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Old Nov 18, 2011, 01:01 AM
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What's the difference between meanwell SE and RSP PSU?

As the title says.

Thx
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Old Nov 18, 2011, 09:44 AM
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The biggest difference that you should be concerned with is power factor correction. SE lacks it. RSP has it. Most need it to prevent AC line shutdown due to overcurrent when charging at high power levels. SE will draw significantly greater current from your AC line than RSP at same DC output power.

Mark
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Old Nov 18, 2011, 10:03 AM
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I see. That's why the RSP are more expensive. Hope I dont have issues using SE.
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Old Nov 18, 2011, 10:08 AM
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As discussed yesterday, you might get away with it.

http://www.rcgroups.com/forums/showp...2&postcount=11

http://www.rcgroups.com/forums/showp...2&postcount=14

Only way to know for sure is to try.

Mark
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Old Nov 18, 2011, 12:30 PM
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Hey Mark,

Looking at Tims calculation, it seems it should be ok.

http://www.rcgroups.com/forums/showp...postcount=3402

i am using a 24v PSU and therefore, the max i can run the charger is about 816W. looking at tims calculations, i should be using the following amps

AC amps = (816 / .85) / 110v = 8.72A

im not sure how the PFC is factored into all this.

Quote:
Originally Posted by mrforsyth View Post
As discussed yesterday, you might get away with it.

http://www.rcgroups.com/forums/showp...2&postcount=11

http://www.rcgroups.com/forums/showp...2&postcount=14

Only way to know for sure is to try.

Mark
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Old Nov 18, 2011, 12:42 PM
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i just found this...

OUCH!

http://www.us.tdk-lambda.com/hp/pdfs...93008502rD.pdf

Model 500A 102A 152A 202A
Non PFC 0.65 0.65 0.65 N/A
Active PFC 0.98 0.98 0.98 0.98

IL = Pav / (VL×PF×Eff)

for my case....

IL = 960W / (110v * 0.65 * 0.9) = 13.5A

for a PSU with PFC

IL = 960W / (110v * 0.98 * 0.9) = 9.89A

all rough calculations..
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Old Nov 18, 2011, 01:21 PM
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Correct. As discussed yesterday, power factor and power supply AC-DC conversion efficiency must be considered. At higher currents, AC line power losses and charger input lead power losses also become factors (which are not a part of the above equations).

.65 PF is a reasonable guess for a Meanwell SE series PS. .98 PF is about right for a decent power supply with PFC (my supplies measure .97-.98).

90% conversion efficiency is pretty generous for a power supply. Some might be 90% but I would use no higher than 80% unless empirical data exists that demonstrates 90% efficiency.

Using 80% conversion efficiency and .65 PF, you'll be bumping up against the limit of a 15A circuit if you're charging at ~820 watt output with a PL-6 when using a Meanwell SE-1000 as a power source.

Tim's calculations are obviously oversimplified and ignore power factor of the power supply. This can lead to all sorts of trouble.

Mark
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Last edited by mrforsyth; Nov 18, 2011 at 01:29 PM.
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Old Nov 18, 2011, 01:23 PM
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Quote:
Originally Posted by mrforsyth View Post
Correct. As discussed yesterday, power factor and power supply AC-DC conversion efficiency must be considered. At higher currents, AC line voltage losses and charger input losses also become factors (which are not a part of the above equations).

.65 PF is a reasonable guess for a Meanwell SE series PS. .98 PF is about right for a PS with PFC (my supplies measure .97-.98).

90% conversion efficiency is pretty generous for a power supply. Some might be 90% but I would use no higher than 80% unless empirical data exists that demonstrates 90% efficiency.

Using 80% conversion efficiency and .65 PF, you'll be bumping up against the limit of a 15A circuit if you're charging at ~820 watt output with a PL-6.

Mark
so you thinking i should get a PFC unit? running 820W, what currents u think i'll be using from 110VAC?
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Old Nov 18, 2011, 01:32 PM
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Quote:
Originally Posted by teeforb View Post
running 820W, what currents u think i'll be using from 110VAC?
As detailed above, I suspect that you'll be close to 15A, which may be perfectly fine. Your circuit breaker will let you know if it's problematic.

You can always measure power factor and AC current draw as well if you're curious.

Mark
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Old Nov 18, 2011, 01:47 PM
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Quote:
Originally Posted by mrforsyth View Post
As detailed above, I suspect that you'll be close to 15A, which may be perfectly fine. Your circuit breaker will let you know if it's problematic.

You can always measure power factor and AC current draw as well if you're curious.

Mark
i just spoke to a guy from meanwell. Good customer service BTW! He told me the following for full load (actual testing)

SE 1000 (non-cpf) 16.4A
RSP 1000 (cpf) 10.5A

Geez! what a difference! I like a lot of elbow room. I may have just wasted $400..
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Old Nov 18, 2011, 01:53 PM
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The SE-1000 numbers are reasonable.

I would challenge the RSP-1000 numbers however, as it's just not possible that it'll only draw 8.5A at full output of the power supply. To do such, the PS would have to have 100% conversion efficiency and line voltage would have to be very close to 120V. I'm betting that the 'real' number for current draw at full output would be somewhere between 10-11A, assuming 110VAC input.

Mark
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Old Nov 18, 2011, 01:59 PM
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Sorry, 10.5 at full
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Old Nov 18, 2011, 02:02 PM
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10.5A is likely spot-on. I'll buy that.
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Old Nov 18, 2011, 02:09 PM
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what do u think about the rsp 2400 24v PSU?
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Old Nov 18, 2011, 02:17 PM
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Like all Meanwell's, I'm sure that the RSP-2400 is a very nicely made supply and would serve you well for a very long time. Note that it requires 220VAC source. This would hamstring most of us.

As I build all of my own supplies, the probability that I would ever own a Meanwell is extremely remote.

Mark
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