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Good way to get 4V - 1A from a 12V supply? Would LM317 work?

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777funk

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I have tried to get 4V at 1A from a 12-14V source and the regulator is getting pretty hot (LM317). It seems like the data sheet said it could supply 1.5A. But I guess that's also up to an 10V*1A=10W power dissipation in the regulator.

Is there a better way to do what I'm trying to do here?

I also tried a MOSFET and PWM. That worked too but the MOSFET got at least as hot as the regulator and I'm sure it wouldn't have lasted long had I left it in circuit.

Curious what I should do.
thanks!
 

wp100

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Hi,

As you show, the regulator has got to deal with a large power so it needs a suitable heatsink.
Running any power component near its limit will shorten its life considerably if you cannot dissipate the heat fast enough.

Clearly a lower input voltage helps, but what is the souce , transformer , smps ?

What are you trying to drive ?
 

777funk

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I'm dropping it down from 12VDC so a transformer isn't an option in this case.
 

erio

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IanP

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I'm dropping it down from 12VDC so a transformer isn't an option in this case.

Conect several diodes between 12V and the input of the LM317, so the actual voltage at the input will be somewhere around 7-7.5V. Then you will need much smaller heatsink.
Each diode drops roughly 0.7-0.8V @1A ..
:wink:
IanP
 
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tpetar

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Better use LM350 3A or LM338T/K (5A).

Also IanP suggestion with diodes will work nice.
 
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sherazi

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I have used LM317 to get 3.9V from 12V Dc supply , and it works, I used it for a gsm module based design which requires somewhat 2A in bursts when transmitting, and at that time the regulator gets hot, using a heatsink solves the problem and it works fine , have tested it to work for long time.
 

tpetar

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Sherazi this of course works LM317 at 12V, but there is high dissipation of power in voltage regulator, because difference in I/O voltages. Why to put some parts on cooking when we have cooler solution. Heat killing semiconductor.

;-)
 

Tahmid

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I have tried to get 4V at 1A from a 12-14V source and the regulator is getting pretty hot (LM317). It seems like the data sheet said it could supply 1.5A. But I guess that's also up to an 10V*1A=10W power dissipation in the regulator.

Is there a better way to do what I'm trying to do here?

I also tried a MOSFET and PWM. That worked too but the MOSFET got at least as hot as the regulator and I'm sure it wouldn't have lasted long had I left it in circuit.

Curious what I should do.
thanks!

Use a DC-DC buck converter. You can try with something like LM2575 or L4971. With buck converters, since the voltage isn't reduced by power dissipation as heat, but instead by PWM, you get less heat.

If you tried with PWM and MOSFET, and the MOSFET got very hot, there must have been some mistake there. If you want to try and fix that, you should post your circuit so that we can have a look.

Here's one circuit using LM2575 you can try:
97_1328708336.gif


Hope this helps.
Tahmid.
 

papunblg

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Hi,

As you show, the regulator has got to deal with a large power so it needs a suitable heatsink.
Running any power component near its limit will shorten its life considerably if you cannot dissipate the heat fast enough.

Clearly a lower input voltage helps, but what is the souce , transformer , smps ?

What are you trying to drive ?

Yes, The main question is what are you trying to drive?

Conect several diodes between 12V and the input of the LM317, so the actual voltage at the input will be somewhere around 7-7.5V. Then you will need much smaller heatsink.
Each diode drops roughly 0.7-0.8V @1A ..

IanP
This is really an intelligent solution. A very low cost solution without much changing the circuit.

The drawn current, input voltage etc. all are within range. Use good heat sink.
 

jamespond

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Try L4960. Easy to build.
 

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  • l4960.pdf
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Tahmid

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This is really an intelligent solution. A very low cost solution without much changing the circuit.

The drawn current, input voltage etc. all are within range. Use good heat sink.

If heat is the only issue, this is a great idea. But, if efficiency is also a concern (which doesn't really seem like it from the OP's original post), this doesn't really help much as efficiency will be roughly the same.

But one thing should be kept in mind: the same current is going to be flowing through all the components (all the diodes). So, check the diode's current capacity and also thermal characteristics to estimate how hot it might get. Because, after constructing the circuit with say 10 diodes, you don't want to see all 10 diodes heating up.

Hope this helps.
Tahmid.
 

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