1000A Portable Power Center "saves battery" by turning off various sets of ports after 30 secs of non-use. There seems no internal way to bypass this feature, unit isn't programmable & China manufacturer doesn't respond to even share what minimum draw is required. This is the FIRST challenge I've had with an otherwise awesome, flexible Power Center.

Problem is: my new 12VDC Fridge takes 3 min @ power-on to start compressor (ITS safety feature) AND Fridge's other circuitry (LCD display, ect.) doesn't seem to draw enough power to keep the Power Center 12V ports hot. I can 'trick' Power Center by temporarily adding additional 12V load..fridge runs fine as it cools..but as soon as compressor shuts off, same problem again...idle fridge doesn't draw enough power to keep 12V ports hot.

For reference, charging another jump-start battery from 12V Power Center port kept Power Center ports hot..it otherwise uses 110VAC->15VDC@1A adapter to charge.

VERY IRONIC that I have to do this, totally ridiculous anyone has to waste power, but seems I need to make a simple barrel-jack dongle to ADD small 12VDC load to another of the three 12V ports..something I can plug-in & forget when only powering fridge. I'm handy enough to follow circuit diagrams & have access to typical hobbyist parts (resistors, capacitors, pots, diodes, LEDs, etc)...just looking to end-up with simplest, smallest device without concerns for heat or fire (so, sadly, no incand bulb).

If not substantially more complicated, costly or larger..seems more efficient to make a pass-thru adapter that only adds this add'l load when needed (when fridge is idle) to continuously provide min power draw.

I understand the challenge without any data from manufacturer..just hoping there's a simple, flexible circuit where I can swap a component, or adjust a pot, until I find the 'magic' that makes Power Center & Fridge work unattended...minimum unnecessary power waste, minimum heat...and reliable enough to safely transport meds during long roadtrips.

Just to move things along: YES, I've searched various forums for hours without any practical solution. YES, I know required MIN power consumption would certainly be helpful (still trying). YES, its too late to return Power Center (I'm acquiring needed TravelVan items slowly, finally just got fridge). NO, I don't have ready resources to buy a different Power Center (very pricey)...this one powers induction burner, nutibullet & microwave without problem..NEVER thought too little consumption would be THE annoying issue.

Though I haven't tried it yet, I'm considering adding barrel jacks to cheap eBay Power Monitor..maybe my next experiment. Perhaps it'll add enough drain to solve the problem with just It+Fridge, but it might also demystify what IDLE Fridge consumes. It COULD be as silly as needing just a few add'l mAmps.

Ideas are MUCH appreciated.

  • \$\begingroup\$ Does it power a ‘juice loosener’? I’m told they are ‘whisper quiet’. You seem to have a grasp on the issue and you’ll need to determine what is the minimum current required. Unless someone has direct experience with the gadget we’re just as clueless. Once you know what the min current is, then you can choose a suitable resistor to provide the min current draw. \$\endgroup\$
    – Kartman
    Commented Dec 2, 2021 at 5:11
  • \$\begingroup\$ It seems to me that someone didn't do their research on what they were buying then, bought from a supplier who has unreliable service help (which could have been verified beforehand). \$\endgroup\$
    – Andy aka
    Commented Dec 2, 2021 at 9:02

1 Answer 1


A power resistor is your best bet for simple. Or a bunch of leds of you want it to be useful. It would be turning it into heat but you could just mount it on the fridge as a heat sink. You can use a timer to pulse it at some frequency which works for some smaller USB power banks, but it depends on how often or how long your controller monitors the line for activity.

The first step is to simply keep reducing the resistance until it stops turning off. Start with a 200 ohm resistor. 12V / 200 ohms = 60 mA and only .72 Watts, so a 1 Watt resistor is enough. If it turns off, add another in parallel. And so on and so on. Or go with a large ceramic 5w resistor etc.

If you want to only turn it on when the compressor is off, you could use a hall effect sensor to sense when the motor is turning.

  • \$\begingroup\$ Or if you can access the terminals of the compressor motor, wire a relay coil in parallel with the motor and use the normally-closed contact of the relay to switch the dummy load. \$\endgroup\$
    – nekomatic
    Commented Dec 2, 2021 at 16:42
  • \$\begingroup\$ BIG Thank You for helpful response. Please be gentle..old-guy trying to learn something here, too. Never thought consuming too little power would be THE problem. Ultimately trying to waste MIN Power Center juice. Just to understand a bit more clearly...COULD I use 200Ohm-1W 20-turn Cement Variable Resistor directly between 12V & Gnd (with 'middle' leg also going to Gnd)? ..start at full 200Ohms & turn a bit at a time until 12V ports remain active? Then remove & measure the resistance to make simple dongle with 1 appropriate Power Resistor between +12V & Gnd? ..and it wouldn't use >1W? \$\endgroup\$
    – Imagr8m8
    Commented Dec 3, 2021 at 0:00
  • \$\begingroup\$ The problem with that is that it's 0 to 200 ohms. So once you get past 83 mA or 144 ohms, you would be exceeding the 1 watt rating. Your better off using a pack of 1 watt resistors in various combinations. \$\endgroup\$
    – Passerby
    Commented Dec 3, 2021 at 0:07
  • \$\begingroup\$ Got it! Intuition screamed not to turn pot even close to 0. A few practical questions: Even though two 200 Ohm resistors in parallel results in 100 Ohms (<144 Ohms you mention), the load is spread between the two resistors, so each doesn't get too hot? Its gonna be about the same amount of heat & waste whether multiple 1W resistors vs one smaller Ohm resistor of much higher watts? \$\endgroup\$
    – Imagr8m8
    Commented Dec 3, 2021 at 1:01
  • \$\begingroup\$ Yes. Roughly equal distribution (based on value tolerances). \$\endgroup\$
    – Passerby
    Commented Dec 3, 2021 at 2:07

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