the cap's temp rise of 13C to this and get 13C + 30C = 43C

==== step 3 work out the capacitors life ===
This goes according to the Arrhenius equation where
life is proportional to exponential of temperature ratio, too complicated for this forum, so just use the doubling for every 10C approximation.
As the capacitor has a 2000hr lifetime at 85C , and it is now 42 degrees lower (this is 4 sets of 10C plus a bit) , it will last for 2 x2 x2 x2 longer (4 sets of doubling) or 16 x 2000hrs or 32000hrs
=== step 4 work out how many years it will last ======
OK , it's only used 20Hrs/week so 32000hrs/20hrs/wk = 1600wks=31years
This is pretty good, and adding a fan will probably double this again.

On the other hand, consider the exact same power supply, delivering 100% rated current, 24/7, mounted horizontally, inside a sealed cabinet, at a steel foundry at 15000' elevation, at 30C air temperature. Then the deltaT of 40C gets doubled by the density effect of high altitude to 80C, which is added to ambient of 30C, which is added to the temp rise inside the sealed cabinet of maybe 30C, for a capacitor temperature of 140C, this is 55C hotter than the capacitor rating (i.e.5.5 sets of 10C) so the life is reduced by a factor of 45, from 2000hrs to 44hrs, i.e. less than 2 days.

Pretty amazing difference, Huh?

This is why you get so many different opinions on the usefulness of cooling, in fact adding a fan (sucking in outside air) to the foundry example will increase the operating life to about the rated life of 2000hrs.

Cheers, Bob T