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[internet name]
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Joined: 01 Apr 2004
Posts: 41
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 Posted: 21 Jul 2004 11:20:53 pm Post subject: |
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One of my friends had a calc that wasn't working for 6 months. I tried to fix it, but couldn't. Whenever I turned it on, It would clear ram, and turn off a second later. I updated the os, cleared all mem, pulled out 5 batteries for 24 hrs and did everything I could think of.
The weird thing (beside none of them working) is that if the batteries were removed, then when they were replaced, the calc would last longer staying on before it turned off again. The longer the batteries were out, the longer it lasted.
24 hrs resulted in about 30 seconds of time between clearing ram, and turning off.
DOES ANYONE HAVE A CLUE ABOUT WHAT THIS IS?!? |
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DigiTan
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| Posted: 22 Jul 2004 12:20:49 am Post subject: |
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I'd say it all depends on where that calculator's been for the past few years. Basically, there are only 4 things that can reset a Z80 like that:
1. Going to "Reset" (so unlikely, it shouldn't have been mentioned)
2. A watchdog timer (does it even have one?)
3. A brownout condition (not likely with new batteries, but it's a possibility)
4. Power spikes/noise (very likely: could be caused by a number of things)
My first suspicion is mis-handling. By this, I mean anything beyond normal wear & tear. Ask your friend if the calculator was ever dropped or exposed to extreme temperatures. This also includes opening the case "just to have a look." Corrosion and loose solder also give tell-tale signs of exposure to moisture.
Anyway, if you have access to one, use a digital multimeter to verify that the supply and ground leads are at the proper voltages. Also test the "tip," "ring," and "ground" of the datalink to verify they are idling at 4.9, 4.9, and 0.0 volts respectively. A logic probe and/or O-scope may be used to detect hazardous noise and voltage spikes.
Last edited by Guest on 22 Jul 2004 12:21:55 am; edited 1 time in total |
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agility
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| Posted: 22 Jul 2004 05:46:40 am Post subject: |
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Quote: Also test the "tip," "ring," and "ground" of the datalink to verify they are idling at 4.9, 4.9, and 0.0 volts respectively.
So the tip and ring are always high unless you're transferring data? Shouldn't they both be low to save battery power? |
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Arcane Wizard
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| Posted: 22 Jul 2004 06:06:36 am Post subject: |
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| I believe those voltages should be 4.7, 4.7, and 0.1. |
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agility
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| Posted: 22 Jul 2004 04:13:01 pm Post subject: |
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| Hmmm... my digital multimeter's reading out 4.47, 4.47, 0.3. |
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Madskillz
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Joined: 02 Jan 2004
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| Posted: 22 Jul 2004 06:45:22 pm Post subject: |
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Actually I have/had this problem too. My calc has been screwed up for about half the school year. I tried everything you did, I also looked inside, nothing wrong. I have no idea what happened to my calc!  It is still screwed up, but not as bad as your friends. I managed to somehow get mine to except calculations (which was very hard, because my ram constantly reset itself). But as soon as I go into memory, phhht nothing ram clears. It says I have some variables, but when I go to delete them, like I said, the ram clears! Strange, it is interesting to see somebody else with my same problem!  |
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agility
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| Posted: 22 Jul 2004 08:22:49 pm Post subject: |
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| Even after re-installing the OS this happens? |
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[internet name]
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| Posted: 22 Jul 2004 10:08:36 pm Post subject: |
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| Yes. By re-installing the os, and clearing all mem, I can't imagine it would be a software problem. |
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DigiTan
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| Posted: 23 Jul 2004 01:46:15 am Post subject: |
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Quote: 4.47, 4.47, 0.3.
That's okay. The main thing to look for is some "balance" between the tip & ring. This just basically rules out loose connections, voltage spikes, and other noise-makers. Although the 0.3 reading makes me a bit suspicious, it may be safe to widen the search. This is just a shot in the dark: but I'm suspecting maybe a corrupt ROM. So many other systems seem to working fairly well (CPU, link, display driver, etc) that I have to put it under the spotlight. But this is pure speculation. Unfortuanantly, the ROM is hard to test without doing a ROM dump (D'oh!).
If I could add one more theory: here's a message I found after searching for similar cases:
Anonymous wrote: I doubt it was new, it was probably refurbished. Back when I had a brand new TI-92 it was broken and kept resetting. T! sent me a refurbished one that did not work as well as the other when new and had a big R scratched into the back.
[Edit]
Almost forgot... About the link voltages. They idle at high voltages mainly because of the open-collector scheme used in their input circuits (to prevent the signal from "floating" randomly when the link is unused). The load on the link is very, very small when it's not used, so it doesn't affect the battery life too badly.
Last edited by Guest on 23 Jul 2004 01:54:29 am; edited 1 time in total |
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[internet name]
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| Posted: 23 Jul 2004 01:26:35 pm Post subject: |
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I don't know what it was, but my friend sent it in. About the link, though. You are right. It only uses electricity when the ground and the 1 are connected. I did some expirimenting, and found that the tip, when connected to the ground sends/recieves a 1, and when the middle is connected to the ground, it sends/recieves a 0. To send a byte F0h, you would connect the tip to ground 4 times (F = 1111b) and the middle to ground 4 times (0 = 0000). What happens is when the tip is connected to ground, you are sending FFh repeatidly (i cant spelll  ). The calculator then thinks a device is communicating with it. I found this out when using the link console in calcsys. I also think that the calculator uses a 1.5 voltage regulator to send 1.5 volts. You only measure the voltage as about .75 because the calc rapidly turns the circuit on and off. That way, with slow voltmeters, the 1 and 0 are combined to be 1/2 of the voltage. The only way to check the battery life is to physically check the batteries. Sorry if this was confusing. I am not an 'A' english student.  |
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agility
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| Posted: 23 Jul 2004 03:31:43 pm Post subject: |
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Quote: (to prevent the signal from "floating" randomly when the link is unused).
Then why don't they just stay at logic zero instead of logic one to prevent floating? Is it too hard or something?
Quote: I did some expirimenting, and found that the tip, when connected to the ground sends/recieves a 1, and when the middle is connected to the ground, it sends/recieves a 0.
You're not connecting it to ground; You're just setting it to logic zero which is about 0.3v, not ground. |
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[internet name]
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| Posted: 23 Jul 2004 10:21:22 pm Post subject: |
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First of all, I don't think anything should be measured in voltage. That only takes the average of all voltages. If you only output 3v at 1/3 the time rapidly, then it will be measured as 1v unless you have a measuring device that graphs the voltage so you can see it go from 3 to 0 to 0 to 3 to 0 to 0 etc.
Second, I am positive (no pun intended) that the farthest from the tip is common ground. One has to be, or else it would be impossible to communicate. Current can't just flow from one calc to the other without returning. That wouldn't be a complete circuit. |
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DigiTan
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| Posted: 24 Jul 2004 01:58:06 am Post subject: |
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Yep. We might to avoid saying "voltage" in this case because that might confuse 'average' voltage with 'instantaneous' voltage.
Well, going back to the link, I mentioned earlier the link uses an open-collector scheme. This is very popular, but the downside is you have to pull the voltage either up or down to prevent data loss (this happened to my IR link for about a week). Given a choice between the two, alot of engineers choose to pull those inputs up to a positive voltage like 5v. Statistically, you're x3 less likely to suffer bit loss of from noise if you pull the voltage up--for subtile reasons that are bit (pun!) off-subject for now. Lets just say the input can handle about x3 more noise if the voltage is constantly pulled high. Of course, it would vary slightly between calc models. |
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agility
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| Posted: 24 Jul 2004 08:04:06 am Post subject: |
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Quote: If you only output 3v at 1/3 the time rapidly, then it will be measured as 1v unless you have a measuring device that graphs the voltage so you can see it go from 3 to 0 to 0 to 3 to 0 to 0 etc.
Oscilloscope?
Quote: Second, I am positive (no pun intended) that the farthest from the tip is common ground. One has to be, or else it would be impossible to communicate. Current can't just flow from one calc to the other without returning. That wouldn't be a complete circuit.
Yes, the base connection is ground, but the signal doesn't flow from one calc's tip to the other calc's ground or anything like that because then you wouldn't get any data transfered. You can't use ground for signals. See what happens is that current will flow for example from calc A's tip to calc B's tip, setting them both high and that signal gets transfered somewhere on the mainboard (probably the processor, not sure), and from there it will eventually make it's way to ground. It doesn't go directly to ground like you said or nothing would get done. |
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[internet name]
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| Posted: 24 Jul 2004 12:08:26 pm Post subject: |
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Quote: It doesn't go directly to ground like you said or nothing would get done.
That is true. All I am saying is that there must be a ground for any data to transfer. |
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agility
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| Posted: 24 Jul 2004 02:59:55 pm Post subject: |
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| Okay then. Your wording in your previous posts was misleading then. |
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[internet name]
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| Posted: 24 Jul 2004 08:05:09 pm Post subject: |
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| Sorry. As I said earlier, I'm not an 'A' english student. |
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agility
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| Posted: 24 Jul 2004 09:35:57 pm Post subject: |
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| Did you ever fix the calc? |
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[internet name]
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| Posted: 25 Jul 2004 09:30:21 pm Post subject: |
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No. They live far away from me. I visited him when I went to a fencing camp. I am pretty sure he sent it in.
The parents were supprised that I knew all the stuff about the calc.  |
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agility
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| Posted: 26 Jul 2004 05:49:13 am Post subject: |
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Hey, if you know how the toaster works any parent will be impressed  . |
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Calculator Tech Support => Technology & Calculator Open Topic
