--- title: "SR equation; mod-5 counter using D flip-flops" subject: "Digital Logic" chapter: 8 key: "semester_3/digital_logic:sr-equation-mod5-d-counter" source: https://www.bctengineer.com/q/semester_3/digital_logic/sr-equation-mod5-d-counter site: BCT Engineering --- # SR equation; mod-5 counter using D flip-flops Digital Logic · Chapter 8: Counters - Asked in: 2081 Baisakh, 3+7 marks - Video: [Digital Electronics: Mod 5 counter using D Flip Flops only](https://www.youtube.com/watch?v=-XtLrskCdkc) — sacademy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/sr-equation-mod5-d-counter - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-8 ## Question Find out characteristic equation for SR flip-flop. Design a mod-5 synchronous counter using D flip-flops with negative edge triggering clock system. ## Short answer SR characteristic: Q+ = S + R'Q, with SR=0. For mod-5 synchronous counter, three negative-edge D flip-flops share the clock. D2 = Q1·Q0, D1 = Q1⊕Q0, D0 = Q2'·Q0'. Counts 000→001→010→011→100→000; unused states lead to valid count. ## Answer #### Characteristic equation of SR flip-flop The characteristic equation gives the next state $Q^+$ in terms of the inputs and the present state $Q$. | S | R | Q | Q+ | |---|---|---|---| | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 1 | | 0 | 1 | 0 | 0 | | 0 | 1 | 1 | 0 | | 1 | 0 | 0 | 1 | | 1 | 0 | 1 | 1 | | 1 | 1 | 0 | X (invalid) | | 1 | 1 | 1 | X (invalid) | | Q+: S / RQ | 00 | 01 | 11 | 10 | |---|---|---|---|---| | 0 | 0 | 1 | 0 | 0 | | 1 | 1 | 1 | X | X | Grouping the row S = 1 (with the X cells) gives $S$; grouping cells 01 of both rows gives $\overline{R}Q$: $$ Q^+ = S + \overline{R}\,Q, \qquad \text{with the condition } SR = 0 $$ #### Mod-5 synchronous counter using D flip-flops (negative-edge clock) **Step 1:** States 000 → 001 → 010 → 011 → 100 → 000. 3 D flip-flops ($Q_2Q_1Q_0$), all clocked by the falling edge. States 101, 110, 111 are don't cares. For a D flip-flop $D = Q^+$. **Step 2: Excitation table** | Present Q2Q1Q0 | Next Q2Q1Q0 | D2 | D1 | D0 | |---|---|---|---|---| | 000 | 001 | 0 | 0 | 1 | | 001 | 010 | 0 | 1 | 0 | | 010 | 011 | 0 | 1 | 1 | | 011 | 100 | 1 | 0 | 0 | | 100 | 000 | 0 | 0 | 0 | **Step 3: K-maps** | D0: Q2 / Q1Q0 | 00 | 01 | 11 | 10 | |---|---|---|---|---| | 0 | 1 | 0 | 0 | 1 | | 1 | 0 | X | X | X | | D1: Q2 / Q1Q0 | 00 | 01 | 11 | 10 | |---|---|---|---|---| | 0 | 0 | 1 | 0 | 1 | | 1 | 0 | X | X | X | | D2: Q2 / Q1Q0 | 00 | 01 | 11 | 10 | |---|---|---|---|---| | 0 | 0 | 0 | 1 | 0 | | 1 | 0 | X | X | X | **Step 4: Equations** $$ \begin{aligned} &D_2 = Q_1Q_0 \\ &D_1 = Q_1\overline{Q_0} + \overline{Q_1}Q_0 \\ &D_0 = \overline{Q_2}\overline{Q_0} \end{aligned} $$ so $D_1 = Q_1 \oplus Q_0$. **Step 5: Circuit diagram** ```text +-----+ +-----+ +-----+ D0>|D Q|-Q0 D1>|D Q|-Q1 D2>|D Q|-Q2 | FF0 | | FF1 | | FF2 | | | | | | | +--o--+ +--o--+ +--o--+ | | | CLK---+-------------+-------------+ D0 = Q2'.Q0' (NOR of Q2 and Q0) D1 = Q1 XOR Q0 D2 = Q1.Q0 ``` **Step 6: Unused states:** 101 → 010, 110 → 010, 111 → 100, so the counter is self-starting. **Timing diagram** ```text CLK __|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾ Q2 ________________|‾‾‾|___________ Q1 ________|‾‾‾‾‾‾‾|___________|‾‾‾ Q0 ____|‾‾‾|___|‾‾‾|_______|‾‾‾|___ Cnt 0 1 2 3 4 0 1 2 ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the characteristic equation of the SR flip-flop? A. Q+ = S + R'Q with SR = 0 B. Q+ = S'R + Q C. Q+ = SR + Q' D. Q+ = S'Q + R Answer: A. From the K-map, Q+ = S + R'Q, valid when the forbidden input S = R = 1 is excluded (SR = 0). (Key point: SR characteristic equation) 2. What is the excitation rule for a D flip-flop? A. D = Q XOR Q+ B. D = Q+ C. D = Q D. D = NOT Q+ Answer: B. A D flip-flop simply copies its input to the output at the clock edge, so D equals the desired next state. (Key point: D excitation) 3. Which set is the correct result for the mod-5 D-flip-flop counter? A. D2 = Q1 + Q0, D1 = Q1Q0, D0 = Q2 Q0 B. D2 = Q1Q0, D1 = Q1 XOR Q0, D0 = Q2'Q0' C. D2 = Q2Q1, D1 = Q1 XOR Q0, D0 = Q0' D. D2 = Q1Q0, D1 = Q1Q0', D0 = Q2'Q0 Answer: B. D2 = 1 only for 011, D1 toggles pattern 01/10 giving the XOR, and D0 = 1 for 000 and 010, giving Q2'Q0'. (Key point: Mod-5 D equations) 4. What happens to the unused state 111 in the mod-5 D counter? A. It goes to 000 B. It stays at 111 C. It goes to 100 D. It goes to 010 Answer: C. With D2 = Q1Q0 = 1, D1 = 1 XOR 1 = 0, D0 = 0, the next state is 100, a valid state. 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