SPI controller implementation in Verilog - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does SPI stand for and what is its main purpose?

SPI stands for Serial Peripheral Interface. It is used to send data between microcontrollers and small peripherals like sensors, SD cards, and displays using a simple, fast, synchronous serial communication.

beginner

Name the four main signals used in SPI communication.

The four main SPI signals are:
1. SCLK (Serial Clock) - clock signal from master
2. MOSI (Master Out Slave In) - data from master to slave
3. MISO (Master In Slave Out) - data from slave to master
4. SS (Slave Select) - selects the slave device

intermediate

In a Verilog SPI controller, what is the role of the shift register?

The shift register holds the data to be sent or received. It shifts bits out on MOSI and shifts bits in from MISO synchronized with the clock, allowing serial data transfer one bit at a time.

intermediate

How does the SPI controller know when a full byte has been transmitted or received?

The SPI controller counts the number of clock pulses. When 8 clock cycles have passed (for 8-bit data), it knows a full byte is done and can signal completion or load new data.

beginner

What is the purpose of the Slave Select (SS) signal in SPI?

The SS signal activates the slave device. When SS is low, the slave listens and communicates. When SS is high, the slave ignores the SPI bus, allowing multiple devices to share the same bus.

Which SPI signal carries data from the master to the slave?

A MOSI

B MISO

C SCLK

D SS

MOSI stands for Master Out Slave In, so it carries data from the master to the slave.

In SPI, what does the clock signal (SCLK) do?

A Selects the slave device

B Synchronizes data transfer

C Sends data from slave to master

D Resets the controller

SCLK provides the timing to synchronize data bits sent and received.

How many bits are typically transferred in one SPI data frame?

A 8 bits

B 32 bits

C 16 bits

D 4 bits

Most SPI transfers use 8-bit frames, but this can vary.

What happens when the Slave Select (SS) line is high?

A Data is transmitted

B Slave device is active

C Slave device ignores SPI bus

D Clock stops

When SS is high, the slave device is not selected and ignores communication.

In a Verilog SPI controller, what component shifts data bits in and out?

A Counter

B Multiplexer

C Decoder

D Shift register

The shift register moves data bits serially for transmission and reception.

Explain how a basic SPI controller works in Verilog, including the role of clock, shift register, and slave select.

Think about how data moves bit by bit with the clock and how the slave is chosen.

Describe the four main SPI signals and their functions in a simple SPI communication setup.

Consider who sends and receives data and how devices know when to listen.

Practice

1. What is the main role of the cs (chip select) signal in an SPI controller?

easy

A. To enable communication with the specific SPI device

B. To provide the clock signal for data transfer

C. To send data from master to slave

D. To receive data from the slave device

Solution

1. Step 1: Understand the function of cs in SPI

   The cs signal is used to select which SPI device the master wants to communicate with by enabling it.

2. Step 2: Differentiate cs from other signals

   The clock signal is sclk, data sent from master is mosi, and data received is miso. So cs is not for clock or data transfer but device selection.

3. Final Answer:

   To enable communication with the specific SPI device -> Option A

4. Quick Check:

   cs = device enable [OK]

Hint: Remember: cs selects device, sclk clocks data [OK]

Common Mistakes:

• Confusing cs with clock signal
• Thinking cs sends or receives data
• Mixing up mosi and miso signals

2. Which of the following is the correct Verilog syntax to declare a 8-bit register named shift_reg?

easy

A. reg shift_reg:8;

B. reg shift_reg[7:0];

C. wire [7:0] shift_reg;

D. reg [7:0] shift_reg;

Solution

1. Step 1: Recall Verilog register declaration syntax

   Registers are declared with reg [width-1:0] name; format. For 8 bits, it is [7:0].

2. Step 2: Check each option

   reg [7:0] shift_reg; uses correct syntax. reg shift_reg[7:0]; reverses the range and name. wire [7:0] shift_reg; uses wire, not reg. reg shift_reg:8; uses invalid colon syntax.

3. Final Answer:

   reg [7:0] shift_reg; -> Option D

4. Quick Check:

   reg [7:0] name; = correct reg declaration [OK]

Hint: Use reg [7:0] name; for 8-bit registers [OK]

Common Mistakes:

• Using wire instead of reg for storage
• Incorrect bit range syntax
• Misplacing brackets around name

3. Given this snippet of an SPI controller shift register update inside a clocked always block:

always @(posedge sclk) begin
  if (cs == 0) begin
    shift_reg <= {shift_reg[6:0], miso};
  end
end

What does this code do on each rising edge of sclk when cs is low?

medium

A. Loads miso into the MSB of shift_reg without shifting

B. Shifts shift_reg right by 1 and inserts miso at MSB

C. Shifts shift_reg left by 1 and inserts miso at LSB

D. Clears shift_reg to zero

Solution

1. Step 1: Analyze the shift operation

   The code concatenates shift_reg[6:0] (bits 6 down to 0) with miso at the right side, effectively shifting left by 1 bit and inserting miso at the least significant bit (LSB).

2. Step 2: Confirm behavior when cs is low

   The shift happens only when cs == 0, which means the device is selected and data is being received bit by bit.

3. Final Answer:

   Shifts shift_reg left by 1 and inserts miso at LSB -> Option C

4. Quick Check:

   shift_reg <= {shift_reg[6:0], miso} = left shift + miso LSB [OK]

Hint: Concatenation {a,b} puts b at LSB, shifting left [OK]

Common Mistakes:

• Thinking it shifts right instead of left
• Confusing MSB and LSB positions
• Ignoring the condition on cs signal

4. Consider this Verilog snippet for SPI clock generation:

always @(posedge clk) begin
  if (cs == 0) begin
    sclk <= ~sclk;
  end else begin
    sclk <= 0;
  end
end

What is the main problem with this code?

medium

A. It toggles sclk on every clk edge, causing sclk frequency to be half of clk only when cs is low

B. It does not reset sclk properly when cs is high

C. It causes sclk to toggle asynchronously to clk

D. It will cause a synthesis error due to missing sensitivity list

Solution

1. Step 1: Understand clock toggling logic

   The code toggles sclk on every rising edge of clk if cs is low, effectively dividing the clock frequency by 2 during active communication.

2. Step 2: Identify the problem

   This behavior is expected for SPI clock generation. The code also resets sclk to 0 when cs is high, which is correct. There is no asynchronous toggling or sensitivity list issue.

3. Final Answer:

   It toggles sclk on every clk edge, causing sclk frequency to be half of clk only when cs is low -> Option A

4. Quick Check:

   sclk toggles at half clk frequency when cs=0 [OK]

Hint: Toggling sclk halves clk frequency during cs low [OK]

Common Mistakes:

• Thinking sclk toggles asynchronously
• Expecting sclk to run continuously
• Confusing sensitivity list with always block usage

5. You want to implement an SPI controller that sends 8 bits from a register tx_data over mosi and receives 8 bits into rx_data from miso. Which approach correctly handles the bit counting and shifting in Verilog?

hard

A. Use an 8-bit counter that decrements on each sclk falling edge when cs is high, shift tx_data right to output MSB on mosi, and shift in miso into rx_data.

B. Use a 3-bit counter that increments on each sclk rising edge when cs is low, shift tx_data left to output MSB on mosi, and shift in miso into rx_data.

C. Use a 4-bit counter that increments on each clk rising edge regardless of cs, shift tx_data right to output LSB on mosi, and ignore miso.

D. Use a 3-bit counter that resets only when cs is high, shift tx_data left to output LSB on mosi, and shift in miso into rx_data.

Solution

1. Step 1: Determine counter size and control

   For 8 bits, a 3-bit counter (0 to 7) is enough. It should increment on sclk rising edge only when cs is low (active communication).

2. Step 2: Understand shifting direction and data lines

   SPI usually sends MSB first, so shifting tx_data left outputs MSB on mosi. Incoming bits from miso are shifted into rx_data.

3. Step 3: Evaluate options

   Use a 3-bit counter that increments on each sclk rising edge when cs is low, shift tx_data left to output MSB on mosi, and shift in miso into rx_data. matches correct counter size, control, and shifting direction. Use a 4-bit counter that increments on each clk rising edge regardless of cs, shift tx_data right to output LSB on mosi, and ignore miso. uses wrong counter size and ignores miso. Use an 8-bit counter that decrements on each sclk falling edge when cs is high, shift tx_data right to output MSB on mosi, and shift in miso into rx_data. uses wrong edge and counter behavior. Use a 3-bit counter that resets only when cs is high, shift tx_data left to output LSB on mosi, and shift in miso into rx_data. incorrectly outputs LSB first and counter reset logic.

4. Final Answer:

   Use a 3-bit counter that increments on each sclk rising edge when cs is low, shift tx_data left to output MSB on mosi, and shift in miso into rx_data. -> Option B

5. Quick Check:

   3-bit counter + sclk rising + MSB first shift = correct SPI send/receive [OK]

Hint: Count 8 bits with 3-bit counter, shift MSB first [OK]

Common Mistakes:

• Using wrong counter size or clock edge
• Shifting in wrong direction for MSB first
• Ignoring chip select signal during counting