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LM317 very hot when use as 300mA constant current

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tong

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I made these circuit to temporary recharge my deep-discharged ni-mh 1.5v sanyo eneloop battery.But LM317 got very hot, can't touch. Did I wrong?

LED is red color. I use LM317 since Vref is only 1.25v then I can use the small resistor under 1W. I measured the current flow across a battery is valid ~320mA.
 
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The LM317 will need to dissipate about 2.4 W, so without any heat-sink it will get hot.
 
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The power dissipated in the LM317 is (9V-1.25V-1.5V) * .32A = 2W (E-design apparently included the power dissipated in the resistor in his estimate).
And that is sufficient to cause the bare LM317 to get quite hot.
You'll need to mount it on a small heatsink.
 
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I think the LED will also be very hot if it is a 1.8V red one: (9V - 1.25V - 1.8V)/150 ohms= 39.7mA.

Note, the currents are even higher when the battery cell is dead.
Your "charger" never turns off which is bad for a Ni-MH cell at such a high charging current.
 
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Maybe adding a dropping resistor in front of the 317 can be an option.
 

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Maybe a 9V battery is used because its voltage drops to only 4.8V (as shown on Energizer's datasheet) when it is almost dead. Then adding a dropping resistor in front of the voltage regulator will cause it not to work.
 

I second AG's warning: you must have a way to sense the battery (temperature, voltage) and cut off the charger when a certain threshold is reached.

Such a large current, if left unattended, may damage the battery.
 

Energizer and a Japanese battery manufacturer say the trickle charge current of a Ni-MH battery should be less than 1/40th its capacity rating. If its rating is 2000mAh then the trickle charge should be less than only 2000/40= 50mA.

These modern Ni-MH batteries hold their charge for 1 year then they do not need a trickle charge.
 

I tried to reduce input voltage to 7.5v, the LM317 didn't start.

Typical LM317 TJunction max = 125 c
Typical LM317 θJunction to Ambient = 50 c/w (not sure)

Tmax - Tamb = 125 - 30 = 95 c

Power Dissipation = ( Vin - Vout ) * Iout = (9 - 1.25 - 1.5) * 0.300 = 1.875 w

θ = 95 / 1.875 = 50.666 c/w
 
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You should not operate an electronic part at its maximum allowed temperature since your ambient might be hotter than 30 degrees in summer or when enclosed and the part might not last very long if it does not get hot enough to go into thermal protection.
 

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