Electronics Manual

Lab Manual Solutions Industrial Control Electronics: Devices, Systems, and Applications 3rd edition Terry L. M. Bartelt Australia • Canada • Mexico • Singapore • Spain • United Kingdom • United States Experiment 1 Operational Amplifiers Experiment Questions 1. analog 2. linear 3. greater 4. 6, – 5. -5V INPUTS V1 +4 +2 +1 +4 0 +3 V2 +1 +3 0 +4 +1 +2 VOUT (V) -5V +5V -5V 0V +5V -5V VIN +0. 2V –0. 4V 0V +0. 32V VOUT -1V +2V 0V -1. 6V VIN VOUT +0. 3V -0. 75V –0. 15V +0. 38V +5V –2. 0V -1V +0. 4V Figure 1-2 b Figure 1-3 b&c Input Voltage V1 V2 V3 +1V +1V +1V +1V –1V –1V +2V –1V –1V –3V –1V +3V +1V +2V –1VOutput Voltage Measured Calculated -3V -3V +1V +1V 0V 0V +1V -2V +1V -2V Figure 1-4b 1 Experiment 2 Schmitt Trigger Procedure Question Answer 1. No. Because the 7476 J-K flip-flop is negative edge triggered, and reacts only to positive-to-negative–going signals that change abruptly. The rectified sine wave does not change fast enough. Step 5 Point 1 . 9 Vth – = _____VDC 1. 7 Vth + = _____VDC Table 2-1 Step 7 Waveform At Point 1 At Point 2 Is the Flip-Flop Toggling (Yes, No) Circuit (a) NO Circuit (b) YES Table 2-2 Experiment Questions 1. Convert electronic signals to square waves. – Perform NAND gate and Inverter logic functions. 2. D. 3. edge 4. Low, High 5. hysteresis 6. Because when sine waves are counted, they must be converted to square waves before being applied to a flip-flop. 2 Experiment 3 Magnitude Comparator Procedure Question Answer 1. If the high order bits are equal, then the output state is determined by comparing the low order bits. Step 2A Input B B2 B1 0 0 1 0 0 0 0 0 1 0 Input A A2 A1 0 0 0 0 1 0 0 0 0 0 Outputs A=B 1 0 0 0 0 B3 0 0 1 0 0 B0 0 0 1 1 1 A3 0 0 1 0 1 A0 0 1 0 0 1AB 0 0 0 1 1 Table 3-2 Step 3B B3 0 0 0 1 0 Input B B2 B1 0 0 0 1 1 1 0 1 1 0 B0 0 1 0 0 1 A3 1 0 0 1 1 Input A A2 A1 1 1 0 1 1 1 0 1 0 1 A0 1 1 0 1 0 Expansion Inputs IAB 1 0 0 0 0 0 0 0 1 0 1 1 0 1 0 AB 0 1 0 1 0 0 1 0 0 1 Table 3-4 Experiment Questions 1. 1111 2. Yes. By connecting a Low to the MSB of inputs A and B, and applying the three binary bits to the remaining inputs. 3. IA > B = 0 IA = B = 0 IA < B = 1 4. 4 5. When A is greater than B, or B is greater than A, the circuit would operate normally. When A is equal to B, however, output A

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