VLSI (4361102) - Summer 2024 Solution

Question 1(a) [3 marks]

#

Draw the structure of FinFET and write its advantages.

Answer:

graph TD
    A[Source] --> B[Gate]
    B --> C[Drain]
    D[Fin Structure] --> E[Multiple Gates]
    F[Silicon Substrate] --> D

Table: FinFET Advantages

Mnemonic: “BCR - Better Control Reduces leakage”

Question 1(b) [4 marks]

#

Explain depletion and inversion of MOS structure under external bias

Answer:

Table: MOS Bias Conditions

Diagram:

• Depletion: Positive gate voltage creates electric field pushing holes away
• Inversion: Higher voltage attracts electrons forming conducting channel

Mnemonic: “DI - Depletion Inverts to conducting channel”

Question 1(c) [7 marks]

#

Explain n-channel MOSFET with the help of its Current-Voltage characteristics.

Answer:

Table: MOSFET Operating Regions

graph LR
    A[Gate] --> B[n-channel]
    C[Source] --> B
    B --> D[Drain]
    E[p-substrate] --> B

Key Equations:

• Linear: ID = μnCox(W/L)[(VGS-VTH)VDS - VDS²/2]

• Saturation: ID = (μnCox/2)(W/L)(VGS-VTH)²

• Structure: Gate controls channel between source and drain

• Operation: Gate voltage modulates channel conductivity

• Applications: Digital switching and analog amplification

Mnemonic: “CLS - Cut-off, Linear, Saturation regions”

Question 1(c OR) [7 marks]

#

Define scaling. Compare full voltage scaling with constant voltage scaling. Write the disadvantages of scaling.

Answer:

Definition: Scaling reduces device dimensions to increase density and performance.

Table: Scaling Comparison

Disadvantages:

• Short Channel Effects: Channel length modulation increases
• Hot Carrier Effects: High electric fields damage devices
• Quantum Effects: Tunneling currents increase significantly

Mnemonic: “SHQ - Short channel, Hot carriers, Quantum effects”

Question 2(a) [3 marks]

#

Draw two input NAND gate using CMOS.

Answer:

Table: NAND Truth Table

Mnemonic: “PP-SS: Parallel PMOS, Series NMOS”

Question 2(b) [4 marks]

#

Explain noise immunity and noise margin for nMOS inverter.

Answer:

Table: Noise Parameters

graph LR
    A[VIL] --> B[VIH]
    C[VOL] --> D[VOH]
    E[NML] --> F[NMH]

• VIL: Maximum low input voltage
• VIH: Minimum high input voltage
• Good noise immunity: Large noise margins prevent false switching

Mnemonic: “HILOL - High/Low Input/Output Levels”

Question 2(c) [7 marks]

#

Explain Voltage Transfer Characteristics (VTC) of CMOS inverter.

Answer:

Table: VTC Regions

graph TD
    A[VIN = 0] --> B[VOUT = VDD]
    C[VIN = VDD/2] --> D[VOUT = VDD/2]
    E[VIN = VDD] --> F[VOUT = 0]

Key Features:

• Sharp transition: Ideal switching behavior
• High gain: Large slope in transition region
• Rail-to-rail: Output swings full supply range

Mnemonic: “ASH - A-region, Sharp transition, High gain”

Question 2(a OR) [3 marks]

#

Implement NOR2 gate using depletion load nMOS.

Answer:

Table: NOR2 Truth Table

Mnemonic: “DPN - Depletion load, Parallel NMOS”

Question 2(b OR) [4 marks]

#

Differentiate between enhancement load inverter and Depletion load inverter.

Answer:

Table: Load Inverter Comparison

• Enhancement: Requires positive gate voltage for conduction
• Depletion: Conducts with zero gate voltage
• Performance: Depletion load provides better characteristics

Mnemonic: “EPDLH - Enhancement Positive, Depletion Lower power, Higher speed”

Question 2(c OR) [7 marks]

#

Explain Depletion load nMOS inverter with its VTC.

Answer:

Circuit Operation:

• Load transistor: Always conducting (VGS = 0, VT < 0)
• Driver transistor: Controlled by input voltage
• Output: Determined by voltage divider action

graph TD
    A[VIN Low] --> B[Driver OFF]
    B --> C[VOUT = VDD]
    D[VIN High] --> E[Driver ON]
    E --> F[VOUT ≈ 0V]

Table: Operating Points

VTC Characteristics:

• VOH: VDD (better than enhancement load)
• VOL: Lower due to depletion load characteristics
• Transition: Sharp switching between states

Mnemonic: “DLB - Depletion Load gives Better high output”

Question 3(a) [3 marks]

#

Implement EX-OR using Depletion load nMOS.

Answer:

Table: XOR Truth Table

Implementation: Y = A⊕B = A’B + AB'

Mnemonic: “XOR - eXclusive OR, different inputs give 1”

Question 3(b) [4 marks]

#

Explain design hierarchy with example.

Answer:

Table: Hierarchy Levels

graph TD
    A[System Level] --> B[Module Level]
    B --> C[Gate Level]
    C --> D[Transistor Level]
    E[CPU] --> F[ALU]
    F --> G[Adder]
    G --> H[MOSFET]

Benefits:

• Modularity: Independent design and testing
• Reusability: Common blocks used multiple times
• Maintainability: Easy debugging and modification

Mnemonic: “SMG-T: System, Module, Gate, Transistor levels”

Question 3(c) [7 marks]

#

Draw and explain Y chart design flow.

Answer:

graph TD
    A[Behavioral Domain] --> D[System Specification]
    B[Structural Domain] --> E[Architecture]
    C[Physical Domain] --> F[Floor Plan]
    D --> G[Algorithm]
    E --> H[Logic Design]
    F --> I[Layout]
    G --> J[RTL]
    H --> K[Gate Level]
    I --> L[Transistor Level]

Table: Y-Chart Domains

Design Flow:

• Top-down: Behavioral → Structural → Physical
• Bottom-up: Physical constraints influence upper levels
• Iterative: Multiple passes for optimization

Mnemonic: “BSP - Behavioral, Structural, Physical domains”

Question 3(a OR) [3 marks]

#

Implement NAND2 - SR latch using CMOS

Answer:

Table: SR Latch Operation

Mnemonic: “SR-HRI: Set, Reset, Hold, Invalid states”

Question 3(b OR) [4 marks]

#

Which method is used to transfer pattern or mask on the silicon wafer? Explain it with neat diagrams

Answer:

Method: Lithography - Pattern transfer using light exposure

graph TD
    A[UV Light Source] --> B[Mask with Pattern]
    B --> C[Photoresist on Wafer]
    C --> D[Exposed Pattern]
    D --> E[Developed Pattern]

Process Steps:

Applications: Creating gates, interconnects, contact holes

Mnemonic: “CED - Coating, Exposure, Development”

Question 3(c OR) [7 marks]

#

Which are the methods used to deposit metal in MOSFET fabrication? Explain deposition in detail with proper diagram.

Answer:

Table: Metal Deposition Methods

graph TD
    A[Target Material] --> B[Ion Bombardment]
    B --> C[Ejected Atoms]
    C --> D[Substrate Coating]
    E[Wafer] --> D

Sputtering Process:

• Ion bombardment: Argon ions hit target material
• Atom ejection: Target atoms knocked off
• Deposition: Atoms settle on wafer surface
• Control: Pressure and power determine rate

Advantages:

• Uniform thickness: Excellent step coverage
• Low temperature: Preserves device integrity
• Variety: Multiple materials possible

Mnemonic: “IBE-DC: Ion Bombardment Ejects atoms for Deposition Control”

Question 4(a) [3 marks]

#

Implement Z= ((A+B+C)·(D+E+F). G)’ with depletion nMOS load.

Answer:

Logic Implementation:

• First level: (A+B+C) and (D+E+F) OR functions
• Second level: AND with G
• Output: Inverted result due to nMOS structure

Mnemonic: “POI - Parallel OR, Inversion at output”

Question 4(b) [4 marks]

#

List and explain the design styles used in VERILOG.

Answer:

Table: Verilog Design Styles

Characteristics:

• Behavioral: Describes what circuit does
• Structural: Shows how components connect
• Mixed: Combines multiple styles for complex designs

Mnemonic: “BDSG - Behavioral, Dataflow, Structural, Gate-level”

Question 4(c) [7 marks]

#

Implement NAND2 SR latch using CMOS and also implement NOR2 SR latch using CMOS.

Answer:

NAND2 SR Latch:

module nand_sr_latch(
    input S, R,
    output Q, Q_bar
);
    nand(Q, S, Q_bar);
    nand(Q_bar, R, Q);
endmodule

NOR2 SR Latch:

module nor_sr_latch(
    input S, R,
    output Q, Q_bar
);
    nor(Q_bar, R, Q);
    nor(Q, S, Q_bar);
endmodule

Table: Latch Comparison

Key Differences:

• NAND: Set/Reset with low inputs
• NOR: Set/Reset with high inputs
• Feedback: Cross-coupled gates maintain state

Mnemonic: “NAND-Low, NOR-High active”

Question 4(a OR) [3 marks]

#

Implement Y= (ABC + DE + F)’ with depletion nMOS load.

Answer:

Implementation Logic:

• ABC: Series connection (AND function)
• DE: Series connection (AND function)
• F: Single transistor
• Result: Y = (ABC + DE + F)’ due to inversion

Mnemonic: “SSS-I: Series-Series-Single with Inversion”

Question 4(b OR) [4 marks]

#

Write Verilog Code to implement full adder.

Answer:

module full_adder(
    input a, b, cin,
    output sum, cout
);
    assign sum = a ^ b ^ cin;
    assign cout = (a & b) | (cin & (a ^ b));
endmodule

Table: Full Adder Truth Table

Logic Functions:

• Sum: Triple XOR operation
• Carry: Majority function of inputs

Mnemonic: “XOR-Sum, Majority-Carry”

Question 4(c OR) [7 marks]

#

Implement Y =(S1’S0’I0 + S1’S0 I1 + S1 S0’ I2 + S1 S2 I3) using depletion load

Answer:

Note: Assuming S2 in last term should be S0.

// 4:1 Multiplexer implementation
module mux_4to1(
    input [1:0] sel,  // S1, S0
    input [3:0] data, // I3, I2, I1, I0
    output Y
);
    assign Y = (sel == 2'b00) ? data[0] :
               (sel == 2'b01) ? data[1] :
               (sel == 2'b10) ? data[2] :
                                data[3];
endmodule

Table: Multiplexer Selection

Circuit Implementation:

• Decoder: S1, S0 generate select signals
• AND gates: Each input ANDed with corresponding select
• OR gate: Combines all AND outputs

Mnemonic: “DAO - Decoder, AND gates, OR combination”

Question 5(a) [3 marks]

#

Implement the logic function G = (PQR +U(S+T))’ using CMOS

Answer:

Implementation:

• pMOS: Parallel for OR, Series for AND (inverted logic)
• nMOS: Series for AND, Parallel for OR (normal logic)
• Result: De Morgan’s law applied automatically

Mnemonic: “PSSP - Parallel Series Series Parallel”

Question 5(b) [4 marks]

#

Implement 8×1 multiplexer using Verilog

Answer:

module mux_8to1(
    input [2:0] sel,     // 3-bit select
    input [7:0] data,    // 8 data inputs
    output reg Y         // Output
);
    always @(*) begin
        case(sel)
            3'b000: Y = data[0];
            3'b001: Y = data[1];
            3'b010: Y = data[2];
            3'b011: Y = data[3];
            3'b100: Y = data[4];
            3'b101: Y = data[5];
            3'b110: Y = data[6];
            3'b111: Y = data[7];
        endcase
    end
endmodule

Table: 8:1 MUX Selection

Mnemonic: “Case-Always: Use case statement in always block”

Question 5(c) [7 marks]

#

Implement 4 bit full adder using structural modeling style in Verilog.

Answer:

module full_adder_4bit(
    input [3:0] a, b,
    input cin,
    output [3:0] sum,
    output cout
);
    wire c1, c2, c3;
    
    full_adder fa0(.a(a[0]), .b(b[0]), .cin(cin), 
                   .sum(sum[0]), .cout(c1));
    full_adder fa1(.a(a[1]), .b(b[1]), .cin(c1), 
                   .sum(sum[1]), .cout(c2));
    full_adder fa2(.a(a[2]), .b(b[2]), .cin(c2), 
                   .sum(sum[2]), .cout(c3));
    full_adder fa3(.a(a[3]), .b(b[3]), .cin(c3), 
                   .sum(sum[3]), .cout(cout));
endmodule

module full_adder(
    input a, b, cin,
    output sum, cout
);
    assign sum = a ^ b ^ cin;
    assign cout = (a & b) | (cin & (a ^ b));
endmodule

Structural Features:

• Module instantiation: Four 1-bit full adders
• Carry chain: Connects carries between stages
• Hierarchical design: Reuses basic full adder module

Table: Ripple Carry Addition

Mnemonic: “RCC - Ripple Carry Chain connection”

Question 5(a OR) [3 marks]

#

Implement logic function Y = ((AF(D + E) )+ (B+ C))’ using CMOS.

Answer:

Logic Breakdown:

• Inner term: AF(D + E) = A AND F AND (D OR E)
• Outer term: (B + C) = B OR C
• Final: Y = (AF(D + E) + (B + C))'

CMOS Implementation:

• PMOS network: Implements complement of function
• NMOS network: Implements original function
• Result: Natural inversion provides Y

Mnemonic: “PNAI - PMOS Network Applies Inversion”

Question 5(b OR) [4 marks]

#

Implement 4 bit up counter using Verilog

Answer:

module counter_4bit_up(
    input clk, reset,
    output reg [3:0] count
);
    always @(posedge clk or posedge reset) begin
        if (reset)
            count <= 4'b0000;
        else
            count <= count + 1;
    end
endmodule

Table: Counter Sequence

Features:

• Synchronous reset: Reset on clock edge
• Auto rollover: 1111 → 0000
• 4-bit range: Counts 0 to 15

Mnemonic: “SRA - Synchronous Reset with Auto rollover”

Question 5(c OR) [7 marks]

#

Implement 3:8 decoder using behavioral modeling style in Verilog.

Answer:

module decoder_3to8(
    input [2:0] select,
    input enable,
    output reg [7:0] out
);
    always @(*) begin
        if (enable) begin
            case(select)
                3'b000: out = 8'b00000001;
                3'b001: out = 8'b00000010;
                3'b010: out = 8'b00000100;
                3'b011: out = 8'b00001000;
                3'b100: out = 8'b00010000;
                3'b101: out = 8'b00100000;
                3'b110: out = 8'b01000000;
                3'b111: out = 8'b10000000;
                default: out = 8'b00000000;
            endcase
        end else begin
            out = 8'b00000000;
        end
    end
endmodule

Table: 3:8 Decoder Truth Table

Key Features:

• Behavioral modeling: Uses always block and case statement
• Enable control: All outputs disabled when enable = 0
• One-hot output: Only one output active at a time
• 3-bit input: Selects one of 8 outputs

Applications:

• Memory addressing: Chip select generation
• Data routing: Channel selection
• Control logic: State machine outputs

Mnemonic: “BEOH - Behavioral Enable One-Hot decoder”