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Everything posted by woolyhead

  1. No luck so far so I will try to buy an installation disc. What are the chances of finding one? I sometimes have a problem with my Epson sx118 printer's scanner but when I run its installation disc the problem goes away. I'm hoping the same thing would happen with an HP installation disc. Am I being reasonable with this? I used the HP printer doctor free download and with its help I managed to get the scanner to work but as soon as I tried to scan without the doctor it didn't work. I was hoping the doctor would fix the problem but it didn't. Any idea why not?
  2. Is it possible to obtain another installation disc for an HP C5280 printer ? Has anyone got one to sell me?
  3. Thanks everyone. But my HP printer C5280 isn't on this list andysym so I'm trying to download a driver from the HP website Ghozer. Vuescan thinks it downloaded something to me yesterday so I'm looking for it swarfendor437.
  4. My HP C5280 has been working as printer and scanner for two years now. Suddenly my pc tells me it cannot open the scanner software because there is no TWAIN source. What does this mean? I noticed that the lack of TWAIN occurred just after Adobe Flash stopped supporting HP. Are these things connected? What must I do to recover the TWAIN source?
  5. My tenor trombone is old but it plays. However, as the slide approaches its maximum length it starts to judder. It has to be moving but it doesn't make any difference whether I'm blowing or not. Why might it judder like this?
  6. If a parcel or a letter with no franking on the stamp arrives, you can get the stamp off without leaving the small oval shape behind by soaking the stamp in water then carefully unpeeling it, being especially careful with the small oval part and not letting it separate from the rest of the stamp. A penknife or a scalpel can be useful here. Stick it on the next letter you post using a clear adhesive. I'm not sure but reusing stamps like this may be illegal, although how anyone would prove it had been used before I don't know.
  7. See Reset HP Printer on this forum posts, page 2 by Waldo . Although this refers to the printer, not the cartridge, I did wonder whether my HP printer's refusal of a certain cartridge was because the printer needed resetting. There is also a method shown on page 1 for overcoming the printer's refusal of a cartridge. I also wondered whether that works on the cartridge chip's stored information. Does anyone know how this cartridge refusal or acceptance works? What chip(s) make the decision, the printer's chips and/or software or the cartridge chip? If so where do they get their information from? And how do they know when a cartridge is running out of ink? In my experience they are often wrong about that, especially if the cartridge hasn't been used for a while. So what is the system and how does it work?
  8. I'm told that cartridges have a chip inside. What information do these chips carry?
  9. Thanks Waldo. That has cleared up some of my questions. I didn't imagine the HP chip could be flexible and as thin as you say but I can see it could be. My printer's contacts form a pattern that almost fully occupies the allocated overall area, although there is a small gap around the edges. I mean there are no large spaces on it, apart from the distances between contact pins. The thin ribbon you talk about would therefore spread right across the allocated area and the actual chip on it would fit between the printer's contacts. These contacts are a bit wider than a pin head and rise up about 1.5 mm and the gap between them is about 2.5mm. I know that chips can be smaller than 1mm in diameter so a chip could go between contacts ok. What about the frame. How thick is it? If it goes around the space allocated for printer contacts it has both elbow room (x.y) and height (1.5 mm in z). The only thing is the electrical situation: A compatible's contact's press on the HP chip's contacts and at the same time these HP chip contacts press on the printer's contacts. So at least some of the pins will have contradictory voltages on them. Electrical circuits can't have that situation. One or both the sets of electrical information has to alter so as to become just one set of values. If the chip's output impedances are considerably smaller than that of the sources internally connected in the compatible and then taken out to a compatible's contacts, the HP chip will win, as it were. And this has to happen for the trick to work. Otherwise if they are roughly equal impedances the result at every pin could well be that the voltage on it is the average of the two. So since the trick does work it must be the former situation regarding output impedances. Good old HP, giving their chips low output impedances! It would also suggest that the trick could not be played by using another compatible's chip instead of an HP one. Maybe only HP chips can do this. You've certainly made me think. a lot about this. Thanks Waldo. 6 hours later ... I just realised something about this question of two different voltages at the same point on every one of the HP contact pins. If the compatible's contact with the HP chip doesn't go right through the chip tape but only into the chip then coming from this there need only be one voltage on the HP printer's contacts. In case anyone is reading this let me say that I realise there doesn't have to be an actual voltage coming from the compatible because that would occur if the compatible had a different input impedance from the HP chip at every point. That would generate a different voltage at these points if the HP printer injected a constant current into each of these points, a different voltage from those of the HP chip. The net result would be the same.
  10. About this transplanted chip: why doesn't it still record the fact that it's cartridge is empty and thus fail to work as required when transplanted into your printer? About the frames: I suppose they fill the space surrounding the printer's contacts right up to its 4 edges and thereby position the contacts of the compatible. In other words act like a jig? That deals with the x and y dimensions anyway. One more thing I don't understand Waldo. In what plane are the leads on the chip positioned relative to the chip's body and how thick is the chip's body? What I'm asking about here is the situation in the z direction ie how the compatible's contacts can reach down to the chip's contacts, past the chip's body. After all, the compatible's contacts group is flat and it was designed to press down on the printer's contacts with no space between it and the printer's contacts. You're saying there is enough space for the compatible to go in and for its contacts to still press down on the chip whose own contacts are pressing down on the printer's contacts, aren't you? The chip must have a very thin body to make this possible I would have thought. Please don't misunderstand me. I believe what you have told us but just can't quite imagine this aspect of it. Could you please explain a bit more. Last question: how many times can you use this chip?
  11. Thanks guys. All done as you said and all working.
  12. Yes it does a bit, Waldo. I've got a pretty clear picture of what you've said now. Presumably there is enough space for the compatible cartridge and the chip, both in the space allowed for just the compatible. And I suppose the chip gets its power via the cartridge contacts in the printer. Your idea is very good indeed. Thank you for posting it to us. As it happens the supplier of my colour cartridge replaced the no-go one and the replacement works ok. But I will eventually buy an original HP cartridge, use it and then extract its chip and try out your idea for myself. Why does the chip work, seeing as how it will have come from an empty cartridge by then. Why doesn't it still record the fact that it's cartridge is empty and thus fail to work as required when transplanted into your printer? In your case, did the chip come from an empty cartridge? Also, what do the plastic holders do? I suppose their dimensions must be fairly critical? Are they used to hold the chip at the right place on the printer's contacts and to stop it sliding down below those contacts? I suppose they fill the space surrounding the contacts and get the position of the contacts by filling this space, right up to its edges. In other words like a jig? How many times can you use this chip? Sorry to ask so many questions but unless I understand exactly what I'm going to do it might not work out for me.
  13. That sounds ingenious! Could you tell me where in my printer I would find these contacts. Are they the ones which a cartridge touches when installed? Should I place the HP chip face up or face down? Presumably face up otherwise how could I place the compatible cartridges over the HP chip and make contact on the right pins?. Would I have to solder the HP chip in place? Could you draw a sketch of what you do, please, to make it easier to understand. I would like to follow your idea when I understand it a bit more.
  14. I tried to return it to one particular supplier but what an embarrassment ... it wasn't one of theirs. But I still have two more delivery notes showing two other suppliers so I'll try them. I like your suggestion of using a continuous ink system. I'd never hear of this type before. Also I'll start studying Epson printers for the reason you suggest. Thank for your help.
  15. Hi Max. No. I hadn't tried cleaning the heads but when I followed your suggestion exactly the heads were not cleaned and the error light flashed throughout the attempt. The printer just refused to clean its heads. I think Zach has the answer ie the cartridge is not accepted because the printer is very fussy about who made it. Some lookalikes work, others don't.
  16. Where can I get a replacement for the installation disc for my HP C 5280 printer?
  17. No Zach, it's not a bad one because it worked for some time before it stopped. I've heard of that service Zach but I don't do much printing. Do you think that may be the trouble due to the long elapsed time between attempts to print. Also I was given this printer some years ago and it wasn't new even then. But when it works the printing quality is good. I was not given the installation disc however. Do you happen to know if I can get an installation disc and where from?
  18. No, they are lookalikes but in the past they have worked ok
  19. Thanks swarfendor 437. I'll give it a go right now...... Well that didn't work for me. I keep getting the message "Cartridge error. Refer to documentation." But thanks for trying to help me.
  20. I put some new HP inks into my HP C5280 printer and it still refused to print. It gave the message "Cartridge error. Refer to documentation." I think this error message is false but the problem can be solved by resetting the printer. But how should I do this? I looked it up on the net and all I could find was a video with no sound and it was too difficult to follow. Could someone please help me.
  21. Thanks the bloke and medusa. I will take a better look at it.
  22. My electric oven has a fan inside and when cooking, hot air comes out into the room. Are fan ovens supposed to do that?
  23. Looking at many of the circuits shown on the net for giving constant currents, they mean by this term that the CIRCUIT APPLIES A VOLTAGE to the battery and whatever current tries to flow, the circuit limits it, even if the battery voltage drops. If the battery voltage rises the "constant" current no longer stays constant but reduces. That's all very well but the term constant current really means that THE CIRCUIT APPLIES A CONSTANT CURRENT to the battery and this is maintained constant whatever the battery voltage does/is. So whether the battery voltage rises or falls the current remains constant. with a proper constant current charger. There is a clear difference between these two types of charger. To monitor the negative dV/dT that occurs when the battery is fully charged the current source must be one of the latter kind, a genuine constant current type. Pulse charging is something else and I don't know much about it yet but I think it offers advantages over the dc type in that the battery responds better. I look forward to seeing any ideas you come up with. Since writing this I found the following comment on Battery University about pulse charging:- "Interspersing discharge pulses between charge pulses is known to improve charge acceptance of nickel-based batteries. Commonly referred to as a “burp” or “reverse load” charge, this method assists in the recombination of gases generated during charge. The result is a cooler and more effective charge than with conventional DC chargers. The method is also said to reduce the “memory” effect as the battery is being exercised with pulses. While pulse charging may be valuable for NiCad and NiMH batteries, this method does not apply to lead- and lithium-based systems as these batteries work best with a pure DC voltage."
  24. Thanks Ghozer. Your idea would produce a charge current of about 100 mA which is C/10 for a 1Ahr cell. This is a rather slow charge rate because NiCads can take a higher proportion of C and this makes the charging quicker and more efficient. Since I'm using D cells which have a C value of 4Ahrs I was really looking for a constant current charger giving about 2A and also a circuit for detecting the negative deltaV point. I looked up those references, apelike, but couldn't find either such circuit. Most of those shown were not proper constant current supplies. Ghozer's circuit is almost constant current, independent of the cell voltage, but I need 2A so if I increase the dc voltage to say 100V and use a 50 ohm resistor I should get 2A, But the wattage in the resistor would be 200W. That seems a bit extravagant in terms of power. If I reduce the voltage to 50V and use a 25 ohm resistor the wattage is then 100W. If I retain the 25V supply and use a 12.5 ohm resistor the wattage is still 50W. All these options seem like a hammer to crack a walnut. Since the cell voltage is only about 1.6V or so while on charge at most and only changes by about 0.6V from a starting voltage of say 1V, it shouldn't need a very big resistor value to make the current independent of cell voltage. Suppose I use 4V supply and a resistor of 1.35 ohms, the wattage would be about 5.4W. The current would be (4-1)/1.35 = 2.22A at the beginning , falling to (4-1.6)/1.35 = 1.78A. This is 2A plus or minus 10% at the expense of only 5.4W. in the resistor. What do you think of this scheme? Is the current constant enough for an NDV circuit to work? I think the NDV reduction in cell voltage is of the order of 8mV.
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