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#1
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I have asked this on a electronic forum, and no one could help me too, maybe here you guys can be a bit more handy! I am trying to do a simple serial-to-serial shift register, with D-flip-flops, just as simple as This. ![]() And i am using a 74HC374 IC, Octal D-Type flip-flops. This is my diagram: ![]() What am i doing Wrong? I ask this because, when i set LOAD to 1 (5 volts), and give a clock to the circuit, all the outputs of the used flip-flpos becomes 1, it never loads only 1 bit! Yes i have made sure that, at the clock input pin, only one clock is submited, so no several clocks could be imputed with a push of a button. Q0, Q1 and Q2 are all connected to resisted LED's to check their state. Why? i can't understand ! |
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#3
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| At the clock input, i have hadded a 1k resistor to ground, and a huge cap. i connect clock to vcc trought a 500 ohm, and it takes a bit for the LEDs to lit up , so that it absorbs the noise when i connect the wire. Still a no go, when it happens, all of them lit up |
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#5
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| The huge cap will cause a slow rising edge, which edge triggered logic circuits don't like. This may cause each d-type to clock at a slight different time as each will have a very slightly different threshold for switching from 0 to 1. To make a shift register they must all switch together. You need to use one of the circuits here :- http://www.elexp.com/t_bounc.htm Also you should have a decoupling capacitor between VCC and ground close to the chip. 100nF should be fine. |
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#6
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| Back ![]() Well i pick up my lazy a$$ and went to buy a 555 IC and some pieces to make a regulated clocker. I mounted them all and... ...it works. Thanks all, i feel like an idiot but at least it works now and i have learned something |
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