Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

LM317T power dissipation problem

Status
Not open for further replies.

isaac12345

Member level 2
Member level 2
Joined
Mar 24, 2010
Messages
47
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Activity points
1,634
Hi all,

I am trying to make a NiMH constant current battery car charger which receives an input of 24V from a step up switching regulator using a lm2577-adj and charges a 2300mAh battery .This step up regulator is capable of supplying a maximum of 1A and when plugged in to the circuit supplies around 500mA. I have decided to use a slow charging rate of 230mA. My problem is that without a heatsink the lm317t , which I am using as a constant current source with the help of a 5.6 ohm resistor, switches itself off after a few mins. I suppose this is because of thermal shutdown. What I am confused about is whether the power draw of lm317t is 289mW (230mA X 1.25) or something higher. If so, is the higher value due to it being connected to the step up regulator or something else that I am missing?
 

I don't think you said what the battery voltage is. Assuming that is the 1.25V you mentioned, you LM317 dissipation is (24 - 1.25)*0.23 = a lot! 1.25*0.23 is the power going in to the battery.

Keith
 

Its a battery pack of 8 1.2V NiMH batteries. So the multimeter theoretical value of 9.6V but it reads approx 12 V.
@keith - so going by your logic the lm317T is drawing 5.23W! and i reckon my max power dissipation is 1.5W. and I cant decrease the input voltage because the lm317t does not work below 15v(approx). Any solutions?
 

The LM317 power disspipation is samply current * LM317 voltage drop(not including the sense resistor voltage). If your
measurements are correct, it would be 0.23*(12-1.2) = 2.5 W. Reducing the input voltage from 24 to 15 V reduces the power
dissipation to 0.4 W.
 

If you have 24V in and 12V out then you have 12V drop which at 0.23A is 2.76W. However, you have to allow for flat batteries so it will be more.

The maximum power of the LM317 is not given. It is determined by the junction temperature. With a good heatsink you can dissipate quite a lot of power.

Keith
 

@Keith - Yes, I calculated the max power by (Tjmax - Ta)/ThermalResistance = (125 - 40) / 50 = 1.7W . I took 40 as ambient temperature as I will be using it in an enclosure.

@FvM - I didnt quite understand how reducing the input voltage from 24 to 15 reduces the power dissipation to 0.4W. could you please explain it in a bit more detail?
 

The voltage drop across the LM317 determines the power dissipation.
P = V * I
Please calculate yourself.
 

I haven't a calculator handy, but 15V in and 9.6V out at 230mA must be around 1.25W not 0.4W.

A heatsink still works in an enclosure, you just get a lower C/W or higher ambient, depending on the way you look at it. I would guess a moderate clip heatsink is all that is needed.

Keith.
 

    isaac12345

    Points: 2
    Helpful Answer Positive Rating
but 15V in and 9.6V out at 230mA must be around 1.25W not 0.4W.
Without doubt. I was refering to the reported measurement however:
So the multimeter theoretical value of 9.6V but it reads approx 12 V.
If the output voltage is 9.6V rather than 12V, the input voltage could be further decreased. Also, you didn't consider
the 1.2V sense resistor voltage drop, that doesn't count for LM317 power dissipation.
 

FvM said:
Also, you didn't consider
the 1.2V sense resistor voltage drop, that doesn't count for LM317 power dissipation.

True. I don't have a circuit or data sheet at the moment, but if I knock off 1.25V I still don't get anywhere near 0.4W.

The practical problem is 24V input, a battery which will be only 8V or less when requiring charging (the fact it measures 12V when charged is irrelevant). So either the circuit can dissipate the power that results at 230mA or something has to change. Either the current must be reduced, a heatsink used or a switching regulator.

Keith.
 

So why would you not want to use the switcher and control
it for whatever you want at the battery, rather than run step-up
and step-down in tandem?
 

@dick_freebird - I dont know how to set up the switcher for constant current from a variable voltage source (which is 12 - 15V).
 

I don't know specifically which are the most suitable, but I am sure it can be done. I did a quick search for "constant current" on the LT site and this was one thing it came up with:



There were plenty of others

Keith.
 

    isaac12345

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top