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There are many handheld product like laptop or an Handheld measurement equipment that all parts would works fine with 3.3V or lower VCC. However these devices still comes with a 11.1ٰV battery or higher voltage.

My question is, doesn't it more efficient to use 3 cell in parallel instead of series in order to reduce power dissipation on voltage regulator?

As an example in fanless laptop (like Asus UX305), 5v seems to be sufficient for most parts, so probably 2 cell 7.4v battery suits better than 3 cell 11.1V.

Where am I going wrong?

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    \$\begingroup\$ You need a DC/DC in any case, and also likely you hae 1.8V-ish parts in there too \$\endgroup\$ – PlasmaHH Mar 22 '17 at 14:39
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    \$\begingroup\$ it's more efficient in DC/DC converters to step down than it is to step up a voltage source. Boost converters are 5-10% less efficient than Buck converters. When using linear voltage regulators, you want to match voltage supply for in->out to be as close as possible to reduce lost energy. \$\endgroup\$ – KyranF Mar 22 '17 at 14:42
  • \$\begingroup\$ @PlasmaHH yes... but as KyranF said, getting close Vin to Vout on voltage regulator reduce energy loss. So use less voltage still seems better. \$\endgroup\$ – pazel1374 Mar 22 '17 at 14:47
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    \$\begingroup\$ @user41209: not really for buck converters, the rated load is the most important factor to optimize for. Also including headroom even with a possible LDO you need at least 3.6V-ish for those things are quite power hungry. You throw away most of the charge of your battery that usually is discharged down to ~3V. Two in series can discharge down to 6V and still easily run a buck down to 3.3V and all other voltages that are needed. \$\endgroup\$ – PlasmaHH Mar 22 '17 at 14:51
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    \$\begingroup\$ "As far as I know, 3.3V for power supply is sufficient for all parts in this RF Analyzer" - are you sure? Have you reverse-engineered the thing? \$\endgroup\$ – pjc50 Mar 22 '17 at 16:16
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The decision to use a particular voltage input is not necessarily made on the major supply voltage used in an appliance.
Consider your question around the ASUS laptop:

Its input is 19 V; a fairly standard laptop supply voltage, but not likely used for anything internally compute or peripheral related.

@19 V and 45 W you can expect about 2.3 A maximum line current.
@12 V, that would rise to about 3.75 A.
@5 V ... 9 A.
@3.7 V (1 cell voltage) ... 12 A.
Voltage loss would become more critical in the wires and connectors at lower voltages making overall design much more challenging. It's also much more challenging to design an internal battery charger since you now have to separate input/battery at the same voltage with FETs to allow a SM charger to operate.

Internally they may not even charge the batteries at full current when the laptop is on at the same time to control both the power dissipation within the shell and current magnitude in the input supply line. You can see from the specs that the compute/display side of the laptop is only 5 -7 W depending on display usage (resolution/brightness).
The battery in the ASUS is likely a 4 cell 14.8 V battery with active capacity management but no cell balancing. I don't see it specified, but would assume about 1 - 2 hours maximum charge time for the battery at say 2C.

Most of Intel's reference designs are based around a 12 V supply (Core level compute elements). To abandon the reference design (which is fully debugged) and design a new way to do it would be a risky undertaking. You rarely see OEMs such as ASUS, Apple or Microsoft stray far from the reference platform in anything other than peripheral devices.

So back to your question ...could the laptop be powered by a single cell paralleled battery pack ....sure, but it would be significantly harder to design the appliance.

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I would put the cells in series. The buck converter can be designed to work well over a fairly large input voltage range, so you can draw the cells down to such a low voltage that there is no energy left.

With multiple cells in parallel, I'd worry about a small mismatch and one cells trying to drive other cells backwards when everything is supposed to be off.

Generally, higher voltage at lower current is easier to deal with and will have less loss than lower voltage at higher current. This is assuming everything is at a "low" voltage, like under 20 V or at least under 30 V. You'd rather not have 100 V in to drive your 3.3 V device, but at 11 V versus 7.5 V, the 11 V is probably easier to use efficiently.

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A common arrangement, especially in power intensive applications like modern CPUs, is called point of load regulation.

This feeds a higher voltage at a lower current around thinner traces on the main board to feed buck regulators at the places where the lower voltages are needed. At that point it becomes lower voltage with higher current and thicker traces.

This has the advantage that the power routing is easier, takes less trace room, the traces are thinner, and you get reduced losses in the power circuitry.

It's the exact same idea as having high voltage power lines to transfer power around the country and then transform it down to lower voltages for consumption: less transmission losses and thinner cables.

And of course for that you want a higher voltage, not a lower voltage. So you start with a high voltage battery arrangement to make life easier for you. After all, boosting to 12v from 3.7v to then buck back to 3.3v or 1.8v would be very wasteful.

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