Question: Charging device specified for 15V 3A with 15V 4.6A

[Bicurico]

Hello,

I bought two field meter, which came without the PSU.

They run on DR35 batteries (NiMH, 10.8V 4000mAh).

The manual says:


EXTERNAL POWER SUPPLY
The instrument can be powered by an external DC source
The voltage provided must be from 10 to 18 volts.
Nevertheless, the charge of the battery is possible only if the voltage is greater than 15 Volts.


While the devices work will with a 12V 3A PSU, it won't charge their batteries.

I only have a 15V 4.6A PSU. I used it to charge the batteies and it worked just fine. However, I wonder if using 4.6A instead of the rated 3A will cause harm to the device and/or battery?

Will it charge "faster" and thus develop more heat or strain some components?

Thanks,
Vitor

[pcprogrammer]

That power supply is most likely rated to deliver a maximum of 4.6A, not a constant 4.6A. It depends on the batteries and the regulator, if it has one, how much current will be drawn.

If the circuit draws more current then things will heat up more, but again it depends on the circuit if it will do harm.

[Bicurico]

Thanks.

I forgot to mention that the port says "3A max".

I didn't notice that until today, so I left my 4.6A charger connected over night. Nothing bad seemed to have happened, I just wonder if it will strain the charging circuit or the battery cells. Especially because I ordered a new DR35 battery for one of the two devices and don't want to ruin it due to wrong charging current.

[pcprogrammer]

It depends on the charging circuit and the battery charge level on how much current is drawn. It might well be that the mentioning 3A max on the port is telling that it can't draw more then 3A due to the circuit in it. Next time when the battery needs charging hook in a current meter and just see how much it draws.

[tunk]

10.8V probably means it's a 9S battery, which should be 12-14V fully charged.
Can you measure the fully charged battery? As for charging current, maybe you
could open the device and see what charging circuit it has? As it says 15-18V
for charging so it should have some circuit. And if you have something like a
"kill-a-watt", then you should have an idea about how much power it draws.

[Brumby]

I only have a 15V 4.6A PSU. I used it to charge the batteies and it worked just fine. However, I wonder if using 4.6A instead of the rated 3A will cause harm to the device and/or battery?

A 15V 4.6A PSU will mean it can supply 15V at up to 4.6A.

4.6 amps is the MAXIMUM current the PSU is rated to be able to provide.

The actual amount of current drawn will be determined by the circuit you attach to it.

Will it charge "faster" and thus develop more heat or strain some components?

It should not.

However, if you have charging circuitry that doesn't have any intelligence, then extended connection could cause issues.  To answer that, refer to the manual for your device.

[capt bullshot]

For a device that says: "Use with any external source, as long as the voltage is within a reasonable range", you won't expect the internal charger to rely on the current limiting function of the external source. Your device will run from that 15V / 4.6A supply and use a maximum of 3A from it.

[Bicurico]

Thanks so far.

The manual does not make reference to maximum current. It only mentioned, what i already copy&pasted: 10-18V input and minimum of 15V to charge the battery (with device switched off).

I cannot get access to the charging circuit, as I would have to disassemble the whole device. I won't do that, because it would take a long time and I would have to remove a rather fragile assembly of boards soldered vertically on the main board.

Right now I am charging the field meter with a 13.8V 3A PSU I have lying around. I think it does charge the battery, but I started a new test with the other device and waiting to see the charging LED switch off, to be sure.

What gives me most satisfaction is that there doesn't seem to be a 100% agreement in the replies so far. It means that my qyestion is not THAT stupid...

Regards,
Vitor

[mariush]

I would assume the device has a battery charging IC which could basically be a glorified linear regulator with current limiting features and/or battery discharge features.

Because of this, it may need 1-2v ABOVE the maximum voltage of the battery pack in order to regulate the output voltage sent to the batteries, and the circuit probably also puts a current sense resistor in series with the battery to be able to monitor the current going into the battery, and this current sense resistor will also cause a bit of voltage drop.

It's probably 9 cells x 1.2v nominal voltage making it 10.8v nominal,  but the voltage could be as high as 9 x 1.35 = 12v  so not surprising to see 15v being required.

Being Ni-Mh, the charge current is probably around 1/10th of the battery pack capacity - for example, if the DR35 battery is rated for 4000mAh, I'd assume it would charge the battery pack at 200-500 mA.

Shouldn't be a problem if you use a power adapter that can supply more current than previous power adapter.

[Bicurico]

Thanks! Much appreciated.