# Power bank with low discharge load

Discussion in 'General Electronics Chat' started by MixXx2005, Jun 3, 2016.

1. ### MixXx2005 Thread Starter New Member

Jun 3, 2016
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0
Hello, I have built an electronics project, that uses 0.003A@5V 95% and 0.090A@5V 5% at the time.
I powered it by 2200miliamp/h Chinese power bank.
By my calculations it should keep my project running for about around 200 hours. But from the testing results I get only around 75 hours.

When I measured current before power bank's circuit - it showed me 0.020A@5v 95% time and 0.150A@5V 5% of the time.
So biggest increase caused by power bank circuit is at 95% time , when there is small load.

Is there any way to get around this problem (simple way) , for example with capacitors and transistors to turn off load from power bank and power it by capacitor ,when small load is needed?

Or I just need more efficient battery circuit?

Thanks!

2. ### kubeek AAC Fanatic!

Sep 20, 2005
4,670
804
Do you need the 5V or could you do with 3.3V? You´d be better off if you could run straight from the battery instead of converting it up to 5V and then possibly back down again.

3. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,386
1,605
What do you mean by power bank? Is that a battery or something else?

Why do you have two sets of current numbers? I would need a sketch of your setup to have any rational reply.

4. ### HW-nut Member

May 12, 2016
45
11
Based on your description, I’m assuming your circuit is sleeping most of the time and only wakes up periodically.

My first thought is how you are measuring the current. If the load (i.e. current draw) is brief pulse and you are using a DMM for the current measurements. The reading may be inaccurate due to the crest factor of the current waveform. Basically, the current pulse is too fast to measure with the meter which in turn measures a lower value.

5. ### BobTPH Active Member

Jun 5, 2013
782
114
The power bank is probably using a boost converter which has a quiescent current higher than your total current draw during the low current phase.

Bob

6. ### MixXx2005 Thread Starter New Member

Jun 3, 2016
3
0
Two current measurements are because I measure before charging/voltage regulator circuit and after it.

And yes, my guess is that regulator circuit use more than device in sleep mode, how can I avoid this?

My example would be to use some sort of another current accumulating device - Capacitor? that I can charge quickly and when device is in sleep mode, voltage regulator also is not used, but it takes power from accumulating device.

How can I do this king of thing?

Here are basic schematic.

7. ### kubeek AAC Fanatic!

Sep 20, 2005
4,670
804
Like I said before, do you really need the full 5V? If you could use 3.3V everything will get a lot easier.

8. ### MixXx2005 Thread Starter New Member

Jun 3, 2016
3
0
I do not know how much V 1 cell will give me, but for safety I think it is not good to take out charging circuit, because of overdischarging risk and also I need and easy way to charge cell.

9. ### kubeek AAC Fanatic!

Sep 20, 2005
4,670
804
I didn´t say you should take the charging circuit out, that has to stay. What I said is you will be better off using the 3.5 to 4.2 V raw battery voltage to power your circuit in standby via a linear regulator, or even for the whole cycle.

10. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,386
1,605
Ok, so the "power bank" has 5v at te input from the battery, and 5v on the output for your device. What is it doing for you? Same output as input... I can do that with a piece of wire.

If it is a charger then you can find jacks to switch it in and out when charging and not charging so it s disconnected during te 200 hour run time.