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Living Room / Re: power bank DC 19V 5A/12V 2.5A Charge Ports External Battery Charger for Laptops
« Last post by 4wd on February 14, 2019, 04:14 PM »50,000mAh @ 3.7V = 185Wh = Not likely to be allowed on a passenger aircraft, (anything over 100Wh is usually only by permission of the airline these days - anything over 160Wh is a definite no according to the few airline T&Cs I've looked at).
Power banks are always quoted with reference to the primary cell voltage they use, in this case Li-Poly which is 3.7V.
It also looks good to the marketing muppets if it says 50,000mAh (@ 3.7V) rather than ~30,000mAh (@ 5V * 0.8) because a it's a bigger number.
Power = Voltage * Current (of the primary cell in this case)
You also have to take into account converter efficiency which will be around 80-85% for a very good power bank and is usually somewhere between 70-80% for the majority.
So your 185 Wh might be 185*0.8 in real life after conversion to a higher output voltage.
There's also probably a reduction in efficiency, (greater losses), the higher the output voltage so while 5V output may be 80% efficient, 19V might only be 70% efficient.
It's also why they can't quote figures based on theoretical efficiencies.
Why they don't give real values ?-Contro (February 14, 2019, 02:32 PM)
Power banks are always quoted with reference to the primary cell voltage they use, in this case Li-Poly which is 3.7V.
It also looks good to the marketing muppets if it says 50,000mAh (@ 3.7V) rather than ~30,000mAh (@ 5V * 0.8) because a it's a bigger number.
I think is 185 Wh independently the voltage.
Power = Voltage * Current (of the primary cell in this case)
So if 19 volts : 185 Wh / ?? Ah = 19 volts ----> we obtain 9737 mAh
You also have to take into account converter efficiency which will be around 80-85% for a very good power bank and is usually somewhere between 70-80% for the majority.
So your 185 Wh might be 185*0.8 in real life after conversion to a higher output voltage.
There's also probably a reduction in efficiency, (greater losses), the higher the output voltage so while 5V output may be 80% efficient, 19V might only be 70% efficient.
It's also why they can't quote figures based on theoretical efficiencies.
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