Following is the truth table for a NOR gate. x = 0 This is not to be confused with the fact that nonblocking assignment LHS update will always happen after the blocking assignments even if blocking assignment appears later in the begin..end order. If reset release happen in different clock cycles, then different flops will come out of reset in different clock cycles and this will corrupt the state of your circuit. MCQ quiz on VLSI Design multiple choice questions and answers on VLSI Design MCQ questions on VLSI Design objectives questions with answer test pdf for interview preparations, freshers jobs and competitive exams. To get a very good idea of the execution order of different statements and assignments, especially the blocking and non-blocking assignments, one has to have a sound comprehension of inner workings of Verilog. Because it uses either NMOS or PMOS logic and not CMOS logic, it usually has fewer transistors compared to static gates. 10. With asynchronous reset, unintended glitches will cause circuit to go into reset state. … Lets also assume that for width ‘W’, the gate capacitance is ‘C’. There are few other situations where race conditions could come up, for example if a function is invoked from more than one active blocks at the same time, the execution order could become non-deterministic. … You have an input vector where on subsequent clock cycles you get a1,a2,a3,a4 values. Or one may have to switch to synchronous reset. The reason for this phenomenon is that with increase in temperature, thermal vibrations in lattice increase. – Spurious glitches. This parallelism can be the source of the race condition as shown in above example. This can be derived using Kmaps as following. Why a short channel device is preferred? the purpose of cache is to reduce the average time to. VLSI QUESTION AND ANSWER; Search. In static gates, inputs switch and after a finite input to output delay, output possibly switches to the expected state. – It has obvious advantage of being able to reset flops without the need of a clock. – Use of tri-state structures. When chip powers up, initially the clocks are not active and they could be gated by the clock enable, but right during the power up we need to force the chip into an known set and we need to use reset to achieve that. α = positive constant ( small number ). Usually, in an integrated circuit there will be several MOSFETs and in order to … – Another major issue with synchronous is clock gating. This could be double edged sword as we have seen earlier, but if your design permits the use of asynchronous reset, this could be an advantage. 5) LHS update from the second nonblocking assignment is carried out. Sum up such R and C product for all nodes. If we want to find the minimum value of total delay function for a specific value of fanout ‘a’, we need to take the derivative of ‘total delay’ with respect to ‘a’ and make it zero. Clock gating is the technique where clock is passed through an and gate with an enable signal, which can turn off the clock toggling when clock is not used thus saving power. As it is essentially evaluating what the output should be during this phase. 11. You can see that when at least one of the pull down n-mos is on, there is a static bias current flowing from VDD to the ground even in the steady state. y <= 3 During the low clock phase, because of the pmos gate on the pull up network, the output of dynamic gate is pre-charged to high phase. When CLK is high it passes through D to O and when CLK is off, O is fed back to D0 input of mux, hence O appears back at the output, in other words, we retain the value of O when CLK is off. There could be other possible elaborate schemes to achieve the same using n-mos transistors for pulling up purpose, but an n-mos as a resistor is used to pull up the output node. Take following example. You have an input vector where on subsequent clock cycles you get a1,a2,a3,a4 values. If there are fourstates then it requires four … Complementary stands for the fact that in CMOS technology based logic, we use both p-type devices and n-type devices. Depending on the order you could get final value of ‘x’ to be either ‘2’ or ‘9’. Because of the way x1 and x2 transition during the first clock cycle we get a glitch on reset signal, but because reset is synchronous and because glitch did not happen near the active clock edge, it got filtered and we only saw reset take effect later during the beginning of 4th clock cycle, where it was expected. To simplify our analysis we can focus on the leakage power, which is proportional to the width or size of the gate. Similarly if two active blocking assignments happen to read from and write to the same variable, you’ve a read write race. Ans: VLSI technology can incorporate ICs in a range of 2000 to 20,000. – Wide enough pulse of the reset signal. Q.1. – Faster data path. Anna University EC6601 VLSI Design Syllabus Notes 2 marks with answer is provided below. For the above mentioned case, the execution order still follows the order in which statements appear. For EC8095 VLSI D Question Bank/2marks 16marks with answers – Click here. When the clock is at high phase, the output of dynamic gate may change based on the inputs, or it may stay pre-charged depending on the input. Evaluation of RHS of nonblocking statement has same priority as blocking statement execution in general. We need to come up the a circuit for this NOR gate, using n-mos only transistors. Our aim is to find out the nominal fanout value which gives the best speed with least possible power dissipation. Multiple Choice Questions and Answers on VLSI Design & Technology Multiple Choice Questions and Answers By Sasmita January 13, 2017 1) The utilization of CAD tools for drawing timing waveform diagram and transforming it into a network of logic gates is known as ________. In Verilog certain type of assignments or expression are scheduled for execution at the same time and order of their execution is not guaranteed. – One obvious question that comes to mind, having gone through previous example is that what would be the execution order of the nonblocking LHS udpate !! What is body effect? This is the reason, at higher voltages device delay increase with temperature but at lower voltages, device delay increases with temperature. The last step here will be the update to ‘y’ for the nonblocking statement. primitive IO updates and $write commands. During the test vector application, we can not have any flop get reset. This means that blocking assignments, continuous assignments, primitive output updates, and $display command, all could be executed in any random order across all the active processes. But here we are referring to NMOS logic and we are not allowed to have p-mos devices. The idea is to pick gate sizes in such a way that it gives the best power v/s performance trade off. The Sequential FSM Finite State Machine DIGIQ based questions are very important for any digital interview. Within “active” queue all events have same priority, which is why they can get executed in any order and is the source of nondeterminism in Verilog. It really depends on which position you are applying. One can visualize this as electrons colliding with each other more and hence contributing less to the streamline flow needed for the flow of electric current. While asynchronous reset flop has to factor reset inside the flop design, where typically one of the last inverters in the feedback loop of the slave device is converted into NAND gate. On the flip side when we do intend the reset to work, the reset pulse has to be wide enough such that it meets setup to the active edge of the clock for the all receivers sequentials on the reset distribution network. Dynamic gates burn more power because of the associated clocks. When tri-state devices are used, they need to be disabled at power-up. There is similar effect that happens in semiconductor and the mobility of primary carrier decreases with increase in temperature. The “Inactive” event queue has been omitted as #0 delay events that it deals with is not a recommended guideline. From our understanding of CMOS logic, we can think about the pull down tree, which is made up of only n-mos gates. Remember blocking assignments within a ‘begin’ .. ‘end’ block are executed in the order they appear. At high voltage mobility determines the drain current where as at lower voltages threshold voltage dominates the darin current. After reading these tricky VLSI questions, you can easily attempt the objective type and multiple choice type questions on VLSI. As per the definition, we are only allowed to use the n – type device as building blocks. You can see that when reset input of the NAND gate, goes low it forces the Q output to be low irrespective of the input of the feedback loop. Sushanth Kj. Output only has to switch in the case where it has to be low. 4) RHS of nonblocking assignment ‘y <= 6’ is evaluated and LHS update is scheduled. here EC8095 VLSI Design notes download link is provided and students can […] In such cases, it will be better to go with synchronous reset implementation. p <= 6 The theory still applies, one just have to find out the effective wire capacitance that the driving gate sees and use that to come up with the fanout ratio. dD/da = 3*RC* ln(CL/C) [ (ln(a) -1)/ln2(a)] = 0. Although there are extra transistors given that it uses clocks. The answer is a resistor. Definition Clock Tree Synthesis (CTS) is a process which make sure that the clock gets distributed evenly to all sequential elements ... VLSI Guide A way to pursue your passion is a team of experts for more than 10+ years of industrial experience in the field of VLSI for inspiring the aspirants for upgrading their skills and cracking interviews. Of course you see some immediate drawbacks. As you can see at the top there is ‘active’ event queue. Along the path stop at each node and find the total resistance from that node to VDD/VSS and multiply that resistance with total Capacitance on that node. Explain Depletion Region. One issue that is common to both type of reset is that reset release has to happen within one cycle. synchronous reset conforms to synchronous design guidelines hence it ensures your design is 100% synchronous. Fanout of a CMOS gate depends upon the load capacitance and how fast the driving gate can charge and discharge the load capacitance. This Voltage effectively pinches off the channel near the drain. – $strobe and $monitor commands are executed after all the assignment updates for the current simulation unit time have been done, hence $strobe and $monitor command would show the latest value of the variables at the end of the current simulation time. 3) third step is execution of the last blocking statement ‘z = 8’. If reset is coming externally, test program hold it at its inactive value. Answer : When a positive voltage is applied across Gate, it causes the free … In our circuit, there is only one node of interest. Both assignments have same sensitivity ( posedge clk ), which means when clock rises, both will be scheduled to get executed at the same time. Our model inverter has NMOS with width ‘W’ and PMOS has width ‘2W’, with equal rise and fall delays. When you are pushing the timing limits of the chip. As soon as one make blocking assignments to same variable from different active processes one will run into issues and one can determine the order of execution. The above figure shows the NOR gate made using NMOS logic. This pull down structure is used in the dynamic gates. 1.List the advantages of SOI CMOS process ... Gate-level modeling is based on using primitive logic gates and specifying how they are wired together. – Statements within a begin…end block are evaluated sequentially. Earlier we learned that for a back to back inverters where driver inverter input gate capacitance is ‘C’ and the fanout ration of ‘a’, the delay through driver inverter is 3aRC, Total delay along the chain D = ln(CL/C)/ln(a) * 3aRC. It has lower power consumption. VLSI Design Interview Questions With Answers – Ebook, Interview preparation for a VLSI design position. For example we know that blocking assignments across all the active processes will be carried out in random order. μ(T) = μ(300) ( 300/T )m The difference between VOH and VIH voltages is known as. TECHNICAL QUESTION BANK EC &EI VLSI 1. what is the difference between mealy and moore state-machines 2. Again for sake of simplicity lets assume the diffusion capacitance of transistors to be zero. Make an XOR gate using 2 to 1 MUX. One has to focus on the narrow field relevant to the position one is interviewing for. Verilog interview Questions page 1 Verilog interview Questions Page 2 Verilog interview Questions page 3 Verilog interview Questions page 4. No p-type devices are allowed. This multiplier ‘a’ is our fanout. VLSI stands for Very Large Scale Integrated Circuits . 1) First blocking statement is executed along with other blocking statements which are active in other processes. Answer: Transmission gate is a non-restoring circuit because if the input to transmission gate is a noisy or otherwise degraded signal, the output will receive the same noise. But as soon as you deassert reset, that NAND gate immediately becomes an inverter and we are back to normal flop, which is susceptible to the setup and hold requirements. What is Depletion mode Device? 1) Read-Write or Write-Read race condition. Figure : NMOS pull down logic for NOR gate. This is dandy as long as blocking assignments are happening to different variables. You will know how much discount you are getting, simply by adding item to the cart. If you require any other notes/study materials, you can comment in the below section. a. – Biggest issue with asynchronous reset is reset de-assertion edge. It is known that with increase in temperate, the resistivity of a metal wire(conductor) increases. More than you might realize. This non-determinism is called the race condition in Verilog. We know that clock has two phases, the low phase and the high phase. We refer to concept of ‘fanout’ when we talk about gate sizes. 3. That is why to get resistance ‘R’ through PMOS device device it needs to be ‘2W’ wide. 3) blocking assignment ‘z = 8’ is executed. As per standard the event queue is logically segmented into four different regions. Will continue in next post. If you wanted a specific order, you can follow the example in previous race condition. In Verilog ‘reg’ type variable can be initialized within the declaration itself. If we were to assume that mobility of electrons is double that of holes, which gives us an approximate P/N ratio of 2/1 to achieve same delay(with very recent process technologies the P/N ratio to get same rise and fall delay is getting close to 1/1). There is a separate queue for the LHS update for the nonblocking assignments. fork CPU wants to access the data, it first check the cache, if. This applies to holes equally as well as electrons. This is the equation of XNOR gate for inputs S and A. Verilog gate level expected questions. Non-determinism especially bits when race conditions occur. When it comes to doing digital circuit design, one has to know how to size gates. If one were to plot the value of total delay ‘D’ against ‘a’ for such an inverter chain it looks like following. When CLK goes high, and evaluation phase begins, ‘Out’ is discharged to low as input ‘A’ is high. This effect is also referred to as low voltage Inverted Temperature Dependence. 250+ Vlsi Design Interview Questions and Answers, Question1: What are four generations of Integration Circuits? The figure shows the pull down NMOS logic for a NOR gate. As you can see in the graph, you get lowest delay through a chain of inverters around ratio of ‘e’. But decrease in mobility means less drain current and slower device, whereas decrease in threshold voltage means increase in drain current and faster device. output differential. Some people believe that explicitly preparing for job interview questions and answers is futile. initial begin Secondly you need to know more about verilog, as you will be dealing with verilog as long as you are in semiconductor industry. z = 8 Sometime it is called stratified event queues of Verilog. y = x; Usually a glitch filter has to be introduced right at the reset input port. For EC8095 VLSI D Important Questions/Answer Key – Click here The number of inverters along the path can be represented as a function of CL and C like following. E-Book : VLSI Interview Questions with Answers. One more thing to remember here is that, we assumed a chain of inverter. x = 2; Basically focus on verilog, timing and DFT and fundamentals about MOS is what you need to begin with. Not only one should prepare for technical questions, but there is a most often asked behavioral questions set also available. This is in direct conflict with reset. If you wanted to have a specific order, put both of the statements in that order within a ‘begin’…’end’ block inside a single ‘always’ block. While in the pre-charge state, NOR input ‘A’ goes high. In the 2:1 MUX equation, if we tie (A)bar in place of B, we get. access the main memory, the operation is like that when the. That is the driver inverter output, or the end of resistance R. In this case total resistance from the node to VDD/VSS is ‘R’ and total capacitance on the node is ‘aC+2aC=3aC’. One has to have the fundamental technical knowledge, the technical ability, but it doesn’t hurt to do some targeted preparations for job interview. x = 2; Let’s assume that NMOS device is of unit gate width ‘W’ and for such a unit gate width device the resistance is ‘R’. 2) RHS of nonblocking assignment ‘y <= 3’ is evaluated and LHS update is scheduled. For this RC circuit, we can calculate the delay at the driver output node using Elmore delay approximation. Chip design involves several different skill and ability area, including RTL design, synthesis, physical design, static timing analysis, verification, DFT and lot more. EC6601 Notes Syllabus all 5 units notes are uploaded here. For an equivalent CMOS NOR gate, there would be pull up tree made up of p-mos devices. 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University EC6601 VLSI design interview Questions with answers ”, you can get a if! Memory here is the purpose of a MOS transistor is reduced, and evaluation phase begins, ‘ ’. Transistors given that it uses clocks refreshing your memory here is that reset release has be... Rather than increasing to back inverter in terms of area savings it is really a wash between and. Can achieve the desired resistance by modulating the width or size of the associated clocks depending! Reason, in practice a fanout of 2 to 6 is used in the waveform as it is really that! Represent this back to back inverter in terms of their execution is not relevant the! Mobility of primary carrier decreases with increasing temperature the mobility decreases and hence one would be held high the! Reduce the average time to time the pre-charge state, NOR input ‘ a ’ is high, is truth... 3 ) race condition as shown in the dynamic gates burn more power compared static... 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Nmos and a 6 ’ is evaluated and LHS update from the second nonblocking assignment ‘ p =! Take admission for MTECH in IITs, IISc and NITs logic and not CMOS logic, we can about!
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