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I bought a cheap USB battery pack to power a portable project. It works great, except it shuts itself down if there's a current draw lower than a certain amount (haven't pinned it down, but I think it's around 60mA). Some of the time, my actual load is less than that (it's an arduino and some LEDs), although it can run quite a bit higher, too.

I can keep the battery pack on by just putting a resistor in parallel with the load, but obviously that's just burning power. Is there a simple-ish circuit I can build that would ensure that there's always a minimum current being drawn, but draw less or no current when the real load increases? That is, if the LEDs were off, it'd be drawing 60mA, but if the LEDs were drawing 200mA, it wouldn't draw any current?

Emoses
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  • 60mA is a lot to draw just to waste. Wouldn't you rather modify the usb battery pack to disable the auto-shutdown? – Passerby Jul 30 '13 at 07:13
  • This doesn't sound sensible - is it a definite documented feature of the battery? Do you have a link to the battery? – Andy aka Jul 30 '13 at 07:36
  • Battery pack: https://www.sparkfun.com/products/11360. It's built to charge portable devices like phones/tablets, but it's a good value for the capacity, built-in charging and 5v regulator. It can also supply 2A if needed, which is cool. But I don't really know enough to mod it to not auto-shutdown. – Emoses Jul 30 '13 at 07:45
  • So, let me see if I got your question right. You need to constantly drain a battery, but you don't want to waste that energy... So you want to *store* that energy?... The energy thats *coming from an energy storage device?* – travisbartley Jul 30 '13 at 09:43
  • No, that's not it. I don't want to store the energy. I just want to ensure that the whole circuit is never draining less than a certain amount of energy from the battery. What's actually going to happen is tha most of the time I'm going to have the LEDs on, but at a low enough setting that the current draw will be very close to the shutoff value. I want make sure it doesn't shut off. – Emoses Jul 30 '13 at 15:30

2 Answers2

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According to the data sheet: -

The TP4056 automatically terminates the charge cycle when the charge current drops to 1/10th the programmed value after the final float voltage is reached.

And there is a resistor (\$R_{PROG}\$) which can be set to something like 10k limiting the output current to 130mA or it can be set to 5k for 250mA but only you know where to set it to for normal operation.

If your running current is less than say 250mA and it doesn't drop to less than 25mA this is do-able with a 5k resistor. The trick is not exceeding the 10:1 difference in currents taken by your circuit.

Here is the link to the document - take note of the table at the top of page 3.

Andy aka
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  • If I understand correctly, that chip is only used when you're charging the battery, not when it's being dischared. I'm using battery as a voltage source, I'm not worried about what happens when the battery is being charged. – Emoses Jul 30 '13 at 15:27
  • @Emoses as far as I can tell this chip, when the battery is deemed to be charged will switch off if the current draw is below one tenth of the charge current as per the quote above. It's the switching off situation that arises because your circuit fails to draw enough current once the charger believes it has finished its charge cycle. So no, the chip controls the switch off situation based on "something" and it will keep on providing power if YOUR circuit draws more than one tenth of the charge current else it shuts YOUR circuit down - read the brochure. – Andy aka Jul 30 '13 at 17:25
  • I'll have to pop it open and check, I think. I was under the impression that there are two circuits in the box: one that controls charging of the LiPo batteries in the case via the input mini-USB port, which is controlled by this charge control chip, and another circuit which provides a regulated 5V over the two USB output ports, which doesn't go through the TP4056. There's also definitely greater than 10:1 ratio when discharging: the pack can supply up to 3A, but the cutoff current is << 300mA – Emoses Jul 30 '13 at 17:30
  • @Emoses I can only go on what the data sheet says dude. – Andy aka Jul 30 '13 at 17:34
  • Update: I broke open the box on a defective one, and there's a small programmable microprocessor that I'm pretty sure handles the logic for shutting off battery if there's no current. That makes sense, since there's a few-second delay before it shuts off. – Emoses Jul 31 '13 at 23:32
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The most obvious solution to me is to program your Arduino to drive some resistors when it isn't driving LEDs. If you have a spare GPIO port this should be straightforward.

Keep a count of num_leds_lit and when this drops below n, set resistor_driver_pin high.

If you don't like the idea of wasting power, have an idle loop that calculates pi to a zillion decimal places. A busy microprocessor probably burns more milliwatts than a truly idle one.

RedGrittyBrick
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  • Makes sense. I was trying to come up with something that didn't go through the Arduino, but that was because the maximum current through the LEDs is too high for the Arduino. But since I know how much this current will be, and that's not very high, that should be fine. Thanks. – Emoses Jul 30 '13 at 15:35