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Digital logic circuit design

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Hi everyone.

Im a new member of this great site.I hope this is the best place for me to improve my knowledge.

Im also a newbie to this electronic world.I have a thought in my mind to design a walkie talkie like circuit .But in my idea i want to design a circuit that can send and receive text messages wirelessly between 2 .im good in digital electrnics design n know about decoders,encoders,multiplexers ,demultiplexers,etc., and also how to design on my own.ive done


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Hi Sundar,
Welcome to our forum. ;D
I was going to attach clips from the datasheet of a 7408 IC and of an LED, but I think you should find them and read them:
1) You are correct in thinking that when both inputs are high for an AND gate then its output is high, but your circuit doesn't make the disconnected input low. Use a voltmeter to see.
TTL ICs like the 7408 have inputs that need a fairly low resistance to ground to cause the input voltage to be a logic low. The max input current is 1.6mA and a typical logic low voltage is 0.4V, so a resistance of 0.4V/1.6mA= 250 ohms or less to ground at the input will cause a logic low when the switch is opened. Use 220 ohms.
2) The minimum recommended supply voltage for TTL gates like a 7408 is 4.75V. Usually a regulated supply voltage of 5.0V is used.
3) The LED is like a dead short to the output of a TTL logic gate unless a current-limiting resistor is in series with the LED. With a 5V supply and a 2V LED, a 120 ohm resistor would protect the gate's output and LED and the LED might be bright enough.
4) If the LED needs a voltage higher than about 3V like a blue or white LED, then a TTL gate's output voltage might not go high enough. TTL outputs are designed to conduct 16mA to ground, but a much lower current at a fairly low voltage when they go high. 


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You may also put a resistor from the anode of the LED to GND to make sure the O/P of the AND gate is pulled low when you don't have both I/Ps high.

No. A resistor at the output to ground of a TTL gate is not required. It is guaranteed to sink 16mA to ground with a max voltage of only 0.4V when it is low. The output might not go high enough to light the LED if it had a resistor to ground.
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Hi again

I've completed making the circuit as said by audioguru and it works PERFECTLY FINE .I also checked a NAND gate (7400) and it also works well.i used a 2 x 3v = 6v supply.i had some resistors with me,a 330 ohm and a 120,didnt use a 220 ohm but i will buy one and see the difference in light intensity.

ive attached the picture of the circuit below.(batteries not included in pic



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yes i was thinking about that too.
what im asking is that the theory is feasible or not in your first look .
Also im a little confused with the keyboard thing.is there any specific IC for doing keyboard job ?
i thought of designing the keyboard myself and finally came to demultiplexers(1 input , 2^n outputs,n selection lines).I also think that some ROM chip is available in market for this.

consider an 8 bit number (00000000).as i said that i like designing my own keyboard rather than using EBCDIC and ASCII,i assume the first 2 bits from MSB as a dummy.i plan to use the other 6 bits since 2^6=64.this would  be enough for alphabets,numbers,special characters and control strings .

why im using 8bit is that the other 2 bits may be used for parity , etc.

First designing a custom 6bit binary to 7x5 matrix converter would be the best idea.for that a 6 variable kmap could be used to get the simplified boolean expression .but this somewhat looks like a tiresome process.so i can choose a an IC to do this.i dont know which ic would do this wonder .

help would be appreciated :)

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