Timeline for TTL 7400 NAND gate circuit not functioning [closed]
Current License: CC BY-SA 3.0
14 events
when toggle format | what | by | license | comment | |
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Jul 21, 2016 at 0:14 | vote | accept | zack1544 | ||
Jul 20, 2016 at 11:11 | answer | added | Bruce Abbott | timeline score: 3 | |
Jul 20, 2016 at 11:10 | history | closed |
Scott Seidman user105652 Bence Kaulics dim Daniel Grillo |
Needs details or clarity | |
Jul 20, 2016 at 4:19 | comment | added | zack1544 | I think that that the "in my head inversion " is a good idea for this circuit. In the ALU, I will probably just add a not gate chip to both the inputs before they go to the NAND gate. Thanks for all your help. | |
Jul 20, 2016 at 4:01 | comment | added | Tom Carpenter | Just treat the switch being closed as a 0. You don't need any extra circuitry - the inversion can be built into your head with no electronics required. | |
Jul 20, 2016 at 2:00 | comment | added | zack1544 | Tom thanks for the explanation, I understand what you are saying. I am making a series of small circuits that perform basic boolean logic and combining them to form a functioning ALU. For this reason I don't want to add unnecessary chips. Is the only solution to add a 4 bit bit not gate between the switch's output and the 7400 input? This would require two more IC's. Is there a way to make the circuit function as a NAND gate without these two IC's? Sorry if this question doesn't make sense I'm a newbie. | |
Jul 20, 2016 at 0:06 | comment | added | SamGibson♦ | This seems a near duplicate to the issues explained in "Connecting AND Gate Chip to an Integrated Circuit" (although a different TTL IC) i.e. switches connected to VCC with no pull-down resistors on the TTL inputs (well spotted @TomCarpenter). | |
Jul 20, 2016 at 0:04 | comment | added | Tom Carpenter | Makes sense. If you invert both inputs of a NAND gate, you get an OR gate. Now that when the switches are pressed the input is a zero - so you need to remember that closed = 0, open = 1. I'll write all this up as an answer in a little while, just having dinner. | |
Jul 19, 2016 at 23:59 | comment | added | zack1544 | Thanks for the quick response Tom. I followed your advice and connected the input switches to GND instead of VCC. Now the circuit functions as an OR Gate which is weird. I will update my question with a schematic diagram and a picture of the circuit. It might take me a while though, since I'm not free for the next hour. Thanks. | |
Jul 19, 2016 at 23:37 | review | Close votes | |||
Jul 20, 2016 at 11:10 | |||||
Jul 19, 2016 at 23:20 | comment | added | Tom Carpenter | Also you are not connecting your inputs properly. You are connecting them to switches so that the inputs are either connected to VCC or left floating. A 7400 series TTL device like the one you have has internal pull ups, which means when floating the pins will pull high - basically all your inputs are always high. Your switches should connect to GND. | |
Jul 19, 2016 at 23:18 | comment | added | Tom Carpenter | If your breadboard is a double length one like shown, remember that typically there is a break in the power rail at the centre point, so it is possible that the GND and VCC are not connected to your NAND. But without an actual photo that is a guess. | |
Jul 19, 2016 at 23:17 | comment | added | Tom Carpenter | Draw a proper schematic (a fritzing diagram of a breadboard is not a schematic) and also upload a picture of your circuit. | |
Jul 19, 2016 at 23:14 | history | asked | zack1544 | CC BY-SA 3.0 |