I have a question about uninterruptible power supply (do it yourself).

I have a simple power bank and relay (or light) i want that my relay will be always on even the power adapter are disconnected.

The answer to connect relay from power bank only is unsuitable for. i need to use both power supplies (battery charger and power bank).

Does this connection are correct?

  • 1
    \$\begingroup\$ Depends on the internals of your power bank. It might work as you intended. \$\endgroup\$ – winny Dec 17 '16 at 10:36
  • \$\begingroup\$ What is the intervals? what do you mean? here is a cheapest power bank (1 cell). \$\endgroup\$ – Trainas Dec 17 '16 at 14:51
  • \$\begingroup\$ Internals. You know, how the electronics inside the power bank work. \$\endgroup\$ – winny Dec 17 '16 at 22:13

enter image description here

If you don't know the type of DC/DC converter in supply mode, it is always a good idea to add a diode as shown in the picture.

The benefit is that you will protect your DC supplies from drawing current from another supply.

The drawback is that because there is a voltage drop on your diode (~0.7 V typical), there will be power loss of:

W = 0.7 * I;

So say, your lamp draws 300 mA, then your USB power loses 210 mW on that USB diode (it will get hot - too hot if you don't use a heat sink!).

However, there is a bigger problem in this circuit. If you disconnect from USB, your power bank will be discharging because it is lighting the lamp, right? But, since it is discharging, battery voltage will drop and this will trigger the battery charger to start charging. So, your powerbank will be discharging to charge itself! A cheap powerbank usually has 75% efficiency with their supply/charger units so a "discharge to self charge" situation has:

1 - 0.75 * 0.75 = ~45% Power loss!!

Your powerbank will not be as productive as it is supposed to be. So, this is not a good idea.

However, you can try this circuit:

enter image description here

As you can see, this way, your USB port and powerbank output are protected. Your powerbank is only charged when there is power at USB port. Also it is charged without a series diode on the way, which helps with efficiency.

Also hint: try to pick USB diode with less on resistance a.k.a. ~0.6V while let the powerbank be higher at ~0.7V. This is because you want your USB to both charge the powerbank and light the lamp. If you don't do this, then it is possible that powerbank lights the lamp, and is charged from input by USB. This is inefficient!

  • \$\begingroup\$ Thanks for perfect graphical answer. It was really useful and it works perfect. voltage drop is not so important for my light part, but the second offer is great, because the first one sometimes get stop battery charging. anyway for me ( as a person who do not have much electronics knows and skils is perfect. \$\endgroup\$ – Trainas Dec 21 '16 at 10:56

Add a diode at the positive terminal of the USB power OUT. Anode near the Power out terminal and Cathode near the junction. Hopefully it would work. However, if the input terminal draws current, as in it is charging, it may activate the Short circuit protection.


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