# ee3102 experiment 1

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Formal Lab Report for University of Minnesota EE3102 Experiment 1TRANSCRIPT

EE3102 LAB REPORT EXP#1 FEB 2007 ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ Feedback AmplifiersMatt xxxxx Student ID : xxxxxxx1/16/07 2/26/07Abstract:Experiment one dealt with feedback amplifiers and the four negative feedback topologies associated with them. We investigated the frequency response of the amplifiers using these feedback topologies and addressed the issues associated with feedback stability. Then additionally, positive feedback was employed to create a sinusoidal oscillator.-1-EE3102 LAB REPORT EXP#1 FEB 2007 ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ Introduction:Feedback in circuit analysis is one of the most important concepts in electrical engineering, whether positive or negative. It can be used to create a wide variety of circuits, from amplifiers to oscillators. In this experiment we used the concept of feedback to investigate the frequency response of the four negative feedback topologies, as well as a positive feedback application, the Wein Bridge. We observed for each topology, the behavior of the circuit as it relates to various frequency and load changes. Additionally, we used many circuit concepts to study each topology in detail and to observe its effects given external changes.ExperimentOpen Loop Voltage GainThe first section of the experiment asked us to determine and record the open loop voltage gain for three different 741 Op Amps. To do so we constructed the circuit shown in Figure 1.1.1 using the specified values.-2-EE3102 LAB REPORT EXP#1 FEB 2007 ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ -+U 2L M 7 4 13267 14 5R 12 KR 21 0 KR 31 0 KR 41 0R 51 0 K0V 11 5 V d c 00V 21 5 V d cV 1 I n p u tV oFig 1.1.1Since the input voltage of the Op Amp is obtained by a simple voltage division we had to choose as accurate resistance values as possible. As it turns out these resistance values were provided to us so everyone would obtain as consistent of data as possible.The procedure of finding the DC open loop voltage gain is identical for all three Op Amps. To do so we set our input voltage gV to some arbitrary value, then proceeded to measure 1 iV and01V . With these values recorded we then changed the input voltage gV to a different voltage level and proceeded to find 2 iV and2 oV . Using these values we were then able to able to calculate the DC open loop voltage gain using the following formula.01 21 2*1000ovi iV VAV V| ` =

. ,The collected values and calculations for the three Op Amps using the procedure outlined above are shown below.For Op Amp 1:-3-EE3102 LAB REPORT EXP#1 FEB 2007 ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ 111.352.64ioV VV V For 104gV mV 221.363.91ioV VV V For 95gV mV 2.64 3.91*10001.35 1.36126000vvAA + | `= +. ,=For Op Amp 2:111.344.14ioV VV V For 106.2gV mV 221.355.08ioV VV V For 1.21gV V 4.14 5.08*10001.34 1.35235000vvAA + | `= +. ,=For Op Amp 3:111.263.87ioV VV V For 87.3gV mV 221.264.89ioV VV V For 1.092gV V 3.87 4.89*10001.26 1.26165000vvAA + | `= +. ,=From the analysis of the DC open loop voltage gain we are able to see that the gains for the three different Op Amps were 126000, 235000, and 165000 respectively.The Four TopologiesSection 1.2 Series/ShuntThe second section of the experiment asked us to design for a no-load voltage gain of 15 for the respective Series/Shunt feedback circuit shown in figure 1.2.1. We were then to measure the midband voltage gain for various resistive loads. With this completed we determined the low frequency small-signal input resistance -4-EE3102 LAB REPORT EXP#1 FEB 2007 ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ and found the dominant pole for the sinusoidal steady state response. Finally we determined the location of the dominant pole in the open loop response of the Op Amp itself.-+U 1U A 7 4 132 67 14 5R 11 4 KR 21 KR 3RR 4L o a d000V 21 5 V d cV 31 5 V d c00V 2 V 1V oV 41 V a c0 V d cFig 1.2.1In order to determine the necessary values of the resistors to achieve a gain of 15 we used the following two port network equivalent shown in figure 1.2.2.R 1RR 2R + +- -V 1I 2 I 1V 2Fig 1.2.2Proceeding we know the gain for a series/shunt amplifier is:1ofifVAVAAA+Here we assumed 1 A >> -5-EE3102 LAB REPORT EXP#1 FEB 2007 ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ Therefore1fA, Where 115Solving for beta yields: 10.0666715 Now using H parameters for our two port circuit we were able to determine the resistor values as follows.2 2121 26.667*10RhR R +Choosing21 R k O arbitrarily and solving for 1R yielded114.015 R k O.With the circuit design completed and the circuit built we were able to determine the voltage gain as a function of various resistive loads.Gain vs Load resistanceLoad vi(V) Vo(V) Gain0 0.505 7.69 15.16810 0.502 0.2 0.40950 0.499 1.036 2.076100 0.498 2.06 4.37500 0.487 7.45 15.291000 0.494 7.58 15.3510000 0.488 7.46 15.413100000 0.488 7.46 15.4151000000 0.494 7.56 15.3Fig 1.2.3A plot of Gain vs. Load Resistance is shown below.-6-EE3102 LAB REPORT EXP#1 FEB 2007 ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ 0246810121416181 10 100 1000 10000 100000 1000000GainLoad Resostance (ohms)Gain vs. Load ResistanceSeries1Fig 1.2.4Next we found the low frequency small signal input resistance and the location of the dominant pole of the sinusoidal steady state response.To determine this we placed a 100k ohm resister between the input and V+ so we could determine the current flowing into the terminal as shown in figure 1.2.5.-+U 1U A 7 4 132 67 14 5R 11 4 KR 21 KR 31 0 0 k00V 21 5 V d cV 31 5 V d c00V 2 V 1V 41 V a c0 V d cV o-7-EE3102 LAB REPORT EXP#1 FEB 2007 ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ Fig 1.2.5In order to determine the current flowing into V+ we measured the voltage values at V1, V2 and then used Ohms Law to compute the current.1210.98V VV VThen using Ohms Law.1 21 0.98100V VIR k 200 I nA With this current found we could then calculate ifR.20.982804.9sifsififV VRI IVRnAR M OTo determine the dominant pole we needed to find the closed loop gain for the circuit. We accomplished this by collecting following data:Gain vs FrequencyFreq (Hz) Vi (V) Vo (V) Gain500 0.5 6.94 13.881000 0.5 7 14.122000 0.5 7.7 15.35000 0.5 8 1610000 0.5 7.8 15.615000 0.5 7.8 15.620000 0.5 7.7 15.425000 0.48 7.3 15.230000 0.48 7 14.5840000 0.48 6.6 13.7550000 0.48 5.8 12.0857000 0.48 5 10.3560000 0.48 4.5 10.4170000 0.48 5.5 9.375Fig 1.2.6A plot of Gain vs. Frequency is shown below.-8-EE3102 LAB REPORT EXP#1 FEB 2007 ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ 024681012141618100 1000 10000 100000GainFrequency (Hz)Gain vs. FrequencySeries1Fig 1.2.7Using this data we are able to see that our midband gain is about 15.3V/V. Using this we could then calculate our closed loop gain.15.3*.707 10.8CLA This can be seen to occur at about 57 KHz.Thus 357dBF KHz With this information we were than able to calculate our dominant pole location.33* **CL dB D OLCL dBDOLA F F AA FFASince Op Amp 1 was used for this experiment, CLA was equal to 126000.10.35*571260004.68DDKFF-9-EE3102 LAB REPORT EXP#1 FEB 2007 ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ Section 1.3 Shunt/SeriesThe third section of the experiment asked us to design for a midband short circuit gain of 100 for the respective Series/Shunt feedback circuit shown in figure 1.3.1. Using this circuit we were to measure the midband gain for various resistive loads.-+U 1U A 7 4 132 67 14 5R 1L o a dR 21 0 100V 21 5 V d cV 31 5 V d c00V AV 2V 41 V a c0 V d c0R 11 0 KV 2 V 1V BR 41 MFig 1.3.1In this circuit we approximated the necessary current source as a voltage source in series with a large resistor. We used a 1M ohm resistor that was chosen arbitrarily.Once again we used two port circuit analyses to determine the necessary resistor values to obtain a midband short circuit gain of 100 (See figure 1.3.1 for the two port network.).R 2RR 1R + +- -V 1 V 2I 2 I 1Fig 1.3.2Proceeding we know the gain for a Shunt/Series amplifier is:-10-EE3102 LAB REPORT EXP#1 FEB 2007 ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ ofsIAI1f AAA+Here we assumed 1 A >> Therefore1fA, Where 1100Solving for beta yields: 10.01100 Now using G parameters for our two port circuit we were able to determine the resistor values as follows.2121 20.01RgR R +Choosing110 R k O arbitrarily and solving for 2R yielded2101.01 R O.With the circuit design completed and the circuit built we were able to determine the current gain as a function of various resistive loads as follows: Gain vs Load resistanceV1 (V) V2(mV) Va (mV) V (mV) Rl (Ohms) I (uA) Io (uA) Gain2.31 300 300 300m 0 1.827 0