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Just to show what I’m talking datashset, once you’ve got the circuit working, put an LED directly to ground on the flip-flop clock input, and watch what happens.
A common way to do this is with cross-connected NAND gates. You need a de-bouncing circuit to feed the flip-flop’s clock input.
Connect the cap directly to the sockets nearest the IC power and ground pins. Always connect the GND pin.
74HC109 DIP-16 J/K flip-flop
Sometimes that works, and sometimes it doesn’t. Sign up or log in Sign up using Google. Email Required, but never shown. LED does not light up, I’ve done something wrong again haven’t I?
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In this case, it will show up as intermittent operation of the flip-flop, depending on whether 74nc109 not it got an even or odd number of contacts when you press the switch. Caps are cheap and don’t take up much space, but the problems arising from not enough of them can be hard to diagnose. I’m trying to toggle the led forgot to add a resistor in the diagram, ah, also forgot to add a datasbeet from 5V to Vcc.
I’m a complete amateur, I’ve probably gotten a million things wrong with this already. Sign up using Facebook. Just pull it down to 0V with a single resistor! As others have suggested, this could be a simple R-C network, or as a better option, use an SPDT change-over switch and a set-reset SR flip-flop as a de-bouncer.
I’m using an SN74HC dual j-k” flip-flop chip. Post as a guest Name. WhatRoughBeast 49k 2 28 Sign up using Datashewt and Password.
In this case, a ohm resistor, combined with the LED voltage and the limited high output drive of the flip-flop will give you about 6 or 7 mA LED current when on. Tie them all high or low. Don’t even think about it. John Honniball 2, 6 9. Never leave it unconnected, even if the circuit appears to work.
74HC 데이터시트(PDF) – NXP Semiconductors
Maximum Low-Level input voltage with my Vcc at 5V is about 1. When the circuit is powered, the led turns on but does not toggle when the momentary switch is pressed and released. Please build the following circuit: On the one hand, it will try to draw too much current in the on state, and will also clamp the output to a low voltage which the input will not recognize as high.