2
\$\begingroup\$

I have Nitecore UM2 (2 slot) battery charger whose current sense/shunt resistor of 0.1 ohms R100 (package 1206) blown. I replaced with working slot to see if everything works good and it is.

I am looking to replace this R100 (0.1 ohm) with R005 (0.05 ohms) shunt resistor. I have two of them found on dead laptop BMS. One BMS had three 18650 in series and another had four 18650.

I use this charger to charge Nimh and 18650 batteries. Charger can output 1A max/cell (adjustable). So 4.2v 1A max is what is going to transfer from this shunt.

The two R005 shunt resistor I have are 1206 and 2512 package. And the question is can I use this shunt in replacement of R100? If yes which package? Will there be an issue with charger providing accurate CC i adjust it to?

I have no idea about this so I had to ask. I searched on the internet already but found nothing.

\$\endgroup\$
3
  • \$\begingroup\$ "Will there be an issue with charger providing accurate CC i adjust it to?" - what do you mean by 'adjust it to'? \$\endgroup\$
    – Bruce Abbott
    Commented Jan 9, 2023 at 11:17
  • \$\begingroup\$ The charger allows me to select the charging ampere from 100mA to 1000mA. That's what I mean if I adjust the current to 500mA. Will the charger be still be providing 500mA even if shunt is replaced with different value. \$\endgroup\$
    – Paul
    Commented Jan 9, 2023 at 15:22
  • \$\begingroup\$ R005 would be 0.005 ohms, not 0.05. That's a 20x increase in current compared to the original R100 resistor! \$\endgroup\$
    – Hearth
    Commented Jan 9, 2023 at 19:55

1 Answer 1

Reset to default
5
\$\begingroup\$

Please do not change the resistor value of the shunt.

U=I*R, the voltage drop over the shunt is used by the charger to measure - and regulate - the current.

When resistance is halved, the voltage drop is halved at the same current - or: the current doubles. It would result in wrong capacitance measurements (showing only half of the actual value). It might result in damaging the rechargeable battery (by applying to high current). And it might result in failures or damages at the charger (when it draws much higher current than its power-supply can handle).

\$\endgroup\$
1
  • \$\begingroup\$ Thank you so much for your response. I will buy the new one then. \$\endgroup\$
    – Paul
    Commented Jan 9, 2023 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.