Electronics Manual

Topics: Square wave, Sine wave, Wave Pages: 3 (519 words) Published: August 22, 2013
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 –1V

Output 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 1

AB 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...