I bought a cheap Ni-Mh battery charger from Aliexpress to charge my rechargeable AA and AAA 1.2V batteries but before I plug them in, I checked the voltage on the charger terminals and it is reading 5V.

I am attaching the link to the product page and also will link image containing specs of the charger below.

So, I don't think I should keep the batteries for charging right? or can it have some intelligent voltage switching technique or maybe if I put all four batteries in it will divide the voltage which it should not but I am not an expert so.... Edit: Added internal circuit images

Product link : Charger

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If you know how (simple) this charger is, there is no surprise that you measure 5 V and the 5 V is totally expected. Also there is no "intelligence" going on in this charger, the circuit is as basic as it can get and will be very similar to this:


simulate this circuit – Schematic created using CircuitLab

If the battery is removed and you use a voltmeter (which has a very high resistance) almost no current (about 0.5 uA) flows so almost no voltage (theoretically 7.5 uV) will drop across R1. So basically there will be 5 V across the battery terminals when there is no battery connected.

When you do place a battery it will have a voltage of around 1.2 V, that will give about 5 V - 1.2 V = 3.8 V across R1 so about 3.8 V / 15 ohms = 0.25 A will flow and that will charge the battery. When the battery is full there might be 1.4 V across it and that will decrease the charging current only slightly to 0.24 A.

That 0.24 A when fully charged will make your batteries warm up a bit, it is then better to remove them from the charger!

Edit: after you added photos from the charger's PCB we can see that there is a large (2.2 ohm?) resistor and it is shared by all the charging circuits. It does not change much to the principle of operation of this charger.

  • \$\begingroup\$ Thanks for the quick response with detailed explanation.I am getting 1.27v reading when i check after connecting the battery. So what i understand is because of change of internal resistance the voltage changes. Now I understand that we should monitor voltage with the system completed. Also are there voltmeters that work even in this type of situations? \$\endgroup\$ – Tinted Dec 7 '18 at 10:13
  • \$\begingroup\$ Monitoring the voltage to determine if the cells are full is not a very reliable way, the voltage of the cell is also temperature dependent. Reliably detecting if a NiMh cells is fully charged is quite complex and only needed when fast charging. That 0.24 A is not too much to overcharge the cells for a while, they can handle that. Read more at: en.wikipedia.org/wiki/Nickel%E2%80%93metal_hydride_battery and batteryuniversity.com/learn/article/… \$\endgroup\$ – Bimpelrekkie Dec 7 '18 at 10:21

This is really a cheap knock-off "charger", beyond any consideration. The 5V reading is for no-load, since to charge anything you need to apply higher voltage. The current limit in this charger is set by series resistor(s), and with no load you read the value of its power supply.

NiMH batteries, and nearly all batteries, do not like to be overcharged. The charge termination conditions for NiMH batteries are complicated, see this Battery University article, and follow their recommendations. If the charge will be not timed manually, the lifetime of your cells will be decreased, and all your investments into rechargables will go into drain. I would advise to throw this "charger" away, and get something that is more intelligent and terminates the charge automatically.

  • \$\begingroup\$ i will read some about the Ni-Mh charging and if this charger does not have auto stop then i will for sure ditch it today. \$\endgroup\$ – Tinted Dec 7 '18 at 18:13
  • \$\begingroup\$ just now ordered a charger with an IC that controls the charge and everything. Thanks for the advice. \$\endgroup\$ – Tinted Dec 7 '18 at 18:17

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