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0-30 Vdc Stabilized Power Supply


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The original circuit should work fine up to 15V at 1A if you replace the old opamps with the newer higher voltage ones. You probably should recalculate the resistors that set the maximum voltage and c

Hi, as promised I made an English translation of my working. Maybe there is few mistakes and I am sorry for that ! Good reading. ExplicationEN.pdf

February 23 above on this page has the latest schematic of the revised 3A lab power supply.

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Twentyone century.
Nevermind!
I plan and made my PCB-s with my hand :-)

1. plan the connections on a papper
2. put the paper to the pcb, and sign the holes with a sharp prod
3. clean the brass plate well
4. now, you can see the place of holes on the brass plate, draw the pcb with water resist ink. Or you can use special pen. Water resist ink makes coherent surface, but hard to draw. You can draw easyer with a pen, but the surface will be ragged after the etchwork.
5. put the pcb into Ferric chloride (Fe2Cl3) solute in hot water, with face down
6. 15...30min and your pcb are ready
7. clean the pcb

That's it!
Take care, to read usage instructions for the ferric chloride!!!

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As I read in the topic:
Q1 for pull down the output voltage, if you decrease the voltage adjust potentiometer.

The current flows throught R7, Q1's emitter voltage increased, and T1 won't conduct->U2 can drive T2

If thers no current flow througth R7, then Q1's emiter go down, it's base higher than it's base->T1 conduct->pull down the output throught D10.

But this is not clear to me. Wait for others' answer!

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I am sorry that I made an error with the emitter of Q1. The original schematic is correct.
I corrected my mistake in a post a while ago.
It is not a serious error. The connection location for the emitter could be at the output's 0V terminal without harm and Q1 will still do its job of shorting to ground the output of U2 when the negative supply quickly disappears when the mains is turned off.

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Ok, then Q1 emitter go back to 0V. I will make the modification. Just a new hole  :-X
I made a summarize about the project on my homepage:
http://sala.sallala.hu/elektronika/szab_dc_tapegyseg/tapegyseg.html
This time only for my language, (Hungarian)
I will update this.


negative supply quickly disappears

And then what about the suggested 10..47uF capacitor paralell to D7 ?
When the mains turns off, and there is a capacitor, -5V will go down slowly, not quickly. Isn't it? It's not a problem?


Other question:
Can I charge NiMH and NiCd batteryes with my powersupply?

NiCd and NiMH bateryes doesn't need any special thing, just constant current, when you charge it in slow charge mode. (With 1/10C, 14h)

Can the batteryes damage my supply? For example, when I stay connected it to the output, and switch off the supply? Or other reason?

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And then what about the suggested 10..47uF capacitor parallel to D7 ?
When the mains turns off, and there is a capacitor, -5V will go down slowly, not quickly. Isn't it? It's not a problem?

C3 is only 47uF and will discharge a lot quicker than the huge main positive filter cap.
The problem is caused when the negative supply disappears then U2 loses its negative supply, causing its output voltage to rise if it was fairly low.
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Audioguru: thank you for the quick answers again!

- Yes, I'm thought to use a series diode too! (But I'm sure now because you confirm)

- I think, timer doesn't neccesarry, only once.

1. Just charge the battery pack, with full voltage and adjusted current according to the battery.
2. Measure the fully charged battery's voltage
3. Next time, adjust the voltage to the measured in the 2. point. The constant current charging automatically "switches" to something like drop charging at the end.

Somebody suggested this method. It's right?

(Of course, this method not the same as use a timer. Just for start charging in the afternoon, and stop tomorrow morning.. And sometimes need to repeat to measure the fully loaded voltage)

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A regulator's current remains constant even when the battery is fully charged, causing overcharging.


If I use a 4,8V battery pack for example, and set the voltage to 5,2V the curent can't remain constant. A fully charged 4,8V NiMH batterypack's voltage about 5,2V. No voltage difference between the supply and the battery, -> no current isn't it?

I use several type of batteries, I need the adjustable curent, so I'm happy, that I have an adjustable current batery charger too :-)
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Yes, a fairly high current would flow into the output of U2. Use a series diode for isolation.


Hey, it's neccesarry for everyone isn't it?
For example, if I put a circuit with a bigger puffer condensator on it's input, that can damage the power supply if the mains off. Not?

I remember, on some three terminal stabilizer IC's manufacturer suggest a bypass diode between the leg 1 and 3. So the puffer condensator can discharge over the diode and the transformer.

What can I waste, if I place permanently a series diode into tha supply's case?
0,6V voltage drop, 3W additional dissipation. Any other thing? Because who's care about that 600mV if it can be save your circuit?
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If I use a 4,8V battery pack for example, and set the voltage to 5,2V the curent can't remain constant. A fully charged 4,8V NiMH batterypack's voltage about 5,2V. No voltage difference between the supply and the battery, -> no current isn't it?

My Energizer Ni-MH cells reach 1.4V when I think they are fully charged and the voltage keeps climbing.
In the case for your 4.8V battery it will be about 5.6V.

post-1706-1427914242189_thumb.png

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0,6V voltage drop, 3W additional dissipation. Any other thing? Because who's care about that 600mV if it can be save your circuit?

You will have a lot more than only 0.6V on a diode carrying 3A, unless it is a Schottly.
If it is external to the circuit then it will ruin the superb voltage regulation. You could protect the supply with a diode inside the feedback loop like this:
EDIT -

post-1706-14279142422258_thumb.png

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My Energizer Ni-MH cells reach 1.4V when I think they are fully charged and the voltage keeps climbing.


While the voltage keeps climbing, your batteryes doesn't fully charged. See it's technical manual! When the voltage stops climbing, and start decrease about a few millivolts, at this point your batteryes fully charged.

The newer -dV chargers stops charging at this point too.

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I have got another problem:

- My fuse (5A) at the secunder side of the transformer glowed.
- T3 went out
- Fuse at the output (5A) desn't damaged

I used a 1,5-3Ohm load at 3-4V at 3A. (Cutting polystirol foam with hot wire) I use only one 2n3055 at this time, with heatsink and a big fan. This worked with no poblem.

When I change the foam, I just turn the voltage and the current potentiometer down. (Or maybe just the curent pot?) But later, when I like to continoue the cutting see the problem.

What happend? Why the fuseage burnt?

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