Dobar Or, better yet, can you complete this truth table? You should include a truth table. It is not clear under what conditions you want to enable the Micro-Kim expansion port. What you are doing with the two outputs is not okay. Will that fry the N? Is it OK for the output pins to be high?

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I tried the arrangement you outlined. Since the inputs source current, they must be pulled Low with a small under 1K resistance to be seen as a logic Low. I successfully made an SR FF. Like this for example:. If using a switch to ground you need a pull up resistor again, say, 10K ohms to Vcc. The input of digital circuits needs to be either on or off. The easiest fix but not the prettiest is to connect a capacitor across the button one leg of the capacitor to one 74lw74n leg, and the other leg of the same capacitor to the other leg of the same button.

Note the pull up resistors at the inputs of both NAND gates. If you use too low of a voltage source, then the circuit will not operate. You need to de-bounce your switches in order to get predictable results.

Like this for example: WhatRoughBeast 49k 2 28 As it happens, you can do this cheaply by using the unused half of the chip as a SR flip-flop. Sign up using Email and Password. Adding a 5K or so pull-up resistor is good practice, but usually not necessary. I did the following: Until its sufficiently charged, the logic level will be one for a period defined by the resistor and capacitor values.

Sign up using Facebook. Try increasing the 74ks74n to ohms or so. Also, connect a resistor from the clock datassheet pin 3 to ground. The traditional solution for switch inputs is to put the switch between the input pin and ground.

I tried unsuccessfully to make an SR flip flop out of it too. If using a switch to Vcc you need a pull down resistor say, 10K ohms to ground. The reason is that the input current is much less when high than when low, so less power is wasted on the unused function. There are quite a number of changes you need to make.

The resistor is used to provide a default logic level to the clock, which in this case is logic level zero since the resistor is connected to ground. Home Questions Tags Users Unanswered. Is there a point where I may consider the chip is not functioning properly?

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