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Considering a 4-cell (laptop) lithium-ion battery, the cell voltages add (since they are in series) to 3.7 V * 4 = 14.8 V. However, the adapter says that it has a 19.5 V DC output. I want to know why is it 19.5 V and not some other value like 14.8 V itself or 24 V for instance?

Also, my adapter says Input: 100-240 V ~ 1.6 A and Output: 19.5 V - 3.33 A (65 W). Does it mean that the adapter efficiency is 65 W/(1.6 A*230V) assuming that the mains voltage is 230 V? Am I actually paying bills for 230*1.6 W or is it that 1.6 A is the maximum current that can be drawn and the actual current varies?

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The adapter does not connect to the laptop's battery directly, there is always a charging control circuit in between. In laptops this usually is a switching type charging circuit to avoid excessive power (heat) dissipation.

If the power adapter delivered 14.8 V then it would be difficult to impossible to fully charge the battery as there would be no "voltage headroom" left for the charging circuit.

The choice for 19.5 V instead of 24 V might be historical and practical. 19.5 V did the job and maybe 24 V might require components rated for a higher voltage.

The input rating of the power adapter is just that a rating which means that the current will not exceed the rated 1.6 A. Usually this maximum current is only reached when you plug in the adapter into the mains. During normal operation the current will be much less. How much will depend on how much power is required from the 19.5 V output.

If the adapter would actually continuously consume 1.6 A then it would get extremely hot as it would need to dissipate a lot of power. So no, you will not be paying a bill for 1.6 A.

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  • \$\begingroup\$ I now get the idea. I wanted to ask another question on similar lines. The DC fast chargers for Electric Vehicles have a rating of 200-600 V. Does that mean that this high voltage is applied directly across the battery whose nominal voltage is typically 48 V or 72 V? Wouldn't that heavily degrade the battery? \$\endgroup\$ – Pikachu Jun 12 '18 at 7:18
  • \$\begingroup\$ @PMD The battery voltage is far higher than that but the rated voltage is not directly applied as an stiff voltage source but rather a CC/CV charger with negotiation. \$\endgroup\$ – winny Jun 12 '18 at 7:24
  • \$\begingroup\$ If in a laptop the adapter is connected to the battery via a charging circuit, what makes you think that this is not the case for an electric vehicle? What do you think would happen if we connect 200 to 600 V DC (with high current capability as we want to charge quickly) directly to a 48 or 72 V battery? BOOM !!! \$\endgroup\$ – Bimpelrekkie Jun 12 '18 at 7:29
  • \$\begingroup\$ I did not get this: "via a charging circuit". Isn't it the case that the final voltage applied across the battery is 19.5V? \$\endgroup\$ – Pikachu Jun 12 '18 at 8:14
  • \$\begingroup\$ No, you're still thinking that the 19.5 V is directly applied to the battery, well it isn't and for good reasons. Li-Ion cells have a nominal voltage of about 3.7 V, 4 in series gives 4 x 3.7 V = 14.8 V. A fully charged Li-Ion cell gives about 4.2 V, so 4 cells is then 16.8 V. I you apply 19.5 V instead of that 16.8 V the you will damage the cells. Also unexpected things can happen (heat/fire/smoke etc). It is simply not how Li-Ion cells must be treated. They have certain limits which MUST be observed. 4.2 V per cell is one limit. The charging circuit takes care of this. \$\endgroup\$ – Bimpelrekkie Jun 12 '18 at 8:22

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