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[on:]

• Will I get a constant 1.5A output current (as stated in the datasheet) although having a power source which its current is lower?
• Although having 6V as the power source, can I get the output voltage at 12V and also constant 1.5A output current?

Can anyone answer my questions and tell me more about it? Began to get confuse on voltage regulators.

Datasheet: http://www.national.com/ds/LM/LM117.pdf

[Reply #1 on:]

Will I get a constant 1.5A output current (as stated in the datasheet) although having a power source which its current is lower?

No. I can get only max current what your power supply capable to provide. 1.5 is a upper limit for L317, so 0 - 1.5 A any value in the range is o'k

Although having 6V as the power source, can I get the output voltage at 12V and also constant 1.5A output current?

No. You can get only 6 - drop off voltage, about 4-5 V.
For 12 V you need DC-DC convertor

[Reply #2 on:]

One more Q, if I input a 12V 0.5A (6W) into the LM317 and expect a 6V output from the LM317, will I get a 1A output current? (Because 6W)

[Reply #3 on:]

One more Q, if I input a 12V 0.5A (6W) into the LM317 and expect a 6V output from the LM317, will I get a 1A output current? (Because 6W)

No. LM317 is a linear regulator. but you are on right path, just look for other non-linear chip, google DC-DC converters application notes

[Reply #4 on:]

So, It'll depend on the power source, right?

[Reply #5 on:]

You mixing up 3 things , which should be considered separately:
1. Max output current, data usually you can get from IC specification;
2. Max Power from power supply what you have;
3. Voltage what you need.
I say you are on right path on 2-nd, 6 V / 1 A at input  could be converted to 12 / 0.5 at the output.
(Actually little bit less, as 'd be conversion losses ).

[Reply #6 on:]

Confused again. So if I had a 12V 0.5A (6W) power source, and I want a converted voltage of 6V. I will have my output at 6V 0.5A? or 6V 1A ?(Power=Voltage x Current = 6W)

[Reply #7 on:]

Confused again. So if I had a 12V 0.5A (6W) power source, and I want a converted voltage of 6V. I will have my output at 6V 0.5A? or 6V 1A ?(Power=Voltage x Current = 6W)

The LM317 is a linear regulator.  A characteristic of linear regulators is that CURRENT IN is the same as CURRENT OUT.  The input and output voltages do not matter.  It is how this type of regulator works. 

Since current stays contant, the voltage difference from input to output is dropped across the regulator.

So if your load is drawing 500mA, this is how the math works.

Input:        Vin * Iout = 12V * 0.5A = 6W
Output:      Vout * Iout = 6V * 0.5A = 3W
Regulator's voltage drop:  Vin - Vout = 12V - 6V = 6V
Regulator:  Vreg * Iout = 6V * 0.5A = 3W

Now, consider if your input voltage was 9V and the output was 6V.
Input:         Vin * Iout =  9V * 0.5A = 4.5W
Output:       Vout * Iout =  6V * 0.5A = 3W
Regualtor's voltage drop:  Vin - Vout = 9V - 6V = 3V
Regulator:    Vreg * Iout = 3V * 0.5A = 1.5W


See, the total Power In is equal to the total Power Out PLUS the Power wasted by the regulator.

If you want to get more voltage or current from an input, you need a DC-to-DC boost converter.  A simple linear regulator will not work that way.

[Reply #8 on:]

Is the LM317's maximum output at 1.5A? Can it handle over 1.5A of current (not output, handle)?

[Reply #9 on:]

Is the LM317's maximum output at 1.5A? Can it handle over 1.5A of current (not output, handle)?

If properly heat sinked, the LM317 can provide up to 1.5A, assuming the input source can provide 1.5A. 

[Reply #10 on:]

Without a proper heatsink, the LM317 with shutdown automatically due to built-in thermal protection at closer to 1A or less not 1.5A.    This is one of those  "reading the datasheet" things that trips people up... Maximum ratings are not the same as Nominal ratings.  You should never PLAN to operate a device at it's Maximum ratings... as those values are there for the designer to know "the point at which you can cause the device to stop working permanently".

Another key point: Physics has some pretty tough laws... and one of them...  very oversimplified is:   "You never get work done for free."   Example: If you start with 6V at 1A power source... and build a circuit that in fact DOES give you 12 Volts out... something has to pay the price... and in this case... it's current availability... you will get 12V Volts... but getting it a the 1 Amp available on the "feed" side  is out of the question.  You will get reduced current availability in a voltage boost solution.

Something else to keep in mind is that a regulator that is adjustable... is using an external reference that is based on actual regulator output.  A fixed regulator has an internal voltage reference that it uses to regulate against.

So, keeping that in mind, if you have a 5V regulator and you feed it 3 volts, the reference circuit never gets to tell the regulator what to regulate against... so regulation of voltage cannot occur.