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I hope you will not mind answering. This question has long been on my mind.

Normally, the adapter output voltage is higher than that of the battery. In my laptop's case, the output voltage of the adapter (or charger or power supply) is 19.2V. (That .2V itself is also a big question for me. Is that so sensitive?) But the voltage of the battery is 10.8V.

My question is, why is the adapter output voltage different from the battery's voltage?

This is not about asking how it is regulated by the internal circuitry, but asking why.

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  • \$\begingroup\$ I explained this almost exactly this in a comment to your previous question so why are you asking it here again? The laptop battery needs to be charged by a lower voltage which depends on temperature, how full the battery is etc. \$\endgroup\$ – Bimpelrekkie Feb 20 at 21:50
  • \$\begingroup\$ @bimpelrekkie, this question is specific, why. I didn't see your previous comments have aswer this question. \$\endgroup\$ – AirCraft Lover Feb 20 at 21:56
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    \$\begingroup\$ The adapter voltage is higher probably because it's easier to design the battery charging circuit that way. Solid-state switches need voltage headroom to switch properly and may be located in a circuit somewhere that requires more than the battery voltage to provide such headroom. Some switches may also need a minimum of 10V and 15V-20V is ideal for these switches. 19.2V is just because it matches a common battery chemistry and therefore mass produced and cheap for the manufacturer to buy off-the-shelf. It's the same reason 13.8V power supplies are common. \$\endgroup\$ – DKNguyen Feb 20 at 22:00
  • \$\begingroup\$ @toor, one day I checked in computer store, there some adapter which the voltage is around 18V point something. Mean, different voltage from mine. But however, they are higher than their battery. So, clearly it is not about easier to design. That must be designed by purpose. \$\endgroup\$ – AirCraft Lover Feb 20 at 22:01
  • \$\begingroup\$ Your mobile phone is also fed with a 5V power supply, but it has to charge a 3.7V lithium battery. The laptop is no different. \$\endgroup\$ – Justme Feb 20 at 22:03
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The voltage on your battery "10.8V" is the "nameplate" voltage, some average voltage that your battery delivers over full discharge cycle. The value of "10.8" indicates that this is a battery of 3 Li-Ion cells in series, giving their standard "nameplate" voltage of 3.6V per cell.

Charging the Li-Ion cells requires variable voltage levels, from 2.5-3 V per cell (when in deeply discharged state) to 4.2V (4.35 in some cases) per cell in "constant-voltage" stage of charging process (otherwise the cell won't be charged to full capacity). So the feeding power must have some overhead to provide the charging process (or let internal charger to do so). So, for 3-cell, it comes up to 12.6 - 12.9 V of input. The external power supply must provide this headroom, which includes minimum of "drop-out" (or regulation) voltage for switching electronics inside the external power supply and internal charger, 1-2 V per device, give or take. Eventually it comes up to 12.9+4 = ~ 16-17V.

The "19.2" nameplate is a bit of mystery, since it is not that stable in the first place. It is just an industry standard. Any AC-DC adapter in the range of 18 - 22 V will happily charge your laptop, very likely.

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  • \$\begingroup\$ When I measured it using digital multi-tester, the voltage output is 19.19V. Almost exactly as written on the name plate. \$\endgroup\$ – AirCraft Lover Feb 21 at 0:59
  • \$\begingroup\$ The way they wrote the output voltage 19.2V, the .2V, I believe, they intented to say so. \$\endgroup\$ – AirCraft Lover Feb 21 at 1:01
  • \$\begingroup\$ @AirCraftLover, the specific way of number notation wrt accuracy is a subject for notations in precise sciences as Physics. Some people believe that the 19.2 stands for "24V -20%". Nameplate voltages on laptop adapters vary, 16.5V, 18V, 19V, 19.2, 19.5 for 120W Dell adapters, or even 20V for some HP/Compaq laptops. \$\endgroup\$ – Ale..chenski Feb 21 at 1:11
  • \$\begingroup\$ I don't understand your last comment. \$\endgroup\$ – AirCraft Lover Feb 21 at 1:57
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From your other question, it appears that you have a laptop power supply with power rating of 65 Watts (3.42 A × 19.2 V). As power is voltage times current, it means that if the voltage is higher, the current is lower, so thinner, cheaper, more flexible wires can be used to deliver that 65W to the laptop when charging. That's why it's not a 65 Watt, 5 Amp 13 V power supply.

Also it means that when the battery is being charged, a DC-DC converter in the charging circuit converts the 19.2 V down to match the battery voltage so that suitable amount of charging current flows into the battery. So in this case a nominal 10.3V battery could be charged at over 6 Amps with the same 65 Watts if the battery can handle that amount of charging current safely. The charging current will be limited by what is safe to the cells.

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  • \$\begingroup\$ Even in the laptop label on the bottom side of the laptop written power rating is 65W, 19V (this is 19V, not 19.2V, slightly different). So, it really required 19V. Then come question in my mind. If the required voltage is 19V, why did the battery is 10.8V? What is actually the voltage requirement? \$\endgroup\$ – AirCraft Lover Feb 21 at 1:06
  • \$\begingroup\$ @AirCraftLover, "What is actually the voltage requirement?" - the actual voltage requirement is written on the bottom of laptop. Plus-minus guardbands, usually +-10%. However, internally the laptop uses variety of voltages, 5V, 3.3V, 2.5V, 1.8V, 1.5V for memory, and 0.8 to 1.2V for CPU. Why doesn't it trigger the similar question in your mind? \$\endgroup\$ – Ale..chenski Feb 21 at 1:53
  • \$\begingroup\$ @Ale..chenski, if it is the requirement, so why the manufacturer didn't design the same voltage with their battery? If can be +/-, then 10% minus from 19V will be 17.1V. This is lower than to series 5 cell (which will give 3.6V*5=18V). If due to 18V then the requirement 65W is not achieved, then they can increased the current. That is my question. \$\endgroup\$ – AirCraft Lover Feb 21 at 2:02
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This is because the laptop is designed for a optional battery you don't have.

Laptop batteries usually come in 6 and 8 cell varieties. A 6 cell battery has two strings of 3 series cells. A 8 cell battery has two strings of 4 series cells. The higher cell count provides more power at the expense of weight. (Sometimes a 4 cell option is available consisting of 1 string of 4 series cells)

Most commonly the smaller capacity battery will sit flush with the case and the high capacity battery sticks out the back.

Ale..chenski identifies a max charge voltage of around 4.2 V/cell. So a 6 cell battery needs 12.6 V max to charge, an 8 cell battery needs 16.8 V. Then add cable losses and regulator losses to get 19.2 V.

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  • \$\begingroup\$ In the bottom of the laptop (my laptop is Toshiba), there is written power rating is 65W 19V. Mean, the laptop itself is required 19V. If they really need 19V, then why they didn't design 19V battery or closer voltage to it? \$\endgroup\$ – AirCraft Lover Feb 21 at 1:10
  • \$\begingroup\$ @AirCraftLover They did design a 19 V battery. You just didn't buy it, you got the smaller 12 V battery instead. The charger and adapter is made for the bigger battery if you ever get it. \$\endgroup\$ – user71659 Feb 21 at 1:14
  • \$\begingroup\$ I understand your explanation about cell array. But even the array is 4 series then paralleled with another 4-series, still giving 14.4V. If fully charged, then it will be normally 16.56V. This is still much below 19V. \$\endgroup\$ – AirCraft Lover Feb 21 at 1:14
  • \$\begingroup\$ Btw, output of the charger is 19.19V. It is the voltage at the port, after all the cables around 1m. I measured using digital multi-meter. \$\endgroup\$ – AirCraft Lover Feb 21 at 1:17
  • \$\begingroup\$ @AirCraftLover , "output of the charger is 19.19V" - I repeat, this is not the charger, it is "AC-DC power supply". The actual battery charger is built into your laptops. This internal charger converts input power (whatever it is, it can accommodate certain range of input voltages), and provides proper charging algorithm, with necessary variable voltage. \$\endgroup\$ – Ale..chenski Feb 21 at 2:43

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