Forum Discussion
BFL13
Aug 28, 2016Explorer II
Thanks for posting those results! Very interesting!
I do have an actual "industry" test report on a proto-type design converter they built where they give the test result for "max efficiency" defined by them as "charger output power (current x voltage) / input power (current x voltage) where they got 83% in that case
So looking at these numbers from above:
"DC volts 13.66, DC amps 39, DC watts 533
AC volts 119.4, AC amps 7.45, AC watts 636, AC volt-amps 896, pf 0.7
Crunching the numbers, efficiency in terms of watts is 83.8%, in terms of VA it's 59.5%. So if the "industry" statement were true, the efficiency is 59.5%."
The DC output watts at 533 checks, but I don't understand the AC figures. 119.4 x 7.45 = 889.53 so where does that 636 come from?
Going just by power out/power in, I get 533/890 = 59.9%
That is where I get the notion that " the industry" counts PF in with "efficiency"
How do we resolve these different interpretations of the actual measurements taken on that 9855?
I do have an actual "industry" test report on a proto-type design converter they built where they give the test result for "max efficiency" defined by them as "charger output power (current x voltage) / input power (current x voltage) where they got 83% in that case
So looking at these numbers from above:
"DC volts 13.66, DC amps 39, DC watts 533
AC volts 119.4, AC amps 7.45, AC watts 636, AC volt-amps 896, pf 0.7
Crunching the numbers, efficiency in terms of watts is 83.8%, in terms of VA it's 59.5%. So if the "industry" statement were true, the efficiency is 59.5%."
The DC output watts at 533 checks, but I don't understand the AC figures. 119.4 x 7.45 = 889.53 so where does that 636 come from?
Going just by power out/power in, I get 533/890 = 59.9%
That is where I get the notion that " the industry" counts PF in with "efficiency"
How do we resolve these different interpretations of the actual measurements taken on that 9855?
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