NAND Gate IC problem... plz help

[ak214]

nand gate ic

while making manchester encoding circuit ... i am giving NRZ-biphase input to the nand gate IC (74ls08) ... but when i connect my input (NRZ-bi) to 7408 , my input automatically converts to NRZ unipolar .... i tried removing IC'S Vcc but the problem continues with ground..
When i checked the input pin of 7408 without my nrz biphase ,,,, there is something flying back from pin 2 of 7408 which is spoiling my ckt ...
kindly some one please help me quickly...
its urgent.
Thanks in advance

 

[silvio]

ic 7408

It's somehow confusing, NRZ-biphase input to nand gate 74LS08 (however it's an two input AND gate), input automatically converts to NRZ unipolar when using 7408.
Can you post a schematic of your circuit ?

When i checked the input pin of 7408 without my nrz biphase ,,,, there is something flying back from pin 2 of 7408 which is spoiling my ckt ...

I bet it is.

 

[ak214]

nand 7408

here is the circuit....

 

[ak214]

uses of ic 74ls08

here is the ciruit... of manchester encoding.

 

[ak214]

ic 74ls08

here is the ciruit... of manchester encoding.

 

[silvio]

nrz-l encoding circuit

You don't have to use 74LS08 or 7408. These are AND gates. You must use 74LS00 or 7400 NAND gates.
If you'll use 74LS08 for gates whom output is 3 and 5, then you must use an OR gate for the gate with output marked 5.
Obvious you must keep the D latch and two inverter gates.

However, there is no NRZ-biphase. It's either ordinary NRZ or Manchester biphase.
The unipolar encoding uses positive voltage but no negative voltage to represent 1s and 0s.
What do you mean "my input automatically converts to NRZ unipolar". Can you imagine how "NRZ-bipolar" looks like ?

 

[ak214]

sorry i mistakenly written nand gate ic as 7408 actually i am using 7400 .........
but wht i am doing is this-->

i am giving input to this circuit by a kit which generates nrz-l and after nrz-l i am making this signal to bipolar and then giving input to this circuit....
the problem is that when i check this input on cro its bipolar that is correct,,,
but when i connect this input to nand gate ic input pin 2 then the at the same point the input signal becomes unipolar .... i think it is necessary to give biphase input to this circuit in order to have output as manchester.
wht is this problem
if i use combination of and & or gate .... can ths problem be removed????

 

[silvio]

Unipolar Uses positive voltage but no negative voltage to represent 1s and 0s

Bipolar : Whenever a 1 is encountered, the voltage jumps alternately to the positive voltage or the negative voltage. 0 is always represented by zero voltage

Manchester: in this encoding scheme a positive to negative mid-bit voltage transition denotes a 0 and a negative to positive transition denotes a 1. The Manchester coding scheme is known as "biphase".

after nrz-l i am making this signal to bipolar and then giving input to this circuit....

How is converted to bipolar. What bipolar means for you ? Why don't use NRZ-L directly ? Why you convert it to "bipolar" ?

then the at the same point the input signal becomes unipolar

What means for you "unipolar" ?

i think it is necessary to give biphase input to this circuit in order to have output as manchester.

The Manchester itself is a biphase signal as a result of converting NRZ.

No offence but it's a terrible mess in your mind.

In order to use your circuit two signals are required : a data NRZ and a clock. In the Manchester code the clock signal is hiden and recovered at the target, based on transitions in the middle of the bit.

What do you want to do ? You have two circuits, one delivering NRZ that must be interfaced with other that accepts Manchester code ?

 

[ak214]

why i want to give nrz biphase is that otherwise i will not get the output as manchester.... i get only +v to 0 waveform and not negative one....
now i am using opamp to convert the output signal to make it +v to -v (now i have removed the input nrz biphase i am just giving nrz -l...
but now i cant realize my output its somewhat confusing.
waveform starts rolling .

 

[silvio]

Using opamp to make the output swing between +v to -V is the only solution because 7400 is supplied between +5v and 0v.
I understand now what's your confusion. The Manchester is a coding way. You can have negative/positive transitions in the middle of the bit between +5v and 0 or between +v and -v as well.
It's not the swinging range that counts but the function provided by Manchester coding.

If you look carefully at the input stage of a TTL gate you can see a diode fitted with anode on gnd and cathode on input gate.
His purpose (called clamping diode) is to limit to -0.6v the input signal which swings to values bellow gnd, down to -V like your input.