Anker A1261 power bank

[najrao]

The label on this says:
Input 5V 2A/ 9V2A
Output 5-6V 3A/ 6-9V 2A/ 9-12V 1.5A
There is  only one output USB socket, and the one micro USB for input.

How do I choose to get 9 or 12V? If 'appropriate' resistors to middle pins, what values? Need a diagram.
Can I charge it with input up to 9V?

The manual says nothing.Thanks for help.

[Romualds]

As I found this power bank supports Quick Charge (QC) 3.0 standard. You should start investigating from this point.

[najrao]

Sam Mallicoat in Hackaday says:
easy to set a QC3 supply to 12V with just two resistors and a toggle or push button switch. Here’s how: Take a 10K Ohm and a 2.2K Ohm and solder in series across the Vbus (red) to ground(black wire). The tap between the two resistors will measure about a Volt. Solder D+ (green to this tap. Then wire the D- (white) through a N.O. switch to the same tap.
Apply adapter or power pack supply and wait 1.5 seconds to push the button. Presto, 12V @1.5! No need to hold the button, the supply stays at 12.

Works perfectly on my power bank!

[Romualds]

Just to clarify -- added diagram. Hope I got everything correctly.
[ Specified attachment is not available ]

UPD: I wrongly labelled input as PD (Power Delivery) instead of QC (Quick Chage). Replaced the schematic image.

[najrao]

Thank you, Romualds. Perhaps you know the resistor values for 9V output?

[mariush]

According to this pdf : https://www.ti.com/lit/ug/tidu917/tidu917.pdf

At pages 6-7 it says:

According the description in the CHY100 datasheet, the processes to enter QC2.0 are:
− Apply a voltage between 0.325 V and 2 V to D+ for at least 1.25 seconds
− Discharge the D- voltage below 0.325 V for at least 1ms while keep the D+ voltage above 0.325 V
− Apply the voltage levels in Table 3 to set the output voltage. (must keep the D+ voltage above 0.325 V)

Code: [Select]

D+                 D-                  Output Voltage
0.325 V – 2 V       0.325 V – 2 V        12 V
>2 V                 0.325 V – 2 V        9 V
0.325 V – 2 V       GND 5 V           (default)


With 5v default , you could pick two resistors to form a voltage divider and get a bit more than 0.325v on D+ and D- , but still be under 2 v after the voltage changes to 9v
For example   you could use a 6.8k resistor and a 1k resistor and you'd get ~0.64v with 5v input, and 1.15v at 9v and 1.54v with 12v input
For >2v you could either directly connect to incoming voltage or use a couple 1k resistors  to get half the input voltage

So if you want to switch to 9v using QuickCharge 2 (or higher I guess) you'd need a small something (a 6 pin micro probably could do it) to initially set the D+ voltage  between 0.3v-2v for at least 1.25s while D- is tied to ground (or nor connected to anything), then set D+ above 2v and D- must go up to 0.3v .. 2v


The schematic in the picture right above me would work because D- is initially not connected to anything, so it should be below 0.3v , and the 10k and 2k voltage divider produces around 0.9v with 5v input. After voltage switches to 12v, the voltage divider would produce 2.1v which is a bit above 2v so maybe slightly tweaking the values to something like 10k and 1.5k may be better  (would produce 0.6v with 5v input, and around 1.5v with 12v input, keeping it below 2v)

You could use a slide switch - in the off switch it outputs to default 5v .. so you wait at least 1.25s, then move slide switch in the ON position, and on one side of the slide switch D+ is connected directly to input voltage instead of <2v (from voltage divider of 2 resistors like above)  and the other side of slide switch would connect D- to <2v

[mariush]

I've edited  my post with some comments about how you could use a slide switch (or two push buttons I guess could also work, if the voltages are only required to be momentary available and not all time).

Mentioning just in case you didn't see it.