I am able to see following specs on my laptop adaptor.

65 W
Input AC: 100 - 240 V, ~1.5 A, 50 - 60 Hz
Output DC: 19.5 V,  3.34 A
My HP laptop:

I am bit unclear about, the conversion of AC to DC. While this conversion takes place, Voltage gets reduced, by a significant no. of times, while current increases marginally. Why is it so ?

I am Software Engineer and knows very little about electrical engineering. Sorry for such a silly question.

Thanks in advance.


Small switching power supplies usually first use a diode rectifier to convert the low frequency AC from the wall to high voltage DC. This high voltage DC is then used as power input for main switching power regular. Without a special circuit called power factor correction this input circuit does not draw power continuously, but only when the amplitude of the sine wave AC reaches a higher voltage than the voltage of the internal bypass capacitors for this high voltage DC, current during this time needs to be significantly higher, because no current is flowing during the remaining time of the AC period. The input current of your power supply is likely only used if you run it at 100 V and the input current is not an average but a peak current instead. Average current is much lower, otherwise it would require active cooling to get away all the heat.

Some power is likely lost due to inefficiencies of the power supply, but efficiency at peak output current should be around 80% or better.

  • \$\begingroup\$ So in this example you might find 80W consumed from the mains if the full load of 65W is delivered to the laptop. That is still an average curernt of less than 1A but as Jan writes the current peaks are higher. \$\endgroup\$ – KalleMP May 26 '16 at 5:22

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