Sallala

1n5402 protect the 2n3055 transistors, but how to protect U2? In a switched off PSU, current can be flown from the output througt D10->R15 and into U2 output. Isn't it? What about to use a Shotky diode? It have only 0,3V drop. Or you can limit the maximum voltage with a trimmer potentiometer series to the voltage adjust pot. I see your diode on various circuits. But I don't understand, how it works. For example, how can it discharge a loaded capacitor on the output? I'm interesting this protection, because I use my psu to slow charge my accumlators. I haven't add the diode on my image yet. Just use it in series on the output. But this isn't good for the voltage stability. And what about D11. This is for protect the psu from a reverse polarity charged capacitor fr example? The original schematic, this is a low current diode. It's enough? (I'm not an electircal expert, please forgive me if I asked stupid a question, or just forward me to a beginner forum..) :-X

Dear Bongo! Why do you place the 1n5402 diode to the top of the 2n3055 transistors? To save the transistors, when the PSU is off, and there is a voltage connected to the output? (A charged big capacitor, orr accumlators for eaxmple) Maybe it isn't enough. As Audioguru says in an earlyer post, big current can be flown through U2's input Audioguru suggested a diode in the feedback, like the image below. I' m waiting for your response. (Sorry for bad English)

Audioguru: Something wrong in my workplace's internet access. I can't see any pictures attached to the forum at work. At home everíthing fine ??? But it's doesn't matter. I'm sorry: T3=Q3 (In my country, Transistor=T) Huhh, I checked it, but I don't remember exactly. It wasn't conduct between E-B or B-C. But I'm sure: the LED was bright always. Even if the supply was in constant voltage mode. Hmm, of course. As the diode, that you drawed. I must to modify the PCB. But this is not makes a problem, if the fuse at the output glowed, and it's in the negative feedback? But how can it's temp exceed 200 degress, when I turned down the votlage potentiometer? As I wrote, while I working, tha supply was good, and Q2 wasn't too hot. Then I leave the 2Ohm load at the output, but turned down the voltage potetntiometer. When I go back, fuse at the bridge rectifier burned. :o

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?

Audioguru: My browser (Firefox) can show PNG files. I don't know, what was the problem. Now I can see the images twice... Thank you anyway! I will buy a high current shotky diode, and will build into the pcb or inside to the supply's case. OK, I understand, what you tell about the batteries.

Audioguru! I can't see your pictures attached!

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.

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?

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 :-)

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)

Can I list all of the posts on one page in this topic somehow?

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. 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?

Great tipp Alun! Thank you, I will try this. But you know, to print something to a leser printer, you must to draw it with a computer. ;-)

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!