How to Make a Power Bank Always On

A normal USB power bank does not stay "on" if there is not enough current being drawn from the USB port. This means that powering for example an Arduino Nano does not work; the power bank switches off.

The most common solution to this problem is to add a circuit that draws more current, in short pulses. This circuit would then be placed outside the power bank.

I wanted to know if it was possible to modify a power bank so that it stayed on all the time. Or if it was possible to lower the current threshold.

So I opened up my Linocell power bank (also available as Classone). Using a knife to pry the top and bottom apart.

Lets see what we have.
There are two ICs mounted on top side of the circuit board.

The battery charger IC (LTC4056) handles charging of the internal batteries, it should only be active when the power bank is being charged.

The booster IC (MT5033) is responsible for stepping up the varying (3-4V) battery voltage to a stable 5V for the USB output.
When the booster is enabled it draws 4mA from the battery, even when there is nothing connected to the USB port. This is why power banks are not always on.

On the rear there are two more ICs.

The protection IC (XB8689D) protects the internal batteries from over discharging and short circuit.
It connects ground on the circuit board to the negative battery terminal, and if the battery voltage drops to 2.9V then it breaks the connection. This leaves both the brain and the booster without a connection to battery negative, so they can not draw any current from the battery.

The brain IC is unmarked. The LEDs and the button are directly connected to it.
The brain enables the booster when the button is pressed, and shows the battery charge state with the LEDs. And it disables the booster and goes to sleep if not enough current is flowing.
How does it measure the current?
How can it be tricked to think the current is good?

I didn't figure that out at the time, so I went with another solution.

Battery positive voltage wired to the enable pin of the booster IC, via a switch mounted on the front panel. With the booster enabled we get 5 volts on the USB ports.

Enabling the booster this way does not wake the brain, so the LEDs don't light up to show the charge state. But pressing the button wakes the brain and shows the charge state, and a long press on the button makes the brain sleep again, even though the booster is on.

The enable pin is meant to be connected to input voltage, so I assume this is also what the brain does, so it should be OK. And it does work, without smoke.

A tiny slide switch was glued in place.

If I were to leave the switch on, the booster would slowly discharge the batteries. Fully charged it would take about two months until the voltage drops so low that the protection circuit disconnects the batteries.
This would be no different than charging my phone until the power bank batteries run out.

With the switch off the power bank works the same as before.


A more elegant solution

The brain measures the current by measuring the voltage drop over two small shunt resistors that sit in series with the negative supply to the USB ports.

The resistors are marked R075 which means 0.075 ohms, and two in parallel means 0.038 ohms.
I don't know the brains exact current limit, but if we assume it is 100 mA then it would have to measure a voltage drop of: 0.038 x 0.1 = 3.8 mV. I guess that is doable.

To lower the brains current limit for staying on means increasing the shunt resistance. If we want to lower the limit from 100 mA to 10 mA we have to increase the shunt resistance by a factor 10, so in this example to 0.38 ohms. That is fine for low currents, but if we then try to draw 2 A we get a voltage drop of 0.76 V and a power loss in the resistor of 1.5 W. That is not fine. So better to look at how to disable the current limit all together.

The brain IC has its ground connected to the ground plane, and has one of its other pins connected to the negative path to the USB ports. This way it can measure the voltage over the shunt resistors.

This measurement pin will be the pin directly connected to the USB negative. But there will only be a 0.038 ohms difference compared to a ground pin, so it may be difficult to measure with a multimeter.
One way to make it easier to measure is to remove the shunt resistors. It doesn't matter if they are destroyed; they are not necessary.

Once the measurement pin on the brain IC has been located it must be de-soldered and lifted from its soldering pad. Because we are going to apply voltage to this pin, to make the brain think that there is current flowing. So, very important that the pin does not make contact with its soldering pad.

Then it is just a matter of connecting the measurement pin to a voltage higher that 4 mV. Maybe the 5 V pin of the brain IC.
And if the shunt resistors where removed then solder in a piece of wire in their place.

Then the brain will always think there is current flowing, so it will never go to sleep by itself. But a long press on the button will make the brain shut down the booster and go to sleep.
And the protection IC will still shut everything down if the batteries goes too low.