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### Topic: 100mA battery discharging circuit (Read 1 time)previous topic - next topic

#### fatality

##### Apr 16, 2012, 02:49 pm
im want to build a discharging circuit using arduino to discharge a 300mAh 8.4v NiMh cell. been doing some research on the web for weeks now and stumbled upon a constant current sink circuit using tl431. The circuit is on page 8 of the datasheet
http://www.100y.com.tw/pdf_file/38-ON-TL431,NCV431.pdf

before building the circuit i would like to get some suggestion from fellow arduinians. Do i have to use high wattage resistor for Rs? or is there any better or easier way to discharge a battery at 100mA??

#### MarkT

#1
##### Apr 16, 2012, 05:37 pm
You can calculate the dissipation in the resistor very simply - P = IV, V = IR, therefore P = I^2 R.  Most standard through-hole resistors are rated for 0.25 to 0.6W depending on exact type.  For a 2.5V reference and 0.1A discharge that means a 25ohm resistor which is 0.25W, so should be OK (it will run hot).  You could use 4 100ohm resistors in parallel to get 25 ohms and a lot less heating.

But back to the original issue - this rate of discharge means something has to dissipate 8.4 x 0.1 = 0.84W (in the circuit you give it will be mainly the transistor, necessitating a power transistor with perhaps a small heat sink.

A far simpler solution is an 82 ohm 2 or 3W power resistor (1W would be enough but again would be pretty hot).  Battery voltage is fairly constant (till discharged) and thus a resistor will be a constant current sink in effect.

Be careful not to let the battery over-discharge, this can drastically reduce the life of most rechargable batteries.
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