Let's break down the problem step by step: ### **Question** **How many 2-input NAND gates are needed to design a 4-bit synchronous up-counter?** --- ### **Step 1: Understanding the 4-bit Synchronous Up-Counter** - A 4-bit synchronous up-counter consists of 4 flip-flops (usually JK or T flip-flops). - For synchronous operation, all flip-flops are triggered by the same clock. - The J and K (or T) inputs of each flip-flop are driven by combinational logic. --- ### **Step 2: Flip-Flop Realization Using NAND Gates** If we assume we need to build the flip-flops (usually JK flip-flops) using only 2-input NAND gates: #### **JK Flip-Flop Implementation** - A typical JK flip-flop requires: - **4 NAND gates** for the basic SR latch. - **Additional 4 NAND gates** for the input gating circuitry. - **Total = 8 NAND gates per JK flip-flop** #### **For 4 Flip-Flops:** - \( 8 \text{ gates/flip-flop} \times 4 = 32 \) gates --- ### **Step 3: Additional Logic for Synchronous Up-Counter** - Additional combinational logic (AND gates) is needed to connect the J and K inputs for each stage appropriately. - Each AND gate can be made using NAND gates (using DeMorgan's laws). #### **Logic for J and K Inputs:** - 2nd FF: J and K connected to Q0 (needs 1 AND gate) - 3rd FF: J and K connected to Q0 AND Q1 (needs 2-input AND gate) - 4th FF: J and K connected to Q0 AND Q1 AND Q2 (needs 3-input AND gate) ##### **AND gate using NAND gates:** - 2-input AND = 2 NAND gates (NAND + NOT) - 3-input AND = 3 NAND gates (cascade) #### **Total for Logic:** - 1 AND (2 NAND) for 2nd FF - 1 AND (2 NAND) for 3rd FF - 1 AND (3 NAND) for 4th FF - **Total = 2 + 2 + 3 = 7 NAND gates** --- ### **Step 4: Total NAND Gates Calculation** - **Flip-flops:** 32 NAND gates - **Combinational logic:** 7 NAND gates **Total = 32 + 7 = 39** --- ## **Final Answer** \[ \boxed{39} \] **You need 39 two-input NAND gates to design a 4-bit synchronous up-counter (including gates needed to build the flip-flops and the required combinational logic).**

Let's analyze both solutions carefully to determine which one is correct. --- ### **Given Data and Assumptions** - **Objective:** Design a 4-bit synchronous up-counter using only 2-input NAND gates. - **Key points:** - The counter uses flip-flops (commonly JK or T flip-flops). - These flip-flops can be constructed from NAND gates. - Additional logic (gates) is needed for the toggle conditions. --- ### **Step-by-step Comparison of Both Solutions** --- ### **Solution 1 (Initial detailed reasoning):** 1. **Flip-Flops:** - Assumed **8 NAND gates per JK flip-flop**. - For 4 flip-flops: \(4 \times 8 = 32\) NAND gates. 2. **Additional logic for J and K inputs:** - To generate toggle signals based on Q outputs: - For Q1: AND of Q0 (1 NAND gate) - For Q2: AND of Q0 and Q1 (2 NAND gates) - For Q3: AND of Q0, Q1, Q2 (3 NAND gates) - Total: \(1 + 2 + 3 = 6\) NAND gates. 3. **Total NAND gates:** - \(32 + 6 = 38\) --- ### **Solution 2 (Your provided answer):** 1. **Flip-Flops:** - Assumed **8 NAND gates per JK flip-flop**. - For 4 flip-flops: \(4 \times 8 = 32\) NAND gates. 2. **Additional logic for toggle conditions:** - Logic for toggling: - FF1 toggles when Q0 = 1 (no AND gate needed, direct connection). - FF2 toggles when Q0 and Q1 are both 1 → AND(Q0, Q1), requiring **2 NAND gates**. - FF3 toggles when Q0, Q1, Q2 are all 1 → AND(Q0, Q1, Q2), which requires **3 NAND gates**. - Total: \(0 + 2 + 3 = 5\) NAND gates. 3. **Total:** - \(32 + 5 = 37\) NAND gates. --- ### **Key Discrepancies** - **Number of NAND gates for AND gates:** - The first solution correctly states: - 2-input AND gate: 2 NAND gates. - 3-input AND gate: 4 NAND gates (by cascading two 2-input AND gates, then NAND with inverted signals). - The second solution states: - 3-input AND requires 3 NAND gates, which is **incorrect**. - Correctly, a 3-input AND gate made from NAND gates requires **4 NAND gates**. --- ### **Conclusion:** - The **first solution** (39 gates) **correctly accounts for**: - The flip-flops (assuming 8 NAND gates each). - The AND gates needed for toggle logic, correctly calculating 2-input and 3-input AND gates. - The **second solution** underestimates the number of NAND gates needed for the 3-input AND gate (which is 4, not 3), resulting in a total of 37 NAND gates, which **may be slightly underestimated**. --- ### **Final Answer:** **The first solution's total of 39 NAND gates is correct,** considering the standard implementation of flip-flops and the logic for toggle control. --- ### **Summary:** | Aspect | First Solution | Second Solution | |---------|------------------|----------------| | Flip-flops (4 units) | 8 NAND gates each = 32 | 8 NAND gates each = 32 | | AND gates for toggle logic | Correctly calculated (2- and 3-input) | Slightly underestimated (3-input AND requires 4 NAND gates) | | Total NAND gates | **39** | **37** (likely underestimated) | --- ## **Final Verdict:** **The first detailed approach (39 NAND gates) is the correct and more accurate answer.**

Let's analyze the problem carefully. --- ### **Question Restatement:** **How many 2-input NAND gates are needed to design a 4-bit synchronous up-counter?** --- ### **Step 1: Understand the basic structure** - A **4-bit synchronous up-counter** typically uses 4 flip-flops (commonly JK or T flip-flops). - All flip-flops are triggered simultaneously by the same clock. - The toggle conditions for each flip-flop depend on the previous flip-flops’ outputs. --- ### **Step 2: Implementing flip-flops using NAND gates** **Assumption:** We are to **build JK flip-flops from only 2-input NAND gates**. **Standard JK flip-flop construction with NAND gates:** - A master-slave JK flip-flop usually requires **8 NAND gates**. **Total flip-flops:** - 4 flip-flops × 8 NANDs each = **32 NAND gates** --- ### **Step 3: Logic for toggle control (J and K inputs)** In a **synchronous up-counter**, the **J** and **K** inputs** are set to toggle the flip-flop** based on the previous outputs: - **FF0 (LSB):** toggles every clock pulse: \( J_0 = K_0 = 1 \) - **FF1:** toggles when Q0 = 1 → \( J_1 = K_1 = Q_0 \) - **FF2:** toggles when Q0=1 AND Q1=1 → \( J_2 = K_2 = Q_0 \cdot Q_1 \) - **FF3:** toggles when Q0=1 AND Q1=1 AND Q2=1 → \( J_3 = K_3 = Q_0 \cdot Q_1 \cdot Q_2 \) --- ### **Step 4: Implementing AND gates with NAND gates** - **2-input AND gate using NANDs:** \( A \cdot B = \text{NAND}(\text{NAND}(A,B), \text{NAND}(A,B)) \) - **Number of NAND gates for AND:** - 2-input AND: **2 NAND gates** - 3-input AND: cascade two 2-input NAND gates plus an inverter (another NAND). Total: **4 NAND gates** --- ### **Step 5: Count the gates needed for each toggle condition** - **Q0:** no logic needed (direct tie to logic high or low as per design). - **Q1:** directly connected to Q0, no additional NAND gates. - **Q2:** AND of Q0 and Q1 → **2 NAND gates** - **Q3:** AND of Q0, Q1, Q2 → **4 NAND gates** **Total extra gates for toggle logic:** \( 0 + 0 + 2 + 4 = 6 \) --- ### **Step 6: Total NAND gates** - **Flip-flops:** **32** NAND gates (4 flip-flops × 8 gates each) - **Logic for toggle control:** **6** NAND gates --- ### **Final total:** \[ \boxed{32 + 6 = \mathbf{38}} \] --- ### **### **Summary:** - **Number of NAND gates for flip-flops:** 32 - **Number of NAND gates for logic (ANDs):** 6 - **Total:** **38** --- ## **Final Answer:** **You need _38_ two-input NAND gates to design a 4-bit synchronous up-counter (assuming NAND-based JK flip-flops and necessary combinational logic).** --- ### **Note:** The differences in previous answers (like 37 or 39) arise from assumptions about the implementation details, such as whether the flip-flops are built with 8 NANDs, whether certain signals are tied directly, or slight variations in counting the AND gates. The above is a standard, well-justified count based on typical constructions.