LM317: About the output current capacity of LM317 and LM337

Hi Jason,

Thx for your answer.

Besides, is there alternative component that is pin compatible with the LM317 and has a higher power rate?

Best Regards

Yatao

Hi Yatao, 

Will you help to provide your Vin, Vout, and Iload? 

Regards, 
Jason

Hi Jason,

Sorry for the delay.

Actually, the load is a PCB and it consists of several op amps and push pulls. The power is around 2.5W.

I did not find the "Rja value can be found under section 7.4 in the datasheet" for LM317KCS, can you tell me its value?

Best

Yatao

Hi Prakash,

Thanks for your reply!

Have a nice day!

Best regards

Yatao

Hi Prakash,

I find that I have made a mistake  that I have misunderstood the cencept between the power losses of LM317 itself and the power LM317 can provide to its load.

The 4.2W is LM317's internal power losses. But is it correct that it can provide much higher power for its load? Actually, I care about the power LM317 can deliever to the load when it doesn't suffer from overheat.

For example, 317's input and output voltage are respectively +28V and +24V. To make the full use of 317 and avoid its overheat, the 317 output current is set at 1A. Then 317's internal losses are 4W and it can provide +24V*1A=24W power to its load.

Are my calculation and analysis correct?

Best

Yatao

Yatao,

Please see calculations below.

PD(max) =(TJmax- Ta)/RqJA

Where

TJmax is the maximum allowable junction temperature.

RqJA is the thermal resistance junction-to-ambient for the package

TA is the ambient temperature.

The regulator power dissipation is calculated using;

PD =(Vi-Vo) x Io

PD(max) = (150-25)/23.5 = 5.31W

PD with Vin = 28V, Vout = 24V & Iout = 1A is (28-24) x 1 = 4W which is less than 5.31W and there should be no issues.

Please let me know if this is what you are looking for?

Dear Prakash,

Thanks for your detailed calculations.

Actually, I'm looking for the maximum power LM317 can deliver to the load.

Then I think that in your calculation , such a maximum poweris Vout*Iout=24W. Is it ok that Lm317 can generate 24W power for its load with no overheat within itself?

Best Regards

Yatao

Hello Yatao,

The maximum power that LM317 can withstand would be 5.3W. Anything above that will cause thermal shutdown of the part.

PD(max) =(TJmax- Ta)/RqJA

PD(max) = (150-25)/23.5 = 5.31W

Thanks,

Prakash L

Hi Prakash,

I know what you mean.

But what I want to confirm is that when LM317 is withstanding 4W in your calculaitions, its output power is Vout*Iout=24W, is this calculation right? Then, is this 24W ok for LM317 to output to its load?

Below is your calculation

PD with Vin = 28V, Vout = 24V & Iout = 1A is (28-24) x 1 = 4W which is less than 5.31W and there should be no issues.

Best Regards

Yatao

Hi Yatao, 

The limiting factor for the maximum power an LDO can support is based on the differential between Vin and Vout and the load current. 

PD with Vin = 28V, Vout = 24V & Iout = 1A is (28-24) x 1 = 4W which is less than 5.31W and there should be no issues. 

--Yes, this calculation is correct and yes, the LDO should work for this kind of power delivery. 

Regards, 
Jason Song

Hi Jason,

OK!

Then in this condition, LM317 is outputting Vout*Iout = 24W to its load. is this correct? :-)

Best Regards

Yatao

Yes, this is correct. 

Hi Jason,

Thanks for your help and explanations!

Best Regards

Yatao