Digital logic: gates, K-maps & flip-flops worksheet with answers

Free printable Digital logic: gates, K-maps & flip-flops worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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Name: ____________________   Date: ____________

1.Easy
ABCF000?001?010?011?100?101?110?111?

The function F = A + B′ + C′ is 0 for exactly one combination of the inputs. Click that row of the truth table.

F = A + B′ + C′
  1. 1ABC = 000
  2. 2ABC = 001
  3. 3ABC = 010
  4. 4ABC = 011
  5. 5ABC = 100
  6. 6ABC = 101
  7. 7ABC = 110
  8. 8ABC = 111
2.Easy
BAF

NAND and NOR are universal gates: any logic function can be built from one of them alone. What is the minimum number N of 2-input NAND gates needed to build the OR function shown?

F = A + B
N =
3.Medium
ABCD000111100001111010111312041517161121131150141819011010

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 1, 2, 3, 5, 6, 7, 8, 9, 12, 13, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 1, 2, 3, 5, 6, 7, 8, 9, 12, 13, 15)
N =
4.Medium
ABCD000111100001111010011312140517161120130150140819111010

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 2, 3, 4, 6, 7, 9, 11, 12). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 2, 3, 4, 6, 7, 9, 11, 12)
N =
5.Hard
CLKJK12345678

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.

Q₀ = 0JK
  1. 1clock edge 1
  2. 2clock edge 2
  3. 3clock edge 3
  4. 4clock edge 4
  5. 5clock edge 5
  6. 6clock edge 6
  7. 7clock edge 7
  8. 8clock edge 8
6.Hard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​TQ4​Q4​TQ5​Q5​CLK

A 6-bit ripple (asynchronous) counter is built from flip-flops with a propagation delay of 20 ns each. All outputs must have settled before the next clock edge. Find the maximum clock frequency fmax.

n = 6tpd = 20 ns
fmax = MHz

Answer key

  1. 1. ABC = 011
  2. 2. N = 3
  3. 3. N = 3
  4. 4. N = 4
  5. 5. clock edge 6
  6. 6. fmax = 8.333 MHz

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