07 October 2017, 16:08 | #1 |
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Kickstart 3.1 ROM - wrong type?
Hi all
I received my Kickstart 3.1 ROM today, proceeded to fot it and noticed it has one more pin on either side than the existing one. The one I ordered said: A500/A600/A2000. Have they sent me the wrong one? |
07 October 2017, 18:09 | #2 |
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Did you order from AmigaKit and was it for your Amiga 2000, also what is printed on the ROM you received?
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07 October 2017, 19:09 | #3 |
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I ordered of Ebay but it looks to be the same ROM as sold on AmigaKit. However, I think I know what the issue is. The ROM has 42pins. The one from my 2000 has 40 pins. The picture in the advert shows the pin closest to the "dimple" not inserted. I think it is supposed to be like that.
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07 October 2017, 21:07 | #4 |
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This ROM sounds like a dual kicker.< It has both 1.3 and 3.1 programmed to it. The extra pins are for ground and switch control between the 2 kick-starts.
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08 October 2017, 23:33 | #5 |
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Or they just got a batch of 27C800 EPROMs at a better price than 27C400. I don't think those extra two legs should be left floating, though.
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09 October 2017, 01:15 | #6 |
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Peel the label, and let us know the part number.
Without that, it's pure speculation to say what might, or might not, be programmed into the chip. Will also give a cluse as to what needs to be done with the extra 2 pins, if anything. Red |
09 October 2017, 02:25 | #7 |
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Ask your ebay seller
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09 October 2017, 04:18 | #8 |
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I inserted the ROM as illustrated in the instructions and it does indeed work...although looks a little strange with the two legs hanging out at the dimple end.
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09 October 2017, 19:58 | #9 | |
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Quote:
Assuming the image is doubled it probably won't be a problem... |
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10 October 2017, 01:57 | #10 |
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Yes, assuming it is doubled. It still is courteous not to leave pins floating.
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10 October 2017, 22:57 | #11 |
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(Not specifically talking about the OPs ROMs)
On a 27C800 if you want to tie A18 high or low, can you just tie the pin directly to +5 or GND or should you use a low value resistor or capacitor or something? Although now that's go me thinking about a dual ROM and toggle switch |
11 October 2017, 10:11 | #12 | ||
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Quote:
CMOS work differently internally and can be connected directly to a power rail. Many modern TTL devices actually use CMOS input stages, so the above note about the resistor probably doesn't apply. Bottom line, I would tie it to +5V with a 10K resistor. Quote:
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11 October 2017, 15:00 | #13 |
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Thank you for the info. When I burnt my ROMs I doubled the image twice accordingly to the fact I was using 27C800 but I didn't do anything with A18. Given I have an issue with this A1200 and USB I should probably tie the pin high to rule it out as a culprit. I don't really see how it could be causing my problem but it won't hurt.
Thanks and apologies to the OP for going off-topic. |
11 October 2017, 21:32 | #14 |
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The duel kickers I have used on 500's have the 2 extra pins floating. That stated they did have ground and switch control connected to them. < Actually I had to connect them as they didn't come with connections complete.
They did come with toggle switch though, from memory I think they are DPST type. I am not sure what type of ROM they are as they have a label over the top of the original surface. Shall have to check this and report back. @Daedalus, That 5v and R10K information was interesting where did you get that info from? |
11 October 2017, 21:36 | #15 |
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I think it's just general practice. I did know it from repairing arcade PCBs. If you look at most connections that have the potential to float (controller inputs for example) there will be pull-up resistors, or resistor arrays. However not only could I not remember if it was a resistor or a capacitor, I couldn't remember the values, or if it was relevant to the floating address line of the 27C800 ROM.
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11 October 2017, 22:17 | #16 |
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Yep, it's standard practice for TTL stuff. They require more current to drive the output low than high, so tying high reduces power consumption, which can be substantial if you're talking about a lot of inputs. And the resistor is required because 5V is fairly close to the breakdown voltage of the junctions in the original designs, meaning that tiny high voltage spikes from switching or other electrical noise can, over time, damage the gate resulting in an always on or always off input. A resistor ensures the input can deal with a higher input voltage and so the spikes have less impact. It's not needed when connected to another logic output because the output isn't normally as high as 5V, and carries less noise. For this reason, a common technique is to drive inputs from other, unused outputs.
The value of the resistor isn't hugely important - anything between 1K and 50K is probably fine. Original 74 series logic probably needs 1K, but they're long obsolete, and even the old LS series uses 10K as a rule of thumb for pull-ups. CMOS chips work totally differently and function more like op-amps that are saturated, so their input impedance is extremely high and they don't require much current at all to drive in input to either state. So tying either high or low is fine. And no resistor is needed because there are no bipolar junctions at close to breakdown voltage, so they can tolerate a wide range of input voltages. Many TTL-style chips are actually using CMOS technology in their cores (e.g. 74HC series), so they don't actually need the resistor or the pulling high, but it's no harm to follow those guidelines, especially when you're not sure what technology the chip uses. I'm sure there are plenty of references online for it, but I'm a fan of old-school technical books. I'm pretty sure I got that information from my 20-year-old and very well worn copy of Practical Electronic Design Data. It's a Babani book, and they're really quite excellent references. Edit: Here it is. Out of print but 2nd hand copies appear to be cheap. Can't recommend it enough as a handy reference for all your electronics tinkering needs. Last edited by Daedalus; 11 October 2017 at 22:29. |
12 October 2017, 14:51 | #17 |
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Daedalus, whao !!Holy s*** do you know your stuff or what? You should open a new thread on resistors and capacitors explained. < I could read your replies all day!
Just when you think you have mastered electronics < THINK AGAIN! I do not know whether to laugh or cry! Awesome information indeed! So am I right in thinking this is only using DC signal? <Can this method work on both signals? caps. Incidentally, this book looks mega, think I should invest immediately. Last edited by MigaTech; 12 October 2017 at 15:09. |
12 October 2017, 16:30 | #18 |
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Yep, that's for digital signals. Analogue is a different thing entirely, unused inputs would often be grounded but it really depends on the exact application. Sometimes it's not required at all, depending on the requirements of the particular chip.
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12 October 2017, 21:29 | #19 | |
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Quote:
Could talk to you more about this but I must allow thread to get back to topic. Thanks again, top stuff Daedalus! Sorry to OP for slipping off of topic. |
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12 October 2017, 21:53 | #20 |
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There is no need to apilogise...the thread has diverted in an interesting and beneficial manner!
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