Sirindhorn International Institute of Technology Thammasat University

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1 Sirindhorn International Institute of Technology Thammasat University School of Information, Computer and Communication Technology COUSE : ECS 304 Basic Electrical Engineering Lab INSTUCTO : Dr Prapun Suksompong (prapun@siittuacth) WEB SITE : EXPEIMENT : 08 Operational Amplifiers II I OBJECTIVE To study various functions of operational amplifiers (part II) II BASIC INFOMATION 1 The pin details of op amp 741 are shown in Figure 81 below (1) Offset null (2) Inverting input (3) Noninverting input (4) V O NC (8) V (7) Output (6) Offset null (5) Noninverting input (3) (2) Inverting input V, Positive power supply (7) (4) V, Negative power supply (6) Output Figure 81: Pin details and configuration of IC 741 Two important characteristics of the ideal op amp are 1) The currents into both input terminals are zero: i i 0 2) The voltage across the input terminals is negligibly small: v v emark: Do not assume that v v 0! v i v i V S V S

2 2 In this experiment, we study the following op amp circuits: a) currenttovoltage converter b) voltagetocurrent converter c) integrating amplifier 3 The voltagetocurrent and currenttovoltage converters are used in electronic voltmeters and ammeters, respectively The voltagetocurrent converter, as shown in Figure 82, produces an output current that depends on the input voltage and the resistor In particular, the output current I out = V i / independent of the loading resistance L The currenttovoltage converter, as shown in Figure 83, produces an output voltage that depends on the input current and the resistor In particular, the output voltage V o = I in independent of the size of the loading resistance L V i V I out I in V L V V L V o Figure 82: Voltagetocurrent converter Figure 83: Currenttovoltage converter 4 An integrating amplifier is shown in Figure 84a v C C i C V v i i in X V v o Figure 84a: Integrating amplifier 2

3 Since no current enters the inverting input of an ideal op amp, all input currents must flow through the capacitor Thus, i C = i in Moreover, for ideal op amp, we know that the voltage at the two input terminals must be the same Therefore, v X = 0 This gives i in = v i / ecall the relationship between the timedependent current and voltage for the capacitor: d ic t C vc t dt In this case, the current through the capacitor is and the voltage across the capacitor is C i t i in vi t v t v v t v t C X o o Hence, vi t d C vo t dt The output voltage then has the following form: t 1 vo t vo 0 vi tdt C, where o 0 v is the initial value of the output voltage Note that the change in the output voltage (when considered at two time instants t 1 and t 2 ) is inversely proportional to the integration of the input voltage 2 1 t o 2 o 1 i C t1 v t v t v t dt Suppose the input voltage waveform v i is a square wave with frequency f and peaktopeak voltage 2h as shown in Figure 84b 0 0 h T/2 Figure 84b: Input waveform Figure 84c: Output waveform 2 h fc For half of the period, the input is fixed at h During this time, the output will decrease At the end of this interval, the total decrease is 1 T h h C 2 2fC 3

4 Similarly, the output will increase during the time that the input is fixed at h Because input has equal areas above and below the ground level, the decrease amount is the same as the increase amount and we see a triangular waveform at the output The peaktopeak h 2 fc as shown in Figure 84c In conclusion, when a square wave voltage is drives an op amp integrator, the output is a triangular wave emark: For the circuit in Figure 84a, an input with nonzero mean (DC offset) can saturate the op amp To see this, suppose the range of the square wave input is from 1 to 2 V Then, during each period of the 2 input, the output will have a 2 fc decrease and a 1 2 fc increase Because the amount of decrease is greater, the output will accumulate this difference during each period It will keep decreasing until it saturates the op amp 5 An alternative method to analyze the integrating amplifier in Figure 84a is to perform an AC analysis Suppose the input is sinusoidal with peak V i and frequency f In AC analysis, we use impedance The relationship between the current and voltage for the capacitor is 1 VC IC ZC IC jc For ideal op amp, we again have Vi IC Iin and VC VX Vo Vo Hence, 1 Vi 1 Vo VC IC jc jc Therefore, the gain at frequency f is 1 j2 fc In particular, the gain at f = 0 is unbounded ecall, from your calculus class, that you can decompose a periodic waveform into a sum of weighted sinusoidal waveforms If your waveform has a nonzero average, then you have a constant in your sum as well This constant is the DC offset Our analysis above shows that if the DC offset is nonzero, it will be (theoretically) amplified by an infinite gain! This will saturate the op amp 6 In practical circuit a resistor is usually shunted across the capacitor as shown in Figure 8 5 In this case, Vi 1 V i p VC Iin ZC // p 1 jpc 1 jc p 4

5 So, and the gain is At f 0, the gain is finite V o V V V o i C V i p j C 1 1 p j C 1 p p p C V v i i in X V v o Figure 85: Input waveform Large p is used so that the overall operation of the circuit is not too different from the original integrating amplifier One important effect of p is that the output will not be triangular anymore You will still observe an output that is very similar to a triangular waveform if the product large compared to the halfperiod time T/2 C p is It can be shown that if the input is a zeromean square wave with frequency f and peaktopeak voltage 2h then, the output will be zeromean waveform with peaktopeak voltage p 1 r 2h 1 r, 1 where r exp 2 fpc 5

6 III MATEIALS EQUIED Power supplies: 12V, DC, regulated Variable 015 V Equipment: Oscilloscope Function generator Multimeter esistors: two 1k one 100k one 12k Semiconductor: one op amp 741 Capacitors: one 0001 F one 001 F one 0047 F IV POCEDUE Part A: Op amp converter i) Voltagetocurrent converter 1 Connect the circuit of Figure 85 2 Adjust voltage supply V in according to the value listed in Table 81 ecord the result, i out correspondingly V in 12 V 12 V I out A 1 k Figure 85: Voltagetocurrent converter 6

7 ii) Currenttovoltage converter 1 Connect the circuit of Figure 86 2 Adjust the voltage supply V in such that the current i in according to Table 82 is obtained 3 Measure V out and record the result in Table 82 1 k 1 k I in A 12 V V in 12 V V out Figure 86: Currenttovoltage converter Caution: Do not connect the output directly to the ground Part B: Op amp integrator 1 Connect the circuit shown in Figure 87 2 ead V in and V out of the circuit, by varying the value of C and record the results as listed in Table k C 12 k 12 V square wave 2 V pp 2 khz V in 12 V V out Figure 87: Op amp integrator 7

8 TABLE 81: Voltagetocurrent converter V in, V i out, ma TA s Signature: TABLE 82: Currenttovoltage converter i in, ma V out, V TA s Signature: 8

9 TABLE 83: Op amp integrator C = 0047 F Waveforms: volts/div =, time/div = V in = V pp V out = V pp C = 001 F Waveforms: volts/div =, time/div = V in = V pp V out = V pp C = 0001 F Waveforms: volts/div =, time/div = V in = V pp V out = V pp TA s Signature: 9

10 QUESTIONS 1 Explain the operation of currenttovoltage converter What is meant by the gain of the operational amplifier? efer to your experimental data to support your answer 2 How can you construct an electronic voltmeter and electronic ammeter using an op amp? What are the requirements? efer to your results to support your answer 3 Explain the operation of the op amp integrator efer to your results and explain why integrator output decreases when capacitor C increases? 10