BANGLADESH UNIVERSITY OF ENGINEERING & TECHNOLOGY

Transcription

1 BANGLADESH UNIVERSITY OF ENGINEERING & TECHNOLOGY Electronics Circuits II Laboratory (EEE 208) Simulation Experiment No. 02 Study of the Characteristics and Application of Operational Amplifier (Part B) Introduction: Operational amplifier is one fundamental building block of analog circuits. When used properly in negative feedback configurations, the overall closed loop transfer characteristic can be precisely set by stable passive components such as resistors, capacitors, and diodes, regardless of the potential variation of open loop parameters. Negative feedback amplifier with op amp operating at its core provides key to highly reliable and stable analog functions. In this experiment, we will be simulating and building some basic op amp circuits, including the most common types, i.e., the Inverting and non inverting multiplier, differential and summing amplifiers, inverting integrator and differentiator. Theory: In this lab, we will be simulating basic configurations using the μa741 op amp. You can get the μa741 part from the library. Note that the amplifier has two terminals labeled os1 and os2 besides the regular pins, and you can leave these two pins unconnected. (These pins are used for offset adjustment for the op amp) 1.Inverting amplifier Figure: 2.1 An inverting amplifier is shown in Fig The principal features of this configuration are The amplifier is inverting with the feedback loop closed. The closed loop gain is solely determined by the feedback resistors R f and R i assuming that the loop gain defined as the product of the open loop gain of the op amp and the feedback factor of the circuit is very large.

2 The positive input terminal is always grounded in this configuration. As a result, the negative input must follow the potential of the positive one with feedback loop closed due to the large loop gain developed by the op amp. This is often referred to as the virtual ground property of feedback op amps. The closed loop gain of this amplifier is, G= 2. Non inverting amplifier Figure: 2.2 A non inverting amplifier is shown in Fig The principal features of this configuration are The amplifier is non inverting with the feedback loop closed. The closed loop gain is solely determined by the feedback resistors R f and R i assuming that the loop gain of the circuit is very large. The positive input terminal now is connected to the input voltage source. The feedback path, however, is still connected around the output terminal and the negative input terminal. The closed loop gain of this amplifier is, G=1+ The input resistance of this configuration is very large. 3. Differential amplifier Figure:2.3

3 A differential amplifier is shown in Fig.2.3. The principal features of this configuration are The amplifier can combine two inputs and obtain the difference with the feedback loop closed. The closed loop gain is solely determined by the feedback resistors R f, R i, R 1 and R 2.In this circuit, we set R f = R 1 and R i = R 2. Thus, the output is V out= (V 4 -V 3 ) The gains for the positive and negative inputs can be set differently by choosing different R 1 and R 2 values from above. 4.Summing amplifier Figure:2.4 A summing amplifier is shown in Fig The principal features of this configuration are The amplifier is inverting for all inputs with the feedback loop closed. The summing is performed at the negative input terminal in current domain due to the virtual ground property (note that the positive input is grounded). The negative input terminal thus is often referred to as the summing node in such context. The weight for each parallel input is solely determined by the resistor connecting this input to the summing node. The output voltage is a linear combination of all the input voltages. 5.Inverting integrator Figure:2.5

4 An inverting integrator is shown in Fig The principal features of this configuration are A capacitor C is connected between the output and op amp inverting input terminal. The negative feedback of the op amp ensures that,the inverting input will be held at 0 volts(the virtual ground). If the input voltage is exactly 0 volts, the output voltage will not change. It will remain constant with respect to ground. However, for a constant positive voltage to the input, the op amp output will fall negative at a linear rate. Thus the output voltage is V o ut= 6.Inverting Differentiator Figure:2.6 An inverting differentiator is shown in Fig The principal features of this configuration are The right side of the capacitor is held to a voltage of 0 volts, due to the virtual ground effect. Therefore, current through the capacitor is solely due to the change in the input voltage. The capacitor current moves through the feedback resistor, producing a drop across it, which is the same as the output voltage. Thus the output voltage is V o ut =

5 Lab work: Draw the circuits shown in Fig.2.1~2.6 in PSpice schematics. Set the input voltages as suggested in Table Select transient from Setup Analysis. Run simulation and mark outputs Roughly fill up Table Table: Linear Application Outputs. CIRCUIT 1. Inverting Multiplier R i =1k, R f =1k,10k,100k Vi=2v p-p Sin, 1kHz Draw Output R f =1k R f =10k R f =100k 2. Inverting Summer R f =1k,10k,100k V 1 =2v pp, 1kHz(rec) V 2 =2v pp, 1kHz(tri) R f =1k R f =10k R f =100k 3. Differential Amplifier (as Subtractor) If R 2 /R 1 = R 4 /R 3 then V o = R 2 /R 1 (V 2 - V 1 ) V 1 =2v pp, 1kHz (rec) V 2 =2v pp, 1kHz (tri) Select R s for unsaturated V o Select R s for saturated V o Select R s for Subtracted V o 4. Inverting Integrator Vi= 2v pp, 1kHz Out put For V i =Vsin Out put For V i =Vrec Out put For V i =Vtri 5. Inverting Differentiator Vi= 2v pp, 1kHz Out put For V i =Vsin Out put For V i =Vrec Out put For V i =Vtri

6 Pre-lab work: 1. How can you convert a sine wave into a cosine wave using an op-amp? 2. How should the supply voltages be chosen in an inverting summer? What will happen if the constraint is not met? Homework problem: 1. Generate a pulse train using an op-amp. The maximum and minimum values of the pulse train are correspondingly 5 V and -10 V and their duration times are correspondingly 0.2 sec and 0.5 sec. Give the plots of both input and output signals. 2. Design a rectifier for sinusoidal input without using any diode. The output must have twice the frequency than the input. Use op-amp. Prepared by: Ajanta Saha, Marjana Mahdia.