光电信息学院《模拟电路与脉冲电路》综合训练题.docVIP

2011年光电信息学院模拟电路与脉冲电路综合训练题组长： 成绩： 组员: 组员: 组员: 组员: 组员: 组员: （请选作4题及以上）Project 1Power SuppliesObjective: This project will show some of the basic principles of power supplies using fullwave rectifier, Zener diode, and fixed-voltage regulator circuits.Components: Bridge Rectifier (50 PIV, 1 A), Zener diode (10 V at 500 mW), 7805 regulatorIntroduction:Most of the direct current (DC) power used in electronic devices is derived by converting 60 Hz, 115 V alternating current (AC) power to direct current power. This AC to DC conversion usually involves a step-down transformer, rectifier, filter, and a regulator. The step-down transformer is used to decrease the AC line voltage from 115 VRMS to an RMS value near the DC voltage needed. The output of the step-down transformer is then fed into a diode rectifier circuit that only outputs positive halves of the input sinusoid. A filter is then used to smooth the rectifier output to achieve a nearly constant DC voltage level. A regulator can be added after the filter to ensure a constant output voltage in spite of changes in load current and input voltages.Two different types of voltage regulators will be used in this project. The first involves a Zener diode circuit and the second involves a voltage regulator circuit. A Zener diode can be used as a voltage regulator when the diode is reverse biased and operated in the breakdown region. To maintain voltage regulation, the Zener diode must be operated in the breakdown region at a current greater than the knee current (IZK). For currents greater than IZK, the Zener diode characteristic curve is nearly vertical and the voltage across the diode changes very little. Of course there is a maximum current the diode can tolerate, so good regulation is provided when the diode is reverse biased with currents between IZK and IZMAX. Zener diodes are available with a wide variety of breakdown voltages. Another type of voltage regulator is available with the 7800 series regulators. This series of fixed-voltage regulators is numbered 78xx, where xx corresponds to the value of the output voltage. Output voltages from 5 to 24 volts are available. These regulators are easy to use and work very well.Design:1. Find approximations for the DC voltage level and AC peak to peak ripple voltage for the bridge rectifier and filter circuit of Figure 1-1.2. For the Zener diode regulator circuit of Figure 1-2 assume that the Zener diode will regulate at 10 V over a current range of 5 mA to 25 mA. Assuming that the current flowing through R is always between 5 mA and 25 mA and the Zener diode is regulating at 10 V, find the minimum values of R and RL required. You may assume the forward diode drop for the two diodes is 1 V.Lab Procedure:1. Construct the bridge rectifier circuit of Figure 1-1 without the capacitor. Use the Variac with the step-down transformer for the input voltage to the bridge rectifier. With the transformer plugged into the Variac, adjust the Variac until the secondary voltage from the transformer equals 12 VRMS. BE CAREFUL not to short the secondary terminals! Observe the secondary waveform on the oscilloscope. Put the oscilloscope on DC coupling and observe the load voltage waveform VL. Remember that both the input source and the load cannot share a common ground terminal.2. Remove power from the circuit. Insert the capacitor as shown in Figure 1-1 being sure to observe the correct polarity. Energize the circuit. With the oscilloscope on DC coupling observe VL. Measure the DC voltage level using the digital voltmeter. With the oscilloscope on AC coupling observe the ripple voltage VR. Compare these measured values with the calculated values.3. Observe the effect of loading on the circuit by changing the load resistor from 1 kW to 500 W. Measure the DC voltage level with the digital voltmeter. Observe the ripple voltage with the oscilloscope set on AC coupling. Compare these values with the previously recorded values.4. Record the Zener diode characteristic curve from the digital curve tracer. Note the value of the breakdown voltage in the breakdown region. Also note the value of the knee current IZK.5. After verifying your designed values for R and RL with the instructor, construct the Zener diode regulator circuit of Figure 1-2. Measure the DC voltage level with the digital voltmeter for the minimum value of RL along with several values above and below the minimum value. Be careful not to overload the Zener diode. Comment on the circuits operation for these different load resistances.6. Construct the 7805 regulator circuit of Figure 1-3 being careful to observe the correct pin configuration of the regulator. Measure the load voltage for RL equal to 300 W, 200 W, and 100 W. Calculate the current for each of these cases. Does the value of the load resistor affect the output voltage?7. Using RL equal to 200 W, record the 7805 regulator input voltage (pin 1) and output voltage (pin 3). Decrease the regulator input voltage by decreasing the setting of the Variac. For each decrease in amplitude, record the regulator input and output voltages. Continue decreasing the amplitude until the output of the regulator drops a measurable amount below 5 V. What is the minimum input voltage needed for the 7805 regulator to produce a 5 V output?Questions:1. Why cant the input source and load have a common ground in the bridge rectifier circuit?2. Can the Zener diode be used as a conventional diode? Explain your answer and verify with a curve from the curve tracer.3. Would the value of the output filter capacitor have to increase, decrease, or remain the same to maintain the same ripple voltage if the bridge rectifier were changed to a half-wave rectifier? Explain your answer.4. How would increasing the frequency of the input source affect the ripple voltage assuming all components remained the same?Project 2Analog Applications of the Operational AmplifierObjective: This project will demonstrate some of the analog applications of an operational amplifier through a summing circuit and a bandpass filter circuit.Components: 741 op-ampIntroduction:Figure 2-1 shows a weighted summer circuit in the inverting configuration. This circuit can be used to sum individual input signals with a variable gain for each signal. The virtual ground at the inverting input terminal of the op-amp keeps the input signals isolated from each other. This isolation makes it possible for each input to be summed with a different gain.The bandpass filter shown in Figure 2-2 uses an op-amp in combination with resistors and capacitors. Since the op-amp can increase the gain of the filter, the filter is classified as an active filter. This bandpass filter circuit is extremely useful because the center frequency can be changed by varying a resistor instead of changing the values of the capacitors. The center frequency is given by:The center frequency can be changed by varying the variable resistor R3. Increasing R3 decreases the center frequency while decreasing R3 increases the center frequency. The bandwidth is given by:Notice that the bandwidth is independent of the variable resistor R3 so the center frequency may be varied without changing the value of the bandwidth. The gain at the center frequency of the bandpass filter is given by:Design:1. Find the relationship between the output and inputs for the weighted summer circuit of Figure 2-1.2. Design a bandpass filter with a center frequency of 2.0 kHz and a bandwidth of 200 Hz. Let the voltage gain at the center frequency be 20. Check your design with PSPICE. Use 15 V supplies for the op-amp. Use RL = 2.4 kW.Figure 2 - 1: Weighted SummerFigure 2 - 2: Bandpass FilterLab Procedure:1. Construct the summing amplifier of Figure 2-1. Design for the transfer function to be VO = -2 VIN1 - VIN2. Use 15 V supplies for the op-amp. Use RL = 2.4 kW.2. Let VIN1 be a 1 V peak sine wave at 1 kHz and VIN2 equal to 5 V DC. Verify the amplifiers operation by monitoring the output waveform on the oscilloscope.3. Construct the bandpass filter of Figure 2-2. Use the designed values for the resistors and capacitors. Use 15 V supplies for the op-amp. Use RL = 2.4 kW.4. Record and plot the frequency response (you may want to use computer control for the sweep and data collection). Find the center frequency, corner frequencies, bandwidth, and center frequency voltage gain to verify that the specifications have been met.5. Change R3 to lower the center frequency from 2.0 kHz to 1.0 kHz. Repeat part 4 for the new frequency response. Verify that the new center frequency is 1.0 kHz. What is the new bandwidth? What is the new center frequency voltage gain? Compare with the measurements of Procedure 4.Questions:1. Could the summer circuit be used with the inputs connected to the noninverting terminal and produce the same affect without the inversion? Explain.2. What is/are the benefit(s) of using an op-amp circuit to produce a bandpass filter over using an RLC circuit with a noninverting op-amp at the output of the RLC circuit?Project 3Analog Computer Applications using the Operational AmplifierObjective: This project will focus on the use of the operational amplifier in performing the mathematical operations of integration and differentiation. The design of a simple circuit (analog computer) to solve a differential equation will also be included.Components: 741 op-ampIntroduction:Figures 3-1 and 3-2 illustrate two op-amp based circuits designed to perform differentiation and integration respectively. The operations are performed real-time and can be helpful in observing both initial transients and steady state response. The analysis of the circuits is based on the ideal op-amp assumptions and performed in the time domain. The resistor RI shown in the two circuits is included to help with stability and for general circuit protection. The value for RI is nominally set equal to the feedback resistor (Figure 3-1) or the input resistor (Figure 3-2). The purpose of the optional resistor is left for student investigation in conjunction with the summary questions.The differentiator and integrator circuits may be combined with standard inverting and non-inverting op-amp circuits to provide the building blocks for analog computers. The resultant analog computer circuits are designed to solve differential and/or integral/differential equations in a real time environment. The ability to easily include, and change, initial conditions and forcing functions are additional benefits of the analog computers. Figure 3-3 illustrates a circuit designed to solve the second order differential equation KY - Y = 0 with the initial condition Y(0) = - VX and K = R1R2C1C2. The initial condition is set by using the momentary contact switch to force the output to equal the applied voltage at t = 0 (the time the switch is closed). While the major advances in digital computers and digital signal processing have reduced the use of these three circuits, they are still a fast and relatively inexpensive method for process control and stability/operation analysis for systems that can be represented in terms of differential equations.Design:1. Derive the expressions relating the input and output signals for the circuits shown in Figures 3-1 and 3-2.2. Design an analog computer to solve with y(0) = 2. Solve the differential equation when f(t) = 0 and verify your results using PSpice.Figure 3 - 1: DifferentiatorFigure 3 - 2: IntegratorFigure 3 - 3: Analog Computer (linear, second order, homogenous differential equation)Lab Procedure:1. Construct the circuit shown in Figure 3-1. Use 15 V supplies for the op-amp and a load resistance of 2.4 kW.2. Verify the operation of the circuit using a 500 mV peak, 50 Hz sinewave as the input signal. Be sure to design the gain such that the output does not saturate.3. Repeat step 2 with a sinewave frequency of 500 Hz. Does the circuit still operate correctly? What changes need to be made to prevent output saturation?4. Repeat steps 2 and 3 using a triangle wave and then using a square wave with the same magnitudes and frequencies as used in steps 2 and 3. 5. Construct the circuit shown in Figure 3-2. Again, use 15 V supplies for the op-amp and a load resistance of 2.4 kW.6. Repeat steps 2 through 4 for this circuit. Be sure to adjust your gain as necessary to maintain an output signal within the saturation limits of 12 V.7. Construct the circuit designed to solve the differential equation in part 2 of the design section. Verify the operation of the design using three different input waves (sine, triangle, and square). Determine the operation for at least three different frequencies - 10 Hz, 1 kHz, and 100 kHz. Explain any differences in operation of the circuit. What affect does the initial condition have on the result?Questions:1. How could you use the differentiator to obtain an estimate of the slew rate for the op-amp?2. Why should you include a resistor in parallel with the capacitor in the integrator?3. What is the purpose of the resistor in series with the input capacitor in the differentiator?4. Is it possible to design a circuit to perform the differentiation and integration functions using the non-inverting input? Explain your answer.Project 4Common Emitter Amplifier(designed for two lab periods)Objective: This project will show how the h-parameters for a BJT can be measured and used in an equivalent circuit model for the BJT. A CE small signal amplifier will be biased and designed to specifications along with both low and high frequency response and adjustment. Series-series feedback will also be used to control the bandwidth and input impedance of the CE amplifier.Components: 2N2222 BJTIntroduction:In order for circuits involving transistors to be analyzed, the terminal behavior of the transistor must be characterized by a model. Two of the models often used for a BJT are the hybrid-p and the h-parameter models. The complete hybrid-p circuit model for the BJT is shown in Figure 9-1. This model includes the internal capacitances and output resistance of the BJT. Inclusion of the internal transistor capacitances makes the hybrid-p model valid throughout the entire frequency range of the transistor. Typical data sheet values of Cp and Cm are 13 pF and 8 pF respectively. These values are so small that Cp and Cm may be considered open circuits for midband frequencies. The resistance rx typically has a value in the tens of ohms and can be considered a short circuit while rm and ro are usually extremely large in value and can be considered open circuits. The h-parameter small signal model for the BJT is characterized by the four h-parameters and is shown in Figure 9-2. Unlike the hybrid-p model, the h-parameter model does not ordinarily include frequency related effects and components and is therefore generally valid only at midband frequencies and below . However the h-parameter model is very useful since the h-parameters can be easily measured for a BJT. The value of hre is usually on the order of 10-4 and can be considered a short circuit. The value of hoe is usually on the order of 10-5 S making 1/hoe effectively an open circuit for most circuit configurations and biases. Making the same assumptions, the hybrid-p and h-parameter models are equivalent at midband frequencies.For a transistor to operate as an amplifier, it must have a stable bias in the active region. To bias a transistor, a constant DC current must be established in the collector and emitter. This current should be as insensitive as possible to variations in temperature and b (or hfe). The voltage across the base-emitter junction decreases about 2 mV for each 1 C rise in temperature, therefore it is important to stabilize VBE to ensure that the transistor does not overheat. The circuit shown in Figure 9-3 is the biasing scheme most often used for discrete transistor circuits. For this circuit, the base is supplied with a fraction of the supply voltage VCC through the voltage divider RB1, RB2. For ease of circuit analysis, the Thevenin equivalent circuit shown in Figure 9-4 can replace the voltage divider network. To ensure that the emitter current is insensitive to variations in b and VBE, VBB should be much greater than VBE and RBB should be much less than bRE. RBB is usually 20-30% of the product bRE. The voltage across RE is also usually 2-3 volts for good b stabilization. This same biasing scheme can be used for all three of the BJT amplifier configurations (CB, CC, CE).The BJT CE amplifier is shown in Figure 9-5. The signal source and resistive load are capacitively coupled to the amplifier. The coupling capacitors C1 and C2, emitter bypass capacitor CE, and internal transistor capacitances shape the frequency response of the amplifier. A typical amplifier frequency response curve is shown in Figure 9-6. The low half power corner frequency FL is controlled by the input and output coupling capacitors and the emitter bypass capacitor. The high half power corner frequency FH is controlled by the internal transistor capacitances and any separate load capacitor. The bandwidth is the difference between the high and low corner frequencies (FH - FL). As the signal frequency drops below midband, the impedance of the coupling capacitors C1 and C2 and emitter bypass capacitor CE increases. The coupling capacitors drop more signal voltage and the emitter bypass capacitor begins to open up and causes increased series-series feedback resulting in a reduction of the gain. One method of relating C1,