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I am trying to make a charging circuit to charge 2 NiMH batteries via trickle charging with 0.015A (15mA)

I have used a current regulator and basically J1 is connected to the source of 24V DC and J2 are the 2x9V NiMH batteries in series that are going to be charged.

Circuit in KiCAD

I'm unsure where to connect the '2' from the J2 as previously I just connected to ground. Is ground just the same thing as the negative terminal on a battery?

As previously it looked like this before I got confused. enter image description here

But the '1' from J1 has to connect somewhere right?

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  • \$\begingroup\$ Why does your title say 24 volts when your battery is 18 volt? \$\endgroup\$
    – Andy aka
    Oct 29, 2019 at 8:09
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    \$\begingroup\$ Even at 15 mA the batteries MAY be damaged by trickle charging. The number of cells in a "9V" battery is uncertain. Some use 8 (8 x 1.2 = 9.6V) but others 'cheat and use less (maybe 6 x 1.2 = 7.2V nominal. I recommend that you clamp the battery voltage at (number of cells x 1.4V) eg IF there are 6 per battery Vclamp = 1.4 x (6 + 6) = 16.8V. If each has 8 cells then 16 x 1.4 = 22.4V. \$\endgroup\$
    – Russell McMahon ♦
    Oct 29, 2019 at 9:21
  • \$\begingroup\$ Connect IN-1 in lower diagram to ground (which is essential for operation) and the two circuits are functionally identical. \$\endgroup\$
    – Russell McMahon ♦
    Nov 10, 2019 at 1:45

1 Answer 1

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But the '1' from J1 has to connect somewhere right?

Yep!

Your first diagram is on the right track, much better than the second diagram, regarding the connections at J1 and J2.

Be aware that you've got negative on pin 1 on one side, and on pin 2 on the other side. This could cause confusion when connecting the batteries.

Is ground just the same thing as the negative terminal on a battery?

It's not exactly the same thing, but it's often treated as the same thing. You can consider ground equivalent to the negative terminal in this application.

Regarding your circuit though, I think you should have a resistor voltage divider to set the LM317 output voltage to ~18V. This page might help to get some sample values: https://circuitdigest.com/calculators/lm317-resistor-voltage-calculator

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  • \$\begingroup\$ The LM317 is connected as a current source. Iout = 1.25V/80 =~ 16 mA. \$\endgroup\$
    – Russell McMahon ♦
    Oct 29, 2019 at 9:18
  • \$\begingroup\$ Won't that result in overcharging though? \$\endgroup\$
    – abb
    Oct 30, 2019 at 22:30
  • \$\begingroup\$ " .... overcharging ..." -> Yes. See my comment on the OP's question re clamping the battery at 1.4V x number of cells. This also applies to a voltage source that is not tailored to the number of cells. eg 16 cells x 1.4V = 22.4V . If his 24V is accurate then 24/16 = 1.5V /cell which may be too high and so MAY never stop taking current. If the batteries have < 8 cells (as many do) then it's worse again. \$\endgroup\$
    – Russell McMahon ♦
    Oct 30, 2019 at 23:16
  • \$\begingroup\$ @abb, if I use a resistor voltage divider, wont my current not be constant anymore? Since i'm using the LM317 as a current regulator \$\endgroup\$
    – Po Chen Liu
    Nov 1, 2019 at 3:21
  • \$\begingroup\$ Yes, that's right. You should add some kind of over-voltage cut out or limit to your circuit. \$\endgroup\$
    – abb
    Nov 1, 2019 at 3:37

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