Q1. What Is Difference Between Ram And Fifo?
FIFO does not have address lines
Ram is used for storage purpose where as FIFO is used for synchronization purpose i.e. when two peripherals are working in different clock domains then we will go for FIFO.
Q2. What Is Significance Of Ras And Cas In Sdram?
SDRAM receives its address command in two address words. It uses a multiplex scheme to save input pins. The first address word is latched into the DRAM chip with the row address strobe (RAS).
Following the RAS command is the column address strobe (CAS) for latching the second address word. Shortly after the RAS and CAS strobes, the stored data is valid for reading.
Q3. Convert D-ff Into Divide By
T_setup= 6nsT_hold = 2nS T_propagation = 10nS
Circuit: Connect Qbar to D and apply the clk at clk of DFF and take the O/P at Q. It gives freq/@Max. Freq of operation: 1/ (propagation delay+setup time) = 1/16ns = 62.5 MHz
Q4. What Will Happen If Contents Of Register Are Shifter Left, Right?
It is well known that in left shift all bits will be shifted left and LSB will be appended with 0 and in right shift all bits will be shifted right and MSB will be appended with 0 this is a straightforward wer What is expected is in a left shift value gets Multiplied by 2
e.g.: consider 0000_1110=14 a left shift will make it 0001_110=28, it the same fashion right shift will Divide the value by 2.
Q5. Give Two Ways Of Converting A Two Input Nand Gate To An Inverter?
a) Short the 2 inputs of the nand gate and apply the single input to it.
b) Connect the output to one of the input and the other to the input signal.
Q6. Difference Between Mealy And Moore State Machine?
A) Mealy and Moore models are the basic models of state machines. A state machine which uses only Entry Actions, so that its output depends on the state, is called a Moore model. A state machine which uses only Input Actions, so that the output depends on the state and also on inputs, is called a Mealy model. The models selected will influence a design but there are no general indications as to which model is better. Choice of a model depends on the application, execution me (for instance, hardware systems are usually best realized as Moore models) and personal preferences of a designer or programmer
B) Mealy machine has outputs that depend on the state and input (thus, the FSM has the output written on edges) Moore machine has outputs that depend on state only (thus, the FSM has the output written in the state itself.
Advantage and Disadvantage
•In Mealy as the output variable is a function both input and state, changes of state of the state variables will be delayed with respect to changes of signal level in the input variables, there are possibilities of glitches appearing in the output variables.
•Moore overcomes glitches as output dependent on only states and not the input signal level.
•All of the concepts can be applied to Moore-model state machines because any Moore state machine can be implemented as a Mealy state machine, although the converse is not true.
•Moore machine: the outputs are properties of states themselves… which me that you get the output after the machine reaches a particular state, or to get some output your machine has to be taken to a state which provides you the output. The outputs are held until you go to some other state Mealy machine:
•Mealy machines give you outputs instantly, that is immediately upon receiving input, but the output is not held after that clock cycle.
Q7. What Is A Multiplexer?
Is a combinational circuit that selects binary information from one of many input lines and directs it to a single output line.
(2n =>n). Where n is selection line.
Q8. Differences Between D-latch And D Flip-flop?
D-latch is level sensitive where as flip-flop is edge sensitive. Flip-flops are made up of latches.
Q9. Tell Some Of Applications Of Buffer?
a) They are used to introduce small delays.
b) They are used to eliminate cross talk caused due to inter electrode capacitance due to close routing.
c) They are used to support high fan-out, e.g.: bufg
Q10. How Can You Convert The Jk Flip-flop To A D Flip-flop?
By connecting the J input to the K through the inverter.
Q11. How Can You Convert An Sr Flip-flop To A Jk Flip-flop?
By giving the feedback we can convert, i.e. !Q=>S and Q=>R.Hence the S and R inputs will act as J and K respectively.
Q12. Implement The Following Circuits:
(a) 3 input NAND gate using min no of 2 input NAND Gates
(b) 3 input NOR gate using min no of 2 input NOR Gates
(c) 3 input XNOR gate using min no of 2 input XNOR Gates
Assuming 3 inputs A,B,C?
3 input NAND Connect:
a) A and B to the first NAND gate
b) Output of first Nand gate is given to the two inputs of the second NAND gate (this basically realizes the inverter functionality)4
c) Output of second NAND gate is given to the input of the third NAND gate, whose other input is C ((A NAND B) NAND (A NAND B)) NAND C Thus, can be implemented using ‘3’ 2-input NAND gates. I guess this is the minimum number of gates that need to be used.
Q13. What Is Difference Between Setup And Hold Time. The Interviewer Was Looking For One Specific Reason, And Its Really A Good Answer Too..the Hint Is Hold Time Doesn’t Depend On Clock, Why Is It So…?
Setup violations are related to two edges of clock, i mean you can vary the clock frequency to correct setup violation. But for hold time, you are only concerned with one edge and do not basically depend on clock frequency.
Q14. An Assembly Line Has 3 Fail Safe Sensors And One Emergency Shutdown Switch. The Line Should Keep Moving Unless Any Of The Following Conditions Arise:
(i) If the emergency switch is pressed
(ii) If the senor1 and sensor2 are activated at the same time.
(iii) If sensor 2 and sensor3 are activated at the same time.
(iv) If all the sensors are activated at the same time
suppose a combinational circuit for above case is to be implemented only with NAND Gates. How many minimum number of 2 input NAND gates are required?
No of 2-input NAND Gates required = 6 you can try the whole implementation.
Q15. You Have Two Counters Counting Upto 16, Built From Negedge Dff , First Circuit Is Synchronous And Second Is “ripple” (cascading), Which Circuit Has A Less Propagation Delay? Why?
The synchronous counter will have lesser delay as the input to each flop is readily available before the clock edge. Whereas the cascade counter will take long time as the output of one flop is used as clock to the other. So the delay will be propagating. For E.g.: 16 state counter = 4 bit counter = 4 Flip flops Let 10ns be the delay of each flop The worst case delay of ripple counter = 10 * 4 = 40ns The delay of synchronous counter = 10ns only.(Delay of 1 flop)
Q16. 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 is the amount of time the data should be stable before the application of the clock signal, where as the hold time is the amount of time the data should be stable after the application of the clock. Setup time signifies maximum delay constraints; hold time is for minimum delay constraints. Setup time is critical for establishing the maximum clock frequency.
Q17. The Circle Can Rotate Clockwise And Back. Use Minimum Hardware To Build A Circuit To Indicate The Direction Of Rotating?
2 sensors are required to find out the direction of rotating. They are placed like at the drawing. One of the m is connected to the data input of D flip-flop, and a second one – to the clock input. If the circle rotates the way clock sensor sees the light first while D input (second sensor) is zero – the output of the flip-flop equals zero, and if D input sensor “fires” first – the output of the flip-flop becomes high.
Q18. Why Is Most Interrupts Active Low?
This wers why most signals are active low if you consider the tristor level of a module, active low me the capacitor in the output terminal gets charged or discharged based on low to high and high to low trition respectively. When it goes from high to low it depends on the pull down resistor that pulls it down and it is relatively easy for the output capacitance to discharge rather than charging. Hence people prefer using active low signals.
Q19. How Will You Implement A Full Subtractor From A Full Adder?
All the bits of subtrahend should be connected to the xor gate. Other input to the xor being one. The input carry bit to the full adder should be made @Then the full adder works like a full subtract.
Q20. Design All The Gates (not, And, Or, Nand, Nor, Xor, Xnor) Using 2:1 Multiplexer?
Using 2:1 Mux, (2 inputs, 1 output and a select line)
a) NOT :Give the input at the select line and connect I0 to 1 & I1 to @So if A is 1, we will get I1 that is 0 at the O/P.
b) AND: Give input A at the select line and 0 to I0 and B to I@O/p is A & B
c) OR: Give input A at the select line and 1 to I1 and B to I@O/p will be A | B
d) NAND: AND + NOT implementations together
e) NOR: OR + NOT implementations together
f) XOR: A at the select line B at I0 and ~B at I@~B can be obtained from (a)
g) XNOR: A at the select line B at I1 and ~B at I0
Q21. How To Achieve 180 Degree Exact Phase Shift?
Never tell using inverter
a) DCM an inbuilt resource in most of FPGA can be configured to get 180 degree phase shift.
b) BUFGDS that is differential signaling buffers which are also inbuilt resource of most of FPGA can be used.
Q22. In A 3-bit Johnson’s Counter What Are The Unused States?
2(power n)-2n is the one used to find the unused states in Johnson counter.
So for a 3-bit counter it is 8-6=2.Unused states=@the two unused states are 010 and 101.
Q23. Difference Between Synchronous And Asynchronous Reset?
Synchronous reset logic will synthesize to smaller flip-flops, particularly if the reset is gated with the logic generating the dinput. But in such a case, the combinational logic gate count grows, so the overall gate count savings may not be that significant. The clock works as a filter for small reset glitches; however, if these glitches occur near the active clock edge, the Flip-flop could go metastable. In some designs, the reset must be generated by a set of internal conditions. A synchronous reset is recommended for these types of designs because it will filter the logic equation glitches between clocks.
Disadvantages of synchronous reset:
Problem with synchronous resets is that the synthesis tool cannot easily distinguish the reset signal from any other data signal. Synchronous resets may need a pulse stretcher to guarantee a reset pulse width wide enough to ensure reset is present during an active edge of the clock. if you have a gated clock to save power, the clock may be disabled coincident with the assertion of reset. Only an asynchronous reset will work in this situation, as the reset might be removed prior to the resumption of the clock. Designs that are pushing the limit for data path timing, cannot afford to have added gates and additional net delays in the data path due to logic inserted to handle synchronous resets.
Asynchronous reset:
The biggest problem with asynchronous resets is the reset release, also called reset removal. Using an asynchronous reset, the designer is guaranteed not to have the reset added to the data path. Another advantage favoring asynchronous resets is that the circuit can be reset with or without a clock present.
Disadvantages of asynchronous reset: ensure that the release of the reset can occur within one clock period. if the release of the reset occurred on or near a clock edge such that the flip-flops went metastable.
Q24. How Do You Detect If Two 8-bit Signals Are Same?
XOR each bits of A with B (for e.g. A [0] xor B [0]) and so on. The o/p of 8 xor gates is then given as i/p to an 8-i/p nor gate.
if o/p is 1 then A=B.
Q25. What Is Difference Between Latch And Flip-flop?
The main difference between latch and FF is that latches are level sensitive while FF is edge sensitive. They both require the use of clock signal and are used in sequential logic. For a latch, the output tracks the input when the clock signal is high, so as long as the clock is logic 1, the output can change if the input also changes.
FF on the other hand, will store the input only when there is a rising/falling edge of the clock. Latch is sensitive to glitches on enable pin, whereas flip-flop is immune to glitches. Latches take fewer gates (also less power) to implement than flip-flops. Latches are faster than flip-flops
Q26. What Is Race-around Problem? How Can You Rectify It?
The clock pulse that remains in the 1 state while both J and K are equal to 1 will cause the output to complement again and repeat complementing until the pulse goes back to 0, this is called the race around problem. To avoid this undesirable operation, the clock pulse must have a time duration that is shorter than the propagation delay time of the F-F, this is restrictive so the alternative is master-slave or edge-triggered construction.
Q27. Design A Circuit That Calculates The Square Of A Number?
It should not use any multiplier circuits. It should use Multiplexers and other logic?
1^2=0+1=1
2^2=1+3=4
3^2=4+5=9
4^2=9+7=16
5^2=16+9=25
See a pattern yet? To get the next square, all you have to do is add the next odd number to the previous square that you found. See how 1,3,5,7 and finally 9 are added. Wouldn’t this be a possible solution to your question since it only will use a counter, multiplexer and a couple of adders? It seems it would take n clock cycles to calculate square of n.
Q28. Difference Between One Hot And Binary Encoding?
Common classifications used to describe the state encoding of an FSM are Binary (or highly encoded) and One hot.
A binary-encoded FSM design only requires as many flip-flops as are needed to uniquely encode the number of states in the state machine. The actual number of flip-flops required is equal to the ceiling of the log-base-2 of the number of states in the FSM.A one hot FSM design requires a flip-flop for each state in the design and only one flip-flop (the flip-flop representing the current or “hot” state) is set at a time in a one hot FSM design.
For a state machine with 9- 16 states, a binary FSM only requires 4 flip-flops while a one hot FSM requires a flip-flop for each state in the design FPGA vendors frequently recommend using a one hot state encoding style because flip-flops are plentiful in an FPGA and the combinational logic required to implement a one hot FSM design is typically smaller than most binary encoding styles.
Since FPGA performance is typically related to the combinational logic size of the FPGA design, one hot FSMs typically run faster than a binary encoded FSM with larger combinational logic blocks
Q29. Given Only Two Xor Gates One Must Function As Buffer And Another As Inverter?
Tie one of xor gates input to 1 it will act as inverter.
Tie one of xor gates input to 0 it will act as buffer.
Q30. 7 Bit Ring Counter’s Initial State Is 01000
6 cycles
Q31. Design A Four-input Nand Gate Using Only Two-input Nand Gates.
Basically, you can tie the inputs of a NAND gate together to get an inverter.
Q32. Consider Two Similar Processors, One With A Clock Skew Of 100ps And Other With A Clock Skew Of 50ps. Which One Is Likely To Have More Power? Why?
Clock skew of 50ps is more likely to have clock power. This is because it is likely that low-skew processor has better designed clock tree with more powerful and number of buffers and overheads to make skew better.
Q33. N Number Of Xnor Gates Is Connected In Series Such That The N Inputs (a0, A1, A2……) Are Given In The Following Way: A0 & A1 To First Xnor Gate And A2 & O/p Of First Xnor To Second Xnor Gate And So
If N=Odd, the circuit acts as even parity detector, i.e. the output will 1 if there are even number of 1’s in the N input…This could also be called as odd parity generator since with this additional 1 as output the total number of 1’s will be ODD. If N=Even, just the opposite, it will be Odd parity detector or Even Parity Generator.
Q34. Is It Possible To Reduce Clock Skew To Zero? Explain Your Answer?
Even though there are clock layout strategies (H-tree) that can in theory reduce clock skew to zero by having the same path length from each flip-flop from the pll, process variations in R and C across the chip will cause clock skew as well as a pure H-Tree scheme is not practical (consumes too much area).
Q35. Given The Following Fifo And Rules, How Deep Does The Fifo Need To Be To Prevent Underflow Or Overflow?
RULES:
1) frequency(clk_A) = frequency(clk_B) / 4
2) period(en_B) = period(clk_A) * 100
3) duty cycle(en_B) = 25%
Assume clk_B = 100MHz (10ns)
From (1), clk_A = 25MHz (40ns)
From (2), period(en_B) = 40ns * 400 = 4000ns, but we only output for 1000ns,due to (3), so 3000ns of the enable we are doing no output work. Therefore, FIFO size = 3000ns/40ns = 75 entries
