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1.2 to 12 v adjustable voltage regulator (5k pot is burning)

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in 1500 to 5000 ohm the output voltage is 16.50v
How can the output be 16.5V when you showed the input is only 12V?
Did you buy a fake LM317 IC from ebay?

An LM317 needs a 120 ohm resistor from its output to its ADJ pin. A more expensive LM117 uses a 240 ohm resistor. If the resistor value is too high then the output voltage of some LM317 ICs goes higher than it should when there is no load.

The pins on an LM317 are 123 when you look at the epoxy with the part number printed on it, not when looking at the metal back.
 
i want to use the regulator .. can you tell me about the full security in the complete circuit

Instead of insisting with something that evidently is not working as it should, do yourself a favor following these steps:

Get three or four LM317 from a reputable source.

Get the suggested resistors (two or three of each value). Add one more to initially avoid using that wretched pot.

Now, throw the current circuit out the window and forget it!

Create a mimic diagram showing all connections to the different points in the circuit. Even those that you are tempted to call "obvious" ones. Try to improve the one you showed somewhere above.

No matter you use a protoboard, solder things in air or use ugly style construction, make it tidy with short wires unless you are daring to use direct connections.

Now start measuring as suggested in some posts above.
 
How can the output be 16.5V when you showed the input is only 12V?
The LM317 may be oscillating which may give an output voltage higher than the input voltage as seen on a digital multimeter.

An LM317 needs a 120 ohm resistor from its output to its ADJ pin. A more expensive LM117 uses a 240 ohm resistor. If the resistor value is too high then the output voltage of some LM317 ICs goes higher than it should when there is no load.
Correct AG. I was looking at the original National LM317 data sheet which shows a minimum load current of 5mA. The latest Texas Instruments (bought National) data sheet shows 10mA as you say. https://www.ti.com/lit/ds/symlink/lm317.pdf

spec

(my post #14 has been updated in view of your comments)
 
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Almost every schematic in National Semi's datasheet of the LM117/LM317 shows the more expensive LM117 with its 240 ohm resistor. I wonder why this guy is trying to use 330 ohms with his LM317?
 
The potentiometer is not really suitable if it is designed for small signal audio type work. Taking the data sheet for a similar Bourns 5K Ohm potentiometer: https://www.mouser.com/ds/2/54/DB18-777896.pdf.
The max power dissipation = 200mW or 100mW depending on law. For a 5K potentiometer this corresponds to a maximum current of 6.3mA and 4.47mA respectively.

spec
 
The 5k pot is used with an LM117 and a 240 ohm resistor to produce a maximum output of only 27.3V. When the resistor is 120 ohms and the pot is 5k ohms then the output of an LM317 will try to be 53.3V which is much too high.

Here he is using an input of 18VDC and might want a maximum output of 16V so when the resistor is 120 ohms then the pot should be 1.5k and it will not get too hot (15mW).
 
A zero flew away from my wrong calculation. I have always used 1/2W pots.
 
I don't why you guy have 16.5V output when you have 13.6VAC, after rectifiers and capacitors, peak voltage is only 17.23V.
Let use oscilloscope to see output voltage wave form.
If you don't have oscilloscope, I have an idea:
Let soak your pot in pure water (if don't have, you can use 70° alcohol or in worst case: mineral water) to prevent pot from burning. Power up, check output voltage with volt meter, then connect output to a load like 22ohm resistor (the resistor also need to be soak in water) or a suitable incandescent lamp. While the load are connected, use volt meter to check output voltage again. If when load are connected, output voltage drop are too much, your LM317 may oscillate like spec have said.
 
How can the output be 16.5V when you showed the input is only 12V?
Did you buy a fake LM317 IC from ebay?

An LM317 needs a 120 ohm resistor from its output to its ADJ pin. A more expensive LM117 uses a 240 ohm resistor. If the resistor value is too high then the output voltage of some LM317 ICs goes higher than it should when there is no load.

The pins on an LM317 are 123 when you look at the epoxy with the part number printed on it, not when looking at the metal back.
yes this input is output of a rectifier and i think after filtering it ,it gives the peak value rather than rms
 
13467463_1710751502520860_136225411_o.jpg

i have purchased a new high price lm317 and a 5k pot .. now i am getting perfect results with 330 ohm value of r1 ..

but i want to use the adjustable regulator practically, how can i secure the whole system and tell me the cautions
 
Hmm, I think it is don't need to put 1A fuse, searh "LM317 current booster" and you will see many circuit make the regulator work well with current capability up to 5A or more...
 
An LM317 limits the current to about 2.2A so you need a fuse only to prevent a fire if you drop a metal tool on the input wiring.
EDIT: I corrected the part number.
 
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You don't like what we tell you anyway, so why does it matter?

The idea behind a fuse is to prevent fires. So, if a AC cord passes into a box, you usually fuse the AC. Fusing is wierd because you have a transformer. This usually means a time delay or SB fuse is used.
If you use Pac=Pdc(max); then figuring out I on the input side, double it and make it a Slow blow fuse and your probably good. e.g. (0.1)(100 VAC) = (5)(2 A); Wrong numbers, but you get the idea. 100 VAC is use in Japan. Your regulator won't put out 2A. So, a 1/4 A SB fuse should be fine in this case.

If you decide on a metal case, then you would want to electrically insulate the regulator from the case, but have good thermal contact. Sil Pads, Mica washers or Kapton washers are used. Sil pads don't need heat sink goop. (A very fine layer). The white stuff gets everywhere.

We don't like the 330 ohm resistor. The small capacitors near the regulator. The datasheet suggests the types. Usually one is ceramic because of it's high frequency abilities. Sometimes they are parallel to get the best of both worlds.

The two protection diodes definitely need to be added if this is stand-alone.

You have choices for the output connectors. Pilot lamp? A power switch on the DC side, so power cuts instantly. An analog voltmeter or even an analog ammeter All of these are possibilities.

Movements were popular at one time, so you could take a 50 uA meter movement and make a voltmeter or current meter of your choice. These may have appreciable series resistances, so that has to be considered if used as an ammeter.

It COULD be worthwhile to have a way to have fixed outputs such as 5V.

A power supply is a good project. In doing it, you will realize that the case, power supply, switches, cords, jacks and real estate cost the most.

The following is basically an aside. Some of the things I have and why I have them.

There's a really nice supply, **broken link removed** which I have delegated to a portable AM radio. It's basically a linear regulator with a multiple tapped secondary, It's designed to replace batteries.

When you combine it with a cord with Adapt-a-plugs, it's not bad, but it has a specific purpose.

This **broken link removed** is a cool supply too.

I have a couple designed as a "quick back-up" for wall warts that break. Using 5.5/2.1 and 5.5./2.5 input plugs to an adapt-a-jack; I can "Make nearly what I need quickly. Polarity reversal is take apart and re-wire or switch thee adapt-a-plug.

for all of my devices, I pretty much have labled the wall adapters:
What it goes to.
Voltage current: 5V 1A C+ (for center positive)
Connector: e.g. 5.5mm/2.5mm
Then there is also flag labeled to the same effect. The AC side may be labeled as well.

The lowest voltage has been 5 V and the highest 17 V for my devices. 5 and 12 are common. I do happen to have a few odd ones like 9 V and 6V. I do have a 24 VDC supply that I could rig up if needed to get 17 V.

The 5.5/2.1 and 5.5/2.5 coax plugs are pretty common for inputs, so by incorporating both, I double my chances. If the adapter is bad, but not the cord, , you can always cut the output cord and use it.
 
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