LM317 output question

[HagenK]

Lets say I connect vin to 12V DC 0.6A, use 240Ω and 910Ω resistors. Vout should be 6V DC, but what about amps? I need at least 1A.

 

[alexan_e]

You can 't get more current from the regulator output than the current available from your source.

 

[KlausST]

Hi,

look for step down switching regulators. With an efficiency of at least 80% it should be possible.
Not with LM317.

Klaus

 

[HagenK]

So the end result will be 6V DC 0.6ampsMAX?

 

[alexan_e]

So the end result will be 6V DC 0.6ampsMAX?

Yes, LM317 works as a variable resistance dropping a few volts so that the output remains constant to the set voltage, but it can't provide current that is not available from the power source.

 

[HagenK]

An other quick question, what wattage resistors should I use for this scheme?

 

[alexan_e]

The difference between ADJ and OUT is constant to about 1.2v so P=V*I = V * V/I = 1.2v * 1.2v / 240 Ohm = 0.006W

The voltage on the 910 Ohm will be 6v-1.2v = 4.8v
4.8v * 4.8v / 910 Ohm = 0.025W

- - - Updated - - -

Note that 240 Ohm may be high, you need the resistor to draw 10mA as a minimum load so that the regulator operates within specs ( >10mA output).

For proper regulation with any load you should use 1.2v/0.01A = 120 Ohm or lower

 

[HagenK]

But when why do many people chose to use 220-240 ohm rather than 120?

 

[alexan_e]

The typical application shows 240 Ohm but usually it shown LM117 and not LM317

https://www.ti.com/lit/ds/symlink/lm117.pdf


LM117 is specified with a min output current of 5mA
LM317 is specified with a min output current of 10mA

If you know that your load can draw a current so than the current of the 240 Ohm resistor and the load current is >10mA total then there is no problem but if you try to measure the output voltage without load then you may get strange results

 

[HagenK]

The voltage on the 910 Ohm will be 6v-1.2v = 4.8v
4.8v * 4.8v / 910 Ohm = 0.025W

Voltage of value 6 which you used in this equation is equal to vout or vin - vout?

 

[alexan_e]

Sorry, wrong calculation. The ADJ pin voltage referenced to the ground will be (output voltage -1.2v) so for 6v output it will be 7.2v

 

[HagenK]

I just tried it and it worked very well, however there is one big problem. I connect a small light bulb to 6V 0.6A and everything was fine, but when I connected it to 6V 1.5A light bulb burned instantly. Could someone explain to me what happened, because I thought there is no such thing as too many amps only to many volts.

 

[alexan_e]

What were the bulb specs? (voltage , watts)

 

[KlausST]

Hi,

what did you expect?

..explain to me what happened..

some things may happen:

1) you overload the 12V supply. Maybe the voltage drops to 9V (or any other voltage) and the LM317 can output 6V and 1.5A. Not likely.

2) you overload the LM317supply. Maybe the voltage drops to 4V (or any other voltage) and current below 1A. The bulb is dimmed. More likely

3) something inbetween.

4) or occationally all voltages are good, but the current drawn is too much. not likely

either case is out of specification of one or more devices. livetime is shortened, damage may occur. i hear the fire brigade.

Klaus

 

[alexan_e]

I connect a small light bulb to 6V 0.6A and everything was fine, but when I connected it to 6V 1.5A light bulb burned instantly.

Did the bulb burn or LM317?

 

[HagenK]

I have no idea. It was really old incandescent light bulb which ran in series when connected to 220V AC wall socket. When powered by 6V 0.6A DC it was a little bit dim. Btw 12V 1.5A is AC.

 

[HagenK]

Bulb. LM317 and resistors still works perfectly.

 

[Audioguru]

It is normal for an old incandescent light bulb to burn out the moment it is turned on because the sudden jolt of power breaks the brittle filament. If the power was applied slowly then the light bulb might still work.

A light bulb powered from the mains or another very powerful power source burns out the moment it is turned on because when it is cold, its filament is a much lower resistance than when it is 2000 degrees C so the current is 10 times normal and it blows the filament. If the supply is AC and the bulb is turned on the moment the AC voltage is at a peak then the instantaneous power in the bulb is double its normal continuous power times 10 because it is cold.

 

[HagenK]

It is normal for an old incandescent light bulb to burn out the moment it is turned on because the sudden jolt of power breaks the brittle filament. If the power was applied slowly then the light bulb might still work.

A light bulb powered from the mains or another very powerful power source burns out the moment it is turned on because when it is cold, its filament is a much lower resistance than when it is 2000 degrees C so the current is 10 times normal and it blows the filament. If the supply is AC and the bulb is turned on the moment the AC voltage is at a peak then the instantaneous power in the bulb is double its normal continuous power times 10 because it is cold.

It might had happened, however 2 light bulbs were used connected in parallel. So the chance that they both burned out because they were cold is so low, that I consider this impossible.

 

[FvM]

I notice that several contributors to this thread are struggling hard to get sufficient information about the (different?) test setup(s).

Does everybody know what has been the input supply, programmed LM317 output voltage and rating of connected light bulbs(s) in the latest test that resulted in "burned" lamps? A hand-sketched schematic might help.