Problem readings schematics - basic power supply using the LM317

[PCBaby]

Hello to all.

I'm afraid I have a naïve question again : I have problems reading certain schematics - this time around a basic power supply using the LM317 (1). The attached picture shows a 0.01uF capacitor placed immediately after the cathode , connected immediately to ground without any resistance nor any other device in between. To my innocent newbie's eyes, this looks like a massive opportunity for a burning battery.  Now, as the same schematics was re-used by several web sites, I guess I misunderstood a few things but I don't know which ones.

...Still : shouldn't this capacitor be placed in serie between the source and the LM317 ? I tried this way, and it worked like a charm : steady 5.13V   regardless of the source (provided the tension was high enough, of course).  It's only afterwards that I realized I hadn't followed the schematics...

And is there a pdf or an intro somewhere teachings how to understand this type of details in schematics ?

Many thanks in advance.

(1) Just in case : since I'm at square one of my electronic pilgrimage, I don't use 240V AC as a source : only a 9V battery or a 12V, 9V, 7,5V, 6V etc. regulator.

[suicidaleggroll]

Cathode?  What cathode?  I don't see a 0.01uF cap anywhere on the schematic.  I do see a 0.1uF cap that's decoupling the input, and a 1uF cap that's decoupling the output.  Are you referring to one of those?

[Macbeth]

Have you pasted the correct schematic? I don't see any cathode or 0.01uF capacitor.

If you mean the 0.1uF then that is very common as a supply decoupling / smoothing capacitor. A capacitor doesn't short DC to ground, in fact quite the opposite (it's a DC block, AC pass). So the AC ripple from a mains rectifier gets smoothed out. A battery will merely charge the capacitor up and then maintain that charge.

[PCBaby]

<<Cathode>>

  I stand corrected : I meant the positive electrode of the battery.  Missa sorry. Blame my poor English...

And yes, I meant the 0,1uF.

[Macbeth]

Take a look at one of the manufacturers datasheets, the explanation of the components in the circuit is all there.

http://www.ti.com/lit/ds/symlink/lm317.pdf 

[alsetalokin4017]

 

only a 9V battery or a 12V, 9V, 7,5V, 6V etc. regulator.

Capacitors do not pass DC (unless they are faulty). If you have connected a capacitor in series with the input (or output) and you are providing DC power as input, you should not see any output at all. Perhaps you have not connected your circuit as shown in the schematic, or perhaps your capacitor is not really a capacitor (shorted, or wrong component type altogether.)

Please show a photograph of your test circuit so we can see the details.

[PCBaby]

@ alsetalokin4017

Here are 3 photographies I made before dismounting my breadboard yesterday. Hope they'll show you something useful, as they are not overly clear.
The "loose" jumpers are the output, typically connected to my multimeter.

I'll try again tomorrow (it's 22:00+ here...).

To all : I really appreciate your help. Thanks a lot.

[suicidaleggroll]

Your input capacitor isn't doing anything.  Remember, all five contacts in a row are connected to each other, both leads of your capacitor are attached to the same point.

Your output capacitor should be much closer to the regulator to do its job as well.  You should have one lead plugged into the same row as pin 2 of the regulator, and the other lead plugged straight into the ground rail.  Those long wires aren't doing you any favors.

[alsetalokin4017]

Yep, the ceramic capacitor is not doing anything, it's not even really in the circuit at all.

Also, your thick wires are going to damage the breadboard eventually, by springing the contacts out so that they won't make good connections. Look at how thin the resistor leads are, compared to the jumper wires you are using. The breadboard is designed for thin wires like these, not the thick wire you are using for your jumpers.

[PCBaby]

Hello. I tried again with two 1uF electrolytic capacitors (because I only have ceramic version of 0.1uF), and I amended the connections according to your comments - see attached photo. Worked fine.

From a total beginner's perspective, I must say it's impressive to see the output staying at 5.13 even when the regulator 'upstream' is changed from 12V to 7.5V and 6V a couple of times.

I believe I understood the logic behind the two capacitors. The datasheet was useful also. Till now I was under the impression that those papers contained only graphs for the specialist. Wrong, I was.

Thanks again.

[alsetalokin4017]

You can use the 0.1 uF ceramic on the Pin 3 input to ground _as the schematic specifies_ , you just have to connect it correctly. Substituting a 1 uF electrolytic for the specified 0.1 uF ceramic might "work" but it isn't best practice.

Linear regulators are great but be aware that the higher the raw supply voltage, the more power the regulator has to "dump" as heat. So you should probably put the heatsink back on.

[PCBaby]

@ alsetalokin4017

I'll use a heatsink anyway, first of all because I don't know yet how to evaluate the heat that will be generated. But I'm not building my power station yet. Next step will be to integrate protection diodes and (afterwards) current regulation.

BTW, TI's datasheet mentions a 77-pages glossary that will be a very useful reference to me (and maybe to other newbies as well) : see http://www.ti.com/lit/ml/slyz022i/slyz022i.pdf. It was created in 2013, but they seem to keep it updated : last revision was done in Jan 2016.

Cheers.

PCPb