Question about the Hyperion EOS0610I

[BrianG]

When calculating efficiency, I simply measure DC current and voltage at the output and input of the charger and divide the two. This lumps the charger's switching PS and CC losses together. But, judging by the amount of heat produced, out of that ~20% eff loss, I would say that ~12% is the CC part, and ~8% is the switcher part. Measuring at the wall lumps charger (PS and CC parts) along with the 120vAC to 12vDC converted. So, 75% is actually pretty good.

The 80% figure I use when I don't know the real figure is generally worst case.