solderdude
Grand Contributor
Feeding/charging a low-current draw device from a powerbank is not going to work with most powerbanks.
The reason for this is that a powerbank has an internal battery of 3.7V.
In order to get 5V from an output there is a step-up converter present in the powerbank that puts out around 5V.
Such a circuit draws current even when not loaded with anything. This would drain the battery even when not charging anything.
The solution is to switch off the internal DCDC converter when current draw drops below a certain value.
As most phones and tablets draw quite a lot of current when charging (everyone needs fast charging) the DCDC converter will switch on when enough current is drawn.
That charging current will drop when the, to be charged, device is (almost) full.
When that minimum required current is reached the DCDC converter will switch off as the device is charged and the powerbank itself won't be loaded any more.
Depending on the ports(1A and/or 2A) and circuits used the output switches off somewhere between 50mA and 200mA.
When the current drops below that threshold current the output(s) of the powerbank switch off and thus stop supplying 5V.
This is a problem when you want to use the powerbank to charge or power low current circuits.
In my case the to be powered device draws less than 50mA.
The output of the powerbank senses a low current draw and thus either does not switch on or switches off again after 10 to 20 seconds.
A solution would be to add a resistor that 'bleeds' a current just below the minimum required current.
BUT that would draw say 80mA/h and drain the battery a bit more as well as dissipate heat (0.4W).
My solution is a simple circuit that 'draws' very short 130mA pulses every few seconds.
Of course I am not the first to think of this but thought I'd share my solution with DIY'ers that also may (have) run into this.
As these pulses are very short (small duty cycle) there is just a 10 mA/h current drawn and the output stays 'on' so 5V will remain present untill it is removed from Powerbank.
The circuit is shown below:
The reason for this is that a powerbank has an internal battery of 3.7V.
In order to get 5V from an output there is a step-up converter present in the powerbank that puts out around 5V.
Such a circuit draws current even when not loaded with anything. This would drain the battery even when not charging anything.
The solution is to switch off the internal DCDC converter when current draw drops below a certain value.
As most phones and tablets draw quite a lot of current when charging (everyone needs fast charging) the DCDC converter will switch on when enough current is drawn.
That charging current will drop when the, to be charged, device is (almost) full.
When that minimum required current is reached the DCDC converter will switch off as the device is charged and the powerbank itself won't be loaded any more.
Depending on the ports(1A and/or 2A) and circuits used the output switches off somewhere between 50mA and 200mA.
When the current drops below that threshold current the output(s) of the powerbank switch off and thus stop supplying 5V.
This is a problem when you want to use the powerbank to charge or power low current circuits.
In my case the to be powered device draws less than 50mA.
The output of the powerbank senses a low current draw and thus either does not switch on or switches off again after 10 to 20 seconds.
A solution would be to add a resistor that 'bleeds' a current just below the minimum required current.
BUT that would draw say 80mA/h and drain the battery a bit more as well as dissipate heat (0.4W).
My solution is a simple circuit that 'draws' very short 130mA pulses every few seconds.
Of course I am not the first to think of this but thought I'd share my solution with DIY'ers that also may (have) run into this.
As these pulses are very short (small duty cycle) there is just a 10 mA/h current drawn and the output stays 'on' so 5V will remain present untill it is removed from Powerbank.
The circuit is shown below: