A D flip-flop stores a bit of data on a clock edge. The synchronous reset lets you clear the stored data at the same time as the clock, making control easier.
D flip-flop with synchronous reset in Verilog
module d_flip_flop_sync_reset(
input wire clk,
input wire reset,
input wire d,
output reg q
);
always @(posedge clk) begin
if (reset) begin
q <= 1'b0;
end else begin
q <= d;
end
end
endmoduleThe reset is checked inside the clock edge block, so it happens only on the clock's rising edge.
The output q is declared as reg because it holds state.
always @(posedge clk) begin
if (reset) q <= 1'b0;
else q <= d;
endmodule dff_sync_reset(
input clk, reset, d,
output reg q
);
always @(posedge clk) begin
if (reset) q <= 1'b0;
else q <= d;
end
endmoduleThis testbench toggles the clock every 5 time units. It applies reset and data inputs to the D flip-flop with synchronous reset. The monitor prints the signals at every change.
module testbench;
reg clk = 0;
reg reset;
reg d;
wire q;
d_flip_flop_sync_reset uut(
.clk(clk),
.reset(reset),
.d(d),
.q(q)
);
always #5 clk = ~clk; // Clock toggles every 5 time units
initial begin
reset = 1;
d = 0;
#10;
reset = 0;
d = 1;
#10;
d = 0;
#10;
reset = 1;
#10;
reset = 0;
d = 1;
#10;
$finish;
end
initial begin
$monitor($time, " clk=%b reset=%b d=%b q=%b", clk, reset, d, q);
end
endmoduleSynchronous reset means reset only works on the clock edge, unlike asynchronous reset which works anytime.
Make sure the reset signal is stable before the clock edge to avoid unexpected behavior.
D flip-flop stores data on clock edges.
Synchronous reset clears data only on clock edges.
This helps control timing and predict circuit behavior.
Practice
D flip-flop with synchronous reset do when the reset signal is active?Solution
Step 1: Understand synchronous reset behavior
A synchronous reset affects the flip-flop only at the clock edge, not immediately.Step 2: Compare reset types
Asynchronous reset clears data immediately, but synchronous reset waits for clock edge.Final Answer:
It clears the stored data only on the clock edge. -> Option DQuick Check:
Synchronous reset = clear on clock edge [OK]
- Confusing synchronous with asynchronous reset
- Thinking reset clears data immediately
- Assuming reset stores input data
Solution
Step 1: Identify synchronous reset sensitivity list
Synchronous reset triggers only on clock edge, so onlyposedge clkis in sensitivity list.Step 2: Check reset condition inside always block
Reset is checked inside the block, not in sensitivity list, matching synchronous reset behavior.Final Answer:
always @(posedge clk) begin if (reset) q <= 0; else q <= d; end -> Option CQuick Check:
Synchronous reset inside posedge clk block [OK]
- Including reset in sensitivity list (asynchronous reset)
- Using reset in sensitivity list but coding synchronous reset
- Using negedge clk incorrectly
q after the second rising edge of clk if reset is 1 at the first clock and 0 at the second?reg q; always @(posedge clk) begin if (reset) q <= 0; else q <= d; end
Solution
Step 1: Analyze first clock edge with reset=1
At first posedge clk, reset=1, so q is set to 0.Step 2: Analyze second clock edge with reset=0
At second posedge clk, reset=0, so q takes value of d input.Final Answer:
d (input value) -> Option BQuick Check:
Reset active clears q, then q follows d [OK]
- Assuming reset clears q immediately
- Thinking q stays 0 after reset deactivates
- Confusing asynchronous reset behavior
always @(posedge clk or posedge reset) begin if (reset) q <= 0; else q <= d; end
Solution
Step 1: Check sensitivity list for synchronous reset
Synchronous reset triggers only on clock edge, so reset must NOT be in sensitivity list.Step 2: Confirm correct sensitivity list
Onlyposedge clkshould be in sensitivity list for synchronous reset.Final Answer:
Reset should not be in the sensitivity list for synchronous reset. -> Option AQuick Check:
Synchronous reset excludes reset from sensitivity list [OK]
- Including reset in sensitivity list for synchronous reset
- Using blocking assignment instead of non-blocking
- Checking reset condition incorrectly
q to 0 on reset and otherwise stores input d. Which Verilog code correctly implements this with minimal logic and proper timing?Solution
Step 1: Confirm synchronous reset sensitivity list
For synchronous reset, onlyposedge clkis in sensitivity list.Step 2: Check reset logic inside always block
Reset condition sets q to 0 when active; else q follows d input.Step 3: Verify minimal and correct logic
always @(posedge clk) begin if (reset) q <= 0; else q <= d; end matches synchronous reset behavior with minimal logic and correct timing.Final Answer:
always @(posedge clk) begin if (reset) q <= 0; else q <= d; end -> Option AQuick Check:
Synchronous reset inside posedge clk block with correct logic [OK]
- Including reset in sensitivity list (asynchronous reset)
- Using negedge clk incorrectly
- Inverting reset logic
Ready to prove it?
Pass the test and this pattern moves unknown
