Essential 86 Family Interview Questions & Answers:

1. Design a Transmission Gate based XOR. Now, how do you convert it to XNOR?

put one bubble in front of xor gate....

2. Explain Transmission Gate-based D-Latch?

The Transmission-Gate input is connected to the D_LATCH
data input (D), the control input to the Transmission-Gate
is connected to the D_LATCH enable input (EN) and the
Transmission-Gate output is the D_LATCH output (Q)

3. Give truth table for a Half Adder, Give a gate level implementation of the same?

x y c s
0 0 0 0
0 1 0 1
1 0 0 1
1 1 1 0

4. What are different Adder circuits you studied?

Half Adder (for addition of two bits)
Full Adder (for addition of three bits)
Carry propagate adder
Carry save adder
Carry look ahead adder

5. Suppose you have a combinational circuit between two registers driven by a clock. What will you do if the delay of the combinational circuit is greater than your clock signal?

Put even number of not gates between clocks of reg A and
Reg B. The not gates will introduce delay between clock of
reg A and reg B.

6. Design a divide-by-3 sequential circuit with 50% duty circle now?

incoming clock by ODD value as assigned in genric
CLK_DIV_BY generic with

50% duty cycle
------------------------------------------------------------
-----------


library IEEE;
USE ieee.std_logic_1164.all;
USE ieee.std_logic_arith.all;
USE ieee.std_logic_unsigned.all;

entity FDIV is

generic(
CLK_DIV_BY : INTEGER :=15; --Give the
odd value with which you want to divide the clock i.e.
3,5,7,9
COUNTVALUE : INTEGER :=4 --Give the bit
count of division ratio value.Ex upto 3= 2 bits; 5 to 7 =
3; 9 to 15 = 4 and so on..
);

port(
CLK : in std_logic;
CLR : in std_logic;
DIV: out std_logic
);
end FDIV;
--------------------------------------------------
Architecture beh of FDIV is
signal DIV_pos, DIV_neg :std_logic;
signal posedgecounter :std_logic_vector((COUNTVALUE - 1)
downto 0);
signal negedgecounter,test :std_logic_vector((COUNTVALUE -
1) downto 0);
begin
-----------------------------
PROCESS(CLK,CLR)
begin
IF ( CLR = '0') THEN
posedgecounter <= (others =>'0');
ELSIF RISING_EDGE(CLK) THEN
posedgecounter <= posedgecounter + 1;
if posedgecounter = conv_std_logic_vector((CLK_DIV_BY -
1),(COUNTVALUE)) then
posedgecounter <= (others =>'0');
end if;
if posedgecounter <= conv_std_logic_vector(((CLK_DIV_BY -
1)/2),(COUNTVALUE)) then
DIV_pos <= '1';
else
DIV_pos <= '0';
end if;
END IF;
END PROCESS;
------------------------------
PROCESS(CLK,CLR)
begin
IF ( CLR = '0') THEN
negedgecounter <= (others =>'0');
ELSIF FALLING_EDGE(CLK) THEN
negedgecounter <= negedgecounter + 1;
if negedgecounter = conv_std_logic_vector((CLK_DIV_BY -
1),(COUNTVALUE)) then
negedgecounter <= (others =>'0');
end if;
if negedgecounter <= conv_std_logic_vector
(((CLK_DIV_BY -1)/2),(COUNTVALUE)) then
DIV_neg <= '1';
else
DIV_neg <= '0';
end if;
END IF;
END PROCESS;
----------------------------------------
DIV<= DIV_pos and DIV_neg;
----------------------------------------
end beh;

7. Please give a circuit to divide frequency of clock cycle by two?

a T flip flop act as a frequency divider

8. What are set up time & hold time constraints What do they signify Which one is critical for estimating maximum clock frequency of a circuit?

Set up time constraint signifies how late the input signal
can arrive before the active edge of the flip-flop. Smaller
the set up time, the better.
Hold time on the other hand signifies how long the value at
the input needs to be held stable after the the active edge.
Again the smaller the hold time, the better.
For estimating maximum clock frequency, set up time is critical.

9. Explain What is the difference detween ISR & function call?

in isr there is no return value but in function call there
is return value

10. Where can we find the sample ASSEMBLY LANGUAGE programs?

write a c code.
and generate the assembly for it using
cc -S xyz.c -o xyz.S
xyz.Swill contain assembly for your c code with instructions
of the processor of your computer/hardware