Sallala Posted September 18, 2005 Author Report Share Posted September 18, 2005 Spagbol:Oh, I missed. You need to modify the Q1's emitter, not Q2.Now I will correct my post.(at Q1, you know, what to modify?) Quote Link to comment Share on other sites More sharing options...
Spagbol Posted September 18, 2005 Report Share Posted September 18, 2005 Aha, suppose I must read thru the thread again. Nice photos by the way, I've never made a PCB before but maybe I will try it. Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 18, 2005 Author Report Share Posted September 18, 2005 T1's emitter connected to the input voltage's null point. This is wrong!T1's emitter must to connect to the output's 0V, otherwise it doesn't make it's job. Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 18, 2005 Author Report Share Posted September 18, 2005 Twentyone century. Nevermind!I plan and made my PCB-s with my hand :-)1. plan the connections on a papper2. put the paper to the pcb, and sign the holes with a sharp prod3. clean the brass plate well4. 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 down6. 15...30min and your pcb are ready7. clean the pcbThat's it!Take care, to read usage instructions for the ferric chloride!!! Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted September 18, 2005 Report Share Posted September 18, 2005 Why bother with all of that when you can use a laser printer and magazine paper as shown here. Quote Link to comment Share on other sites More sharing options...
Spagbol Posted September 18, 2005 Report Share Posted September 18, 2005 Not sure I understand about Q1s emitter, looks ok in original cct to me. It is a safeguard against there being no -5v rail isn't it? So when there's no -5v the transistor (Q1) conducts and pulls U2 to ground so there's no drive to the output transistors?? Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 18, 2005 Author Report Share Posted September 18, 2005 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 T2If 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! Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 18, 2005 Author Report Share Posted September 18, 2005 Why bother with all of that when you can use a laser printer and magazine paper as shown here.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. ;-) Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 19, 2005 Report Share Posted September 19, 2005 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. Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 19, 2005 Author Report Share Posted September 19, 2005 Ok, then Q1 emitter go back to 0V. I will make the modification. Just a new hole :-XI made a summarize about the project on my homepage:http://sala.sallala.hu/elektronika/szab_dc_tapegyseg/tapegyseg.htmlThis 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? Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 19, 2005 Author Report Share Posted September 19, 2005 Can I list all of the posts on one page in this topic somehow? Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 19, 2005 Report Share Posted September 19, 2005 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. Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 19, 2005 Author Report Share Posted September 19, 2005 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 voltage3. 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) Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 19, 2005 Report Share Posted September 19, 2005 I just use a wall-wart DC supply and a resistor to charge my batteries overnight. A regulator's current remains constant even when the battery is fully charged, causing overcharging. My resistor has less voltage across it as the battery reaches full charge so overcharging isn't as much a problem. Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 19, 2005 Author Report Share Posted September 19, 2005 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 :-) Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 19, 2005 Author Report Share Posted September 19, 2005 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? Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 19, 2005 Report Share Posted September 19, 2005 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. Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 19, 2005 Report Share Posted September 19, 2005 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 - Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 19, 2005 Author Report Share Posted September 19, 2005 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. Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 19, 2005 Author Report Share Posted September 19, 2005 Audioguru! I can't see your pictures attached! Quote Link to comment Share on other sites More sharing options...
Spagbol Posted September 19, 2005 Report Share Posted September 19, 2005 Presumably output taken from top of C7 in new circuit rather than emitter of Q4, just to clarify. Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 19, 2005 Report Share Posted September 19, 2005 I think it would be very difficult to set the proper voltage of this supply to the hiccup point in the charging voltage curve of a battery. If you set the voltage slightly too high, the battery will continue over-charging.I forgot to move the supply output connection on my previous post. I've corrected it. Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 19, 2005 Report Share Posted September 19, 2005 Audioguru! I can't see your pictures attached!Doesn't your viewer show PNG? Here they are as GIF: Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 20, 2005 Author Report Share Posted September 20, 2005 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. Quote Link to comment Share on other sites More sharing options...
Sallala Posted September 27, 2005 Author Report Share Posted September 27, 2005 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 damagedI 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? Quote Link to comment Share on other sites More sharing options...
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