Typically mobile devices that have a mains-powered supply will accept voltage that is multiple of some single battery voltage. This answer focuses more on the broader aspects of your question. Russell's answer ( ) does a great job of looking at the details. Making this sort of change in a system NOT designed to accommodate a lower voltage battery could be injurious to the health of the battery, the equipment and the user. Despite the lower voltage of the new battery the charging circuitry accommodated the change, recognising the battery and adjusting accordingly. The 4 cells version operated in 4S configuration and the 6 cell version in 2P3S. I recently replaced a netbook battery with 4 cells with an extended capacity version with 6 cells. A good buck converter design can exceed 95 % efficient and in this sort of application should never be under 90 % efficient (although some may be). The buck converter accommodates this reduced voltage to maintain charging efficiency. At 3 V/cell 1/2/3/4 cells have a terminal voltage of 3/6/9/12 volt. How low they are allowed to discharge to is subject to technical considerations related to longevity and capacity. When a Lithium Ion cell is close to fully discharged it's terminal voltage is about 3 V. As little drop as possible is desirable to minimise energy wastage. Voltage drop other than in the battery can occur in SMPS switch (usually a MOSFET), SMPS diodes (or synchronous rectifier), wiring, connectors, resistive current sense elements and protection circuitry.
With say 0.7 V of headroom it would notionally be possible to use say 16.8 V + 0.5 V = 17.5 V from the power supply-but using 19 V ensures that there is enough for any eventuality and the excess is not wasted as the buck converter converts the voltage down as required. Most of this is not needed and the difference is accommodated by the buck converter, which acts as an “electronic gearbox”, taking in energy at one voltage and outputting it at a lower voltage and appropriately higher current.
TEST MAC MINI POWER SUPPLY 10 PIN SERIES
In this case, using a buck converter it would be possible to charge up to 4 cells in series.Ħ cells in 2 parallel strings of 3 (2P3S),Ĩ cells in 2 parallel strings of 4 (2P4S)Īnd with a source voltage of 19 V it would be possible to charge 1, 2, 3 or 4 LiIon cells in series and any number of parallel strings of these.įor cells at 16.8 V leave a headroom of (19−16.8) = 2.4 volt for the electronics. In many cases a buck converter can be made more efficient than a boost converter. A SMPS can be a Boost converter (steps voltage up) or Buck converter (steps voltage down) or swap from one to the other as required. It is usual for a charger to use a switched mode power supply (SMPS) to convert the available voltage to required voltage. At the very least about 0.1 V extra might do but usually at least 0.5 V would be useful and more might be used.
To charge a 4.2 V cell at least slightly more voltage is required to provide some “headroom” to allow charge control electronics to function. Each battery consists of at least a number of LiIon cells in a series 'string' and may consist of a number of parallel combinations of several series strings.Ī Lithium Ion cell has a maximum charging voltage of 4.2 V (4.3 V for the brave and foolhardy). Voltages slightly below 19 V can be used but 19 V is a useful standard voltage that will meet most eventualities.Īlmost all modern laptops use Lithium Ion (LiIon) batteries. Various combinations of series and parallel cells can be accommodated. The voltage is slightly more than a multiple of the fully charged voltage of a Lithium Ion battery-the type used in almost every modern laptop.ġ9 V provides a voltage which is suitable for use for charging up to 4 x Lithium Ion cells in series using a buck converter to drop the excess voltage efficiently. This is not a design question as posed, but it has relevance to design of battery charging systems. That isn't a multiple of anything suitable. Now there're laptops that use external power supplies rated at exactly 19 volts.
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