How to sink 100ma from power bank intermittently

The Issue: I am powering 12V leds through an Orico Q1 power bank. The power bank shuts off after about a minute due to a senser that detects the draw and if its below a value it turns off. I believe the leds are not drawing enough current.
I also can not get the datasheet as the company is in china and wont reply to my emails.

I have read a possible solution being to use a micro controller to pulse a signal to sink about 80 ma. from "Pin SINK Current Limitations:" at the link -> Arduino Playground - ArduinoPinCurrentLimitations
I beleive I can choose 3 pins from one group and 2 from another to reach ~100ma without reaching the cap of any group.
The power bank indicates that it will supply 12V at 1.35 amp .

My question is: Is there a way in which I could wire 5 pins and set them all together to LOW to sink roughly 100ma (sum of 5 pins draw) from the powerbank?
also if its possible, considering the powerbank can output 1.35 amps how can i calculate what resistors I would need, if any?

Use a transistor connected to one pin. Calculate the value of the collector resistor by ohms law.
R = 12/0.08

The way I did it (with a 5V bank) was find out the minimum current that will keep the bank alive, I started with a 23 mA load resistor (220 Ohm @ 5V, 560 @ 12V) across the bank output and added additional 220 Rs in parallel until the bank stopped going to sleep, (you will need 560 Ohms @ 12V) ended up with 4 (88 mA) so used a 56 Ohm, 1/2 Watt dummy load switched by a 2N3904 NPN transistor like G-Mike said.
Then to save power, I tried pulsing it on and off with various pulse lengths and intervals 'till I found a minimum the would keep it alive but waste least power. Here’s a sketch I used for that, to set the pulse width, type “p” and a number of milliseconds in the top of Serial monitor and hit enter (p100{ENTER]) do the same for interval with an “i” (i10000[ENTER]). Connect the transistor base to pin 8 with a 1k resistor. I ended up with a 75 mS pulse width and 30 second interval. Note: be sure to use a dummy resistor with high enough wattage rating so it don’t get too hot.

unsigned long iStart; // interval start
unsigned long iEnd = 4800; // interval end
unsigned int pLen = 300; // pulse length
const byte pPin = 8, // pulse pin
           ledPin = 13; // led pin
void setup()
{
  Serial.begin(9600); //set serial monitor to match
  pinMode(pPin,OUTPUT);
  pinMode(ledPin,OUTPUT);
  delay(6000);
  prntIt();
}
void loop()
{
  digitalWrite(pPin,millis() - iStart < pLen);
  digitalWrite(ledPin,digitalRead(pPin));
  if(millis() - iStart > iEnd)
    iStart += iEnd;
  while (Serial.available() > 0) {
    char inChar = Serial.read();
    if(inChar == 'i')
      iEnd = Serial.parseInt();
    else if(inChar == 'p')
      pLen = Serial.parseInt();
    else
    {  
       Serial.println("invalid input");
       while(Serial.available() > 0)
         Serial.read();
    }
    iStart = millis();
    prntIt();
  }  
}
void prntIt()
{
  Serial.print("interval = ");
  Serial.print(iEnd);
  Serial.print("  pulse length = ");
  Serial.print(pLen);
  Serial.println("\n");
}

out.png
Looks like they would make those banks without the phone charging stuff and a 5.5 X 2.1 mm barrel type socket but keep the over discharge protection cutoff.

There are some solutions here: Powering the arduino nano with a Power bank - Project Guidance - Arduino Forum
For the one I have, a 5 volt power bank, I found that switching a load for 20 mS every 2 seconds was enough. I used a 555 timer, a transistor and a 47 Ohm resistor.

for OUTSIDER, I have some clarifying questions.

am I correct that you continued to add resistors in parallel untill you reached the mA needed to prevent shutoff, between the positive side of the powerbank and the dummy load. Then you used a transistor to toggle the dummy load on and off, while also using a 1k resistor on the data pin to prevent to much current from reaching the micro controller?

were you able to test the proper resistance to prevent shutoff before chosing the dummy load? The way I understand this is that you need the dummy load to prevent shutoff (I wouldnt know how to test that its successfully without actually using the dummy load?)

Also since I dont have the datasheet for my powerbank I was going to go with 120 mA draw to be safe which would be 100 ohms at 1.44 watt. I tried searching for a dummy load like that but I might be searching incorrectly. also does the dummy load need to match exactly with the values I have or is there some acceptable fluctuation for the values of the type of dummy load I get?

Sorry its taken me a bit of research to get to a point of comfortable understanding >_< I greatly appreciate your help!!!

while also using a 1k resistor on the data pin

That is the pin on he transistor’s base.

A 1.44 W is the peak load not the continuous load. For a very short pulse once a minute a 0.5W resistor will do.

You can either calculate the value you need or keep altering the resistor until you find the sweet spot of minimum current that works with your power pack.

so this 1 watt 75 ohm dummy load could be used?

https://www.rfparts.com/000-46650-75rfx.html

1 watt 75 ohm would be 8.66025V. does that mean the device can only take 8.6 V input?

No no, just replace "dummy load" with "resistor", if for example you determine that your bank requires, say, 80 mA to keep alive, at 12V you would need a 150 Ohm resistor to maintain that current, if on constantly, it would dissipate (waste) 0.96 (call it 1) Watts (and get very hot). If you don't care about the waste, you won't need the "blinking" circuit. If you do and the ON duty cycle of the pulse is less than 25%, you could get away with a 1/2 Watt R. If you want 120 mA your resistor would be 100 Ohms, 1.5 Watts. I would go with a 5 Watt so it doesn't get so hot.
The 1k resistor limits current into the transistor base so it don't go "POOF" :slight_smile:

The way to test for minimum current that I used was:
For a 5V power bank a 220 Ohm, 1/2 Watt resistor passes a current of 23 mA , connect one across the power bank output and wait a couple of minutes to see if it goes to sleep. If it does, add another 220R and repeat until the bank stays alive, add up the number of 23 mA, say, it turns out to be 6, 138 mA, 36.2 Ohms at 5V, closest standard value under that is 33 Ohm, 152 mA. 33 Ohm, 1 Watt is your fixed resistor, but can be a lesser wattage if pulsed, according to pulse duty cycle.

This may be bloody obvious, but your problem could be dealt with by just using a shittier power bank that doesn't have the autoshutoff funtionality (the cheapo ones often don't)

Wonder why they don't make them with a standard 5.5 X 2.1 mm barrel receptacle and protection for over discharging but without the phone charging stuff, for general purpose power?

I wow, I didnt realize. That ended up being quite easy in the end. I was confused that the leds can take a higher current about 1.2 amps for the strips im using which I thought was a large enough load for the power bank. I had also read that leds LEDs have such a small electrical load that regular transformers do not register that they're wired to a bulb at all. I guess I was unclear about those concept but now I think I get it.

I guess I dont quite understand load vs draw for leds

Using the suggestions you made I was able to fine a decent resistance to prevent the power bank from turning on! Thank you so much for you help!!