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I am supposed to drain a constant current of 2A or 3A from a battery (most probably a NiMH battery). What could be a possible way to do that ? Purpose of doing it is to calculate the battery's Energy (using a micro controller) for the discharging cycle, while keeping its current constant.

Thankyou,

EDITED

What if i am using a buck-boost converter with a micro-controller to charge my battery and then discharge it? (Buck topology will work during the charging cycle and boost topology during the discharging cycle.) Would the following idea be practical ?

I mean if i choose a resistor (lets say for discharging 15V battery @ 20A) and connect it to my circuit given above during the discharging (also disconnecting my circuit from the main power supply), would i be able to drain a constant current for my test purpose ? I dont know exactly how the boost topology will exactly work during the discharging cycle. But looking at the circuit, will i be able to draw a constant current out of my battery using PWM from my micro-controller ? And, would placing a constant value resistor for the discharging be a correct method to do (in this case) ? Please help me if i am thinking totally wrong. Please tell me would my idea be practical or i am just being silly here. (sensors and resistors which will be used to measure the voltage & current supplied to and drained from the battery are yet not drawn here by me) Thankyou.

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  • \$\begingroup\$ A current sink? \$\endgroup\$
    – PlasmaHH
    Commented Feb 24, 2015 at 12:35
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    \$\begingroup\$ Search for eg constant current load \$\endgroup\$
    – Russell McMahon ♦
    Commented Feb 24, 2015 at 12:49
  • \$\begingroup\$ Stack Exchange EE is a good place to look :-) - eg se Abdullah's answer here \$\endgroup\$
    – Russell McMahon ♦
    Commented Feb 24, 2015 at 12:52
  • \$\begingroup\$ You need to think about what tolerances you need to hold. If, for instance, you need to monitor a battery from 1.2 to 1.1 volts (terminating the test when the battery voltage drops below 1.1), and the current can vary by +/- 5%, a simple resistor will do the job. If you need the current specified within +/- 1%, that's a different story. \$\endgroup\$
    – WhatRoughBeast
    Commented Feb 24, 2015 at 14:28

2 Answers 2

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The usual solution to this is a dummy load, or 'active load'. These are pieces of test equipment specifically designed to consume a configurable constant current. Many can also measure and accumulate the total number of amp hours or watt hours consumed.

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  • \$\begingroup\$ Ive edited my orignal question, can you please share your thought on that ? Thankyou. \$\endgroup\$
    – yiipmann
    Commented Apr 9, 2015 at 14:25
  • \$\begingroup\$ There's really no reason to use a switching regulator for the discharge stage; use a constant current load, or just a resistor if your voltage is fairly consistent. \$\endgroup\$
    – Nick Johnson
    Commented Apr 12, 2015 at 17:33
  • \$\begingroup\$ I just looked into constant current loads and found out that they will require designing (additional work). Also i dont think the voltage is fairly consistent, as the maximum voltage is 15V, and minimum would be around 3V. So would a constant resistor will do in this case ? Also, is there any drawback, if i use switching regulator for the discharging stage ? \$\endgroup\$
    – yiipmann
    Commented Apr 13, 2015 at 11:52
  • \$\begingroup\$ No, a resistor won't work if the voltage varies - you're going to have to have an active circuit of some form. There's no disadvantage per-se of using a switching converter to discharge - there's just no real point in doing so, and it will require designing, which you appear to be allergic to. \$\endgroup\$
    – Nick Johnson
    Commented Apr 13, 2015 at 12:51
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I believe the setup in the question should work but I'd probably want to disengage the commercial voltage source when discharging via the boost (buck operating in reverse). There might be some damage done to the commercial voltage source with reverse voltages and, if you use a relay with a changeover contact this is easily achieved.

The value of R depends on how much you are boosting but bear in mind, with the figures you have quoted, the power is going to be 300 watts - it's going to get warm and you may want to remove that heat with a fan.

I would also want to monitor the voltage across the resistor to make the minute changes to PWM duty cycle needed to keep this voltage constant.

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