I've noticed the battery level must reach a minimum level before it can be powered up on some mobile computing devices. This requires waiting several minutes with the device plugged in. I often wonder if the battery is being used for a secondary purpose such as providing some kind of filtering or power regulation regulation.

Is the battery doing more than just providing current? If not, why can't the circuit draw current from the wall wart while charging the battery at the same time?

  • \$\begingroup\$ I suspect it's to ensure that if the thing is immediately unplugged it has enough power to complete the startup/shutdown process. \$\endgroup\$ – pjc50 Jun 9 '18 at 18:46
  • \$\begingroup\$ Everything does, just at different levels \$\endgroup\$ – PlasmaHH Jun 10 '18 at 8:36

When a phone powers up, its boot process is power-consuming. More, during the boot and attempting to connect to network, it will use its radio, which will consume sizable peak currents (0.5-1A). This burst current might be unavailable for general chargers from the wall.

Now, in accord with Li-Ion safety regulations, all Li-Ion cells must be protected, including over-discharge. This is an independent function of the stock primary cell. If the battery senses its output voltage below 2.5-2.7 V, it will automatically shut itself down, disconnect. If you try to boot a phone on a weak battery, the voltage might droop, and the battery might surprise the operating system with self disconnect, and corruption to boot/system files may occur, and the phone might brick out. Therefore, the battery must have enough "juice" to hold the boot process and sustain short radio bursts. That's why the phone's bootstrap processor waits for proper charging status from the system battery gauge before booting full-rate.

When a discharged cell begins charging, the over-discharge protection turns off, and the internal phone charger will see certain voltage on the cell. It could be just ~2 V or something, still too low for normal operations. In this situation the charger usually enters so-called "pre-charge" mode, charging the cell slowly at 20-50-100 mA rate. This slow-current stage is necessary to revive internal chemistry and prevent cell gassing/bloating. So it might take some time when the cell voltage (Charge Status) gets up to the minimum operating voltage (usually ~3V), when the charger can engage full-scale (fast) charging mode, and the battery has enough charge to power radio bursts without sagging below the battery over-discharge threshold and self-disconnect. All these precautions are necessary to ensure safe and reliable phone functionality.

  • \$\begingroup\$ Thanks a bunch. There's quite a few YouTubers claiming you can bypass a dead battery on ANY laptop by cutting off the power adapter connector and soldering it directly to the motherboard's power input terminals. For some reason, their claim didn't sit well with me. The "burst" you mention implies the battery on some devices behaves as a capacitor. \$\endgroup\$ – user148298 Jun 10 '18 at 21:03

Batteries get damaged when they're over-discharged. Your phone might get damaged when it's shut down unexpectedly. (An unexpected shutdown can cause filesystem corruption and data loss, which might even brick the phone.)

This leads to a bad situation when the phone is plugged into a charger and powered on with a very low battery level: When the phone is unplugged from the charger, it needs to shut down, but in order to do that, it needs at least a certain amount of energy which the battery might not have anymore. So when it's unplugged with a very low charge level, the designers of the phone would have to choose between damaging the phone's filesystem (shut down immediately) or damaging the battery (shut down properly with the risk of over-discharging). The third option is to just not let the phone turn on when it's discharged too deeply, and that's what most manufacturers do.

Additionally, a phone battery needs to be very slowly charged up to a certain percentage when it's discharged too deeply in order to avoid damage to the battery. That's why the first few percent of charge take so long.

  • \$\begingroup\$ Hmmm. Is it possible to construct a battery which charges up using 20V, but outputs 3.3V or 5V? Such a battery may eliminate quite a bit of extra power electronics and possibly increase efficiency. I've heard people directly wiring their circuit board from the leads of their power adapters. I would guess this doesn't work for all devices. \$\endgroup\$ – user148298 Jun 9 '18 at 19:16
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    \$\begingroup\$ @user148298 No, batteries always charge and discharge at roughly the same voltage. Any difference in voltage is due to losses. Additionally, the battery voltage changes depending on the charge level, so there's (almost) always a regulator needed after the battery. \$\endgroup\$ – Jonathan S. Jun 9 '18 at 19:20
  • \$\begingroup\$ Thanks for the help. I read some battery packs are charged in series, but discharged in parallel. So, for instance, if you have 10 5V cell pack, you would charge it in series, but the circuit draws 5V because the output is wired in parallel. Perhaps, I read this wrong. \$\endgroup\$ – user148298 Jun 9 '18 at 19:31
  • \$\begingroup\$ @user148298 That does in fact work, I was talking about a single cell. Doing this requires expensive switching devices, however, and an additional step-down converter for when your battery pack is charging and switched into "high-voltage mode". It would also need a battery balancer, which would take away most of the gained efficiency. \$\endgroup\$ – Jonathan S. Jun 9 '18 at 19:34
  • \$\begingroup\$ -1, the trickle charge part is wrong. You trickle charge when the battery is full, when it is empty, or the protection is triggered, you just need to charge it with a very low current, in the range of few mA. \$\endgroup\$ – Vladimir Cravero Jun 9 '18 at 19:42

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