Chip help A-B 314A472 8016M datasheet

I'm trying to troubleshoot a board problem and I can't find any info on this chip. The 8016M is under the 314A472. Can anyone help? Thanks!

-Gerry

Reply to
GerryL
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8016 will be the date code (16th week of 1980). So the 314A472 is the part number. Could just be the 314, for example LM314. How many pins does the IC have? Any idea what it actually is, op-amp, logic gate, etc Rob
Reply to
viking

It's a 14 pin. This board is where a problem keypad connects. This is a Bendix board, used in a CNC machine. The data entry keypad is not functioning properly, several keys do nothing when pressed. (but the keys/buttons are working, I tested for continuity). There are a couple of 14 pin AND chips and a couple 16 pin diode arrays nearby. Any thoughts?

-Gerry

vik> > I'm trying to troubleshoot a board problem and I can't find any info on

Reply to
GerryL

On 12 Jan 2007 17:00:45 -0800, "GerryL" put finger to keyboard and composed:

Plugging "314A472" into the search box at

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brings up a resistor network made by Allen Bradley. The "472" could be 4700 ohms.

- Franc Zabkar

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Reply to
Franc Zabkar

OK, this has been very helpful. The 314A472 is a 5K resistor network, and it is fine. I've narrowed the problem down to one section of a UHP-400. This is described as a quad power and relay driver. I have to say I'm confused by the datasheet though, it's here

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The UHP-400 is listed as an AND driving a switching transistor. In the datasheet the symbol is what I am used to calling a NAND, and the chip is operating like a NAND. Can someone have a look and tell me if I'm missing something? In my circuit, pins 1&2 are tied together, and 4&5, and 9&10, and 12&13. So, if it's an AND, I'd assume driving them high would sink the corresponding pin. But no, I have to take them low to get the corresponding pin to sink. I'm not sure if I'm being clear on this. I'm fairly new to playing with this, so I may be missing something basic. One part of the chip will not operate, so I assume that is the problem. If I bring 1/2 low, 3 will sink. If I bring 9/10 low, 8 will sink. If I bring 12/13 low, 11 sinks. Bringing 4/5 low will not sink pin 6. This appears to be an older chip that is not easily available. I assume I can build an external circuit to replace this part of the chip...any suggestions? Thanks! -Gerry

Reply to
GerryL

The transistor provides an inversion, so the NAND gate symbol plus the inverting transistor results in an AND gate. Note that the outputs are "open collector" - since the output is just the collector of an NPN transistor, the chip can pull an output low, but the external circuit is responsible for pulling the output high.

In an AND gate, a high input should result in a high output (no inversion)

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Reply to
Peter Bennett

On 13 Jan 2007 19:15:51 -0800, "GerryL" put finger to keyboard and composed:

You can still get them here ($36.87):

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Otherwise, if one of your UHP400 chips has a spare gate, then rewire your circuit to use that.

- Franc Zabkar

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Reply to
Franc Zabkar

Franc, thanks a lot for that info. I don't have a spare gate, but that was a good idea. I was hoping to construct the circuit with individual components outside of the UHP400 and tack it on, but I haven't been successful with that. I did find a UHP400 on eBay last night and I bought it, but then I went to their eBay store to see that they ship once or twice a week, expect 8-10 day deliveries through airmail, and if you need it faster, don't order. Great! :-( But the price was better at 12.50 or so. :-) We might buy one from that other link just to get it quick from NJ. I've found it frustrating when doing a search for components to run into all the 'request for quote' sites. Set up account, etc, find out that there is $150 and up minimum order. It's certainly not set up for the home hobbiest who is trying to learn by troubleshooting/repairing this stuff. And the alternative is to send a board out and pay someone $500 to replace a chip, or a capacitor. That hurts when it's a machine at home for hobby work.

-Gerry

Franc Zabkar wrote:

Reply to
GerryL

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