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I have a 20V 2.25A laptop power supply that i want to split 3 ways to charge 3 X NiMH battery packs simultaneously (5000,2000,2000 MAh each @ 7.2V).

My plan is to use 3 of these XL-4015 dc to dc step down modules to regulate the voltage and amperage. My question is how can i stop the battery from charging when its full without me having to monitor them.

I'm a hobbyist not a pro. so don't get too technical with me :-)

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    \$\begingroup\$ Welcome to EE.SE. Lithium ion batteries and some others are charged at their rated voltage until the current drawn drops below 10% of the Mah rating, upon which the charger shuts off until the battery voltage drops 20% to 30%. In other words you need a semi-intelligent charger to be safe. \$\endgroup\$ – Sparky256 Feb 14 at 0:45
  • \$\begingroup\$ @Sparky256, thanks for the comment. I read from one of the other answers here in SE that i can safely charge at one tenth of the C (500 Milliamps for a 5000Mah) for a slower charge with a timer to cut off power at the 5th hour. is that correct? \$\endgroup\$ – Ahsan Feb 14 at 3:08
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    \$\begingroup\$ You need something smarter than a timer. If the battery charges in one hour then keeps charging 4 more hours it may begin to 'cook' a little bit. Better to trickle charge at 1% of C, or try a timer and thermo-disc type temperature sensor that would shut off the charge if battery gets over \$180^oF\$. \$\endgroup\$ – Sparky256 Feb 14 at 3:17
  • \$\begingroup\$ 1% of the C being 50 Milliamps? also if i do that it can be left on charge indefinitely? \$\endgroup\$ – Ahsan Feb 14 at 3:26
  • \$\begingroup\$ If your charger can both shut off with time and temperature then it is safe to leave the battery connected. If the battery voltage drops below 5 volts the charger should come back on at 10% of C. \$\endgroup\$ – Sparky256 Feb 14 at 4:48
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NiMh batteries requires 1.4V per cell to charge full, the data provided indicates it's a 6 cell pack. I assume you have limited component on hand, I give you the below circuit.

It will charge the battery to about 90% full and you can leave it to charge overnight; without damaging anything. Make sure the resistor is 2W and the diode is rated 2A for safety. Do make sure the BAT1 terminal is at 8.3V max after long period of charge.

Technical: Reduce R1 to 1 ohm will increase the charging current, allows faster charging time. D1 is normal diode, not schottky diode; some may spec 0.6V forward voltage. You may want to start the V1 (dc-dc) voltage at 8.7V to have a feel first.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ but why would i need the resistor? if i have the dc to dc step down, i can bring the voltage down to 8.3V with the step down. \$\endgroup\$ – Ahsan Feb 14 at 2:51
  • \$\begingroup\$ @Ahsan. You can omit the resistor if you can confidently limit the current to 0.5A. You may also remove the diode as long as the battery pack that you use are not higher voltage than the dc-dc output. \$\endgroup\$ – Jason Han Feb 14 at 2:59
  • \$\begingroup\$ This is a dangerous circuit as the rising impedance of the battery begins to make D1 behave like a short. 2 ohms is not enough current limiting if no active current limiting is used. \$\endgroup\$ – Sparky256 Feb 14 at 6:10
  • \$\begingroup\$ @Sparky256. Please advise how does a diode become a short in this scenario? Also this is about NiMh battery, not Li ion battery as you mention in earlier post. The diode use is not a zener diode, the only way to short it is apply high enough reverse voltage. Applying 50V to 1N4001 won't even break it. NiMh rated voltage is 1.2V x 6 = 7.2V, I = (8.3-7.2)/2 = 0.55A As the battery voltage rise, the charging current will gradually reduce. This battery charger is not constant current, this is a constant voltage charger. \$\endgroup\$ – Jason Han Feb 14 at 6:45
  • \$\begingroup\$ A series diode is not a voltage or current regulator or limiter. As the battery charges the Vdrop of the diode goes down. \$\endgroup\$ – Sparky256 Feb 14 at 7:15

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