Electronic Circuits. Laboratory 6 - Solution

Transcription

1 Institut für Integrierte Systeme Integrated Systems Laboratory Autumn Semester 2016 Electronic Circuits Prof. Dr. Qiuting Huang Laboratory 6 - Solution and Last Update: Task 1 Schmitt Trigger Task 1 Theory Assume that V sq = +V CC. Set up the condition for the input voltage V tr at which the Schmitt trigger output V sq will switch to V CC? Hint: calculate V + using superposition, i.e. first set V sq to zero and then V tr and add the results. (1) Integrator (2) Schmitt Trigger Figure 1: Schematic of the function generator

2 We assume that V + is positive and the output is V sq = +V CC. In that case the OpAmp behaves like a voltage source which provides a voltage of V CC at the output. We therefore have two voltage sources: the OpAmp voltage source which controls V sq and the input voltage source which determines V tr. To analyze the circuit we can therefore use the superposition principle as suggested in the hint. Superposition principle: V + Vsq=0 = R 1 + V tr therefore the total voltage V + is V + Vtr=0 = R 1 R 1 + V sq V + = V + Vsq=0 + V + Vtr=0 = R 1V sq + V tr R 1 +. The output voltage will switch to V CC when V + 0. Mathematically: V +! = 0 = V tr,1 = R 1 V sq = R 1 V CC At what input voltage V tr will the output switch from V sq = V CC to V sq = +V CC? V tr,1 = R 1 V CC Consider the hysteresis diagram in Fig. 2. What are the values of the four voltages V 1 to V 4? V 1 = V CC V 2 = R 1 V CC V 3 = V CC V 4 = R 1 V CC 2

3 Figure 2: Hysteresis loop of a Schmitt trigger Draw V + (t) and the output V sq (t) of the Schmitt trigger assuming /R 1 = 2 in Figure 14. Label the maximum amplitude and the zero crossings of all signals. Note: the easiest way to draw V + is to first construct V sq and then obtain V + from V + = R 1V sq + V tr R 1 + = 1 3 V sq V tr 3

4 Figure 3: Voltages for Task 1 Task 1 Simulation 4

5 Figure 4: MultiSim schematics Figure 5: Simulation of V tr and V +. 5

6 Figure 6: Simulation of V tr and V sq. Task 1 Hands On Figure 7: Measurement of V tr and V +. 6

7 Figure 8: Measurement of V tr and V sq. Task 2 Function Generator Task 2 Theory Assume that +V CC is applied to the input of the integrator V sq. Calculate the integrator output voltage V tr (t). Assume that the capacitor is initially uncharged. From Lab 3 we know that the output is given by V tr (t) = 1 RC V sq (t) dt = V CC RC t Using the results of Task 1 and the previous question, what is the period of the square wave at V sq? What is the frequency? The square wave V sq switches from V CC to +V CC when V tr = R 1 V CC and it stays at +V CC until V tr = R 1 V CC. This is illustrated in Fig. 9. As can be seen from Fig. 9 Figure 9: Working principle of the relaxation oscillator 7

8 V tr ( t p 2 )! = R 1 V CC where V tr (t) is given by 1 V tr(t) = R 1 V CC V CC RC t and therefore R 1 V CC V CC t p RC 2! = R 1 V CC = t p = 4 R 1 RC f = 1 t p = 1 4 R 1 RC Assume R 1 = 5 kω, = 10 kω, R = 100 kω. What C do you have to choose to achieve a frequency of 100 Hz? 1 C ideal = 4 R = 50 nf 1 Rf During the lab session you will have access the following capacitors: 100 pf, 33 nf, 47 nf, 10 uf. Which one is closest to your calculated value? Calculate f real give C real. C real = 47 nf f real = 1 t p = 1 4 R 1 RC = Hz Using the values from the previous question, draw V tr (t) and V sq (t) in Figure 10. What is the amplitude of the generated voltages V tr and V sq? Which resistor would you replace by a potentiometer to make the frequency tunable? Why? Change R, it is the only resistor which does not affect the amplitude of the triangular signal (it is desirable to control amplitude and frequency independently). 1 Note that the capacitor is not uncharged at t = 0. V tr(0) = R 1 V CC 8

9 Figure 10: Voltages for Task 2 Task 2 Simulation What is the frequency of the output signals? f sim = 104 Hz 9

10 Figure 11: MultiSim schematics Figure 12: Simulation of V tr and V sq. 10

11 Figure 13: Measurement of V tr and V sq. Task 2 Hands On Measure the spectrum of the V tr. How would you extend the function generator to also produce a sine wave in addition to the square and triangular waves? The spectrum (see Fig. 15) shows that the triangular wave could be converted to a sine wave by using a lowpass filter which filters out the harmonics 2. The first harmonic is at 3f, so 3f should be in the stopband of the filter while f is in the passband. As an example, Fig. 19 shows the result of applying a passive second order RC filter with f C = 150 Hz to V tr. 2 Note: instead of inspecting the spectrum we could have calculated the Fourier series to get the location of the harmonics. 11

12 Figure 14: Frequency difference between measurement and simulation stems from the resistor R. Figure 15: Spectrum of the triangular wave (simulation) 12

13 Figure 16: Spectrum of the filtered triangular wave (simulation) Figure 17: Sine wave from filtered triangular wave (simulation) 13

14 Figure 18: Sine wave from filtered triangular wave (measurement) Figure 19: Square wave and triangular wave with different frequency ( 1.8 khz) (measurement) 14