Assignment title: Information


Assignment 1 Description Marks out of Wtg(%) Due date Assignment 1 200 20 19 August 2016 Part A: Astable and Switching (4%) (1) Astable Use a single 4528 dual monostable to construct an astable circuit with a mark-space ratio of 0.33 and times of the order of 1 to 2 seconds. Put indicating LEDs on the circuit to indicate both times. Design hints 1. The final transition of the times pulse from one monostable may be used to trigger the other monostable, producing waveforms as shown: 2. Mark-space means high time to low time. 3. You will need to look up the data for the I.C. you are using since details of timing relationships are different for each (see data sheets). Report Provide all calculations and full final circuit schematic diagram.(2) MOSFET Switching Find out information on the operation of, and configuring of, MOSFETs to be used in switching circuits. In particular note the differences between BJTs and MOSFETs in this role. Draw up a table to highlight the differences and hence the pros and cons on each device for particular situations (eg. Switching high-to-low or low-to-high (ie. P or N type), high or low current switching, low or high voltage switching). Consider the following BJT switching circuit. Analyse the operation of the circuit to understand the parameters involved. Choose suitable replacement MOSFETs to be used instead of the output switching BJTs in the given circuit. Include any necessary circuit changes for the new devices to operate so as to maintain the circuit's required parameters. Calculate the power dissipation of the device(s) when ON. Note that the input logic levels "0" and "1" are 0V and 5V respectively. Where Vcc = 12V and Relay resistance = 18Ω .Part B: Power Transistor amplifier design (7%) Design Design a push-pull amplifier using the circuit arrangement given and with the following specifications: Supply voltages ± X V Output power Y W Voltage gain > Z dB Low frequency roll-off point < Q Hz (Lower bandwidth limit) Rin > J ohms [Specifications (X,Y,Z,Q, J) will be individualised to students on StudyDesk] Show all calculations for component values from DC and AC considerations. Choose a suitable input signal voltage to achieve the specification. [See Design Assignment A, Part C for design tips] Simulation Use MICROCAP (or similar) to simulate your design, using any suitable transistors (e.g. Q1). 1. Plot transient analysis of output (on load resistor) and input (to C1). You should expect to see a voltage gain >5 without distortion to the output sine waveform (ie: no clipping). 2. Short out the diode and R4 and repeat the transient analysis. Crossover distortion should be more visible. 3. Determine the d.c. bias values at the base, collector and emitter of Q3 and at the common emitter of the output stage. If your circuit does not work as expected, this should point to the errors. 4. Run an A.C. Analysis to confirm the gain and bandwidth requirements (ie: correct capacitor choices).ELE2504 – Electronic design and analysis 4 Part C: Voltage controlled oscillator design (5%) Circuit The circuit given below generates a rectangular wave and a triangular wave whose frequency is determined by a voltage applied to the input. Circuit operation Assume the transistor is off, i.e. V o2 is LOW. Vo1 ramps downwards due to the integrator R3C. Meanwhile, since V +2 is a positive voltage, when Vo1 passes V+2, Vo2 changes to HIGH, the transistor turns on and C discharges via R4 and the transistor. Hence Vo1 ramps up and the cycle repeats. Design task Design a VCO (voltage controlled oscillator) using an LM324 quad op. amp.or LM741 op. amp.s (refer to the data sheets): [Specifications will be individualised to student on StudyDesk] Example Design specification V DD 9V Vin range 2V to 6V Frequency range 100Hz – 300Hz Duty cycle D:1 (high to low) 3:1 Vo 1 range 2 to 7V A full analysis of the circuit operation and a design example follow.ELE2504 – Electronic design and analysis 5 1 1 2 2 Test Assemble your design and check that it works. An LM324 (single supply) op. amp. and any small transistor should work. You will also need to provide VDD / 2 with a voltage divider of two 1k ohm resistors or similar. If you have access to equipment, measure the performance of your circuit in regard to each of the design specifications. Analysis Op. amp. A1 operates an integrator with an input voltage Vin / 2 set by the voltage divider R 1 and R2. When TR1 is off, A1 integrates by charging C via R3 (pushing current into the left hand side of C for positive Vin). When TR1 is on, C discharges to ground via R4. By the principle that for an active op. amp., V+ = V–, and since V+ is held at Vin / 2: a. When TR1 is off, current into C, = I1: I  V in Vin / 2 R 3 I  Vin 2R 3 and hence V o1 ramps downwards at a rate of: V in 2R 3C b. When TR1 is on, current out of C via: R 4  I2 I  Vin 2R 4 but in this case since V– is still at V in / 2, current is still also charging C via R3. So we have: I  Vin 2R4ELE2504 – Electronic design and analysis 6 So the net current outflow from C is I 2 – I1. Now A 2 acts as a comparator and its output switches TR1 when Vo1 reaches some fraction (say β) of Vo2 which can be assumed to switch between zero and VDD. Define   R5 R 5  R6 V DD VDD When V o2 is VDD , then V2   ( 2 )  2  1  VDD 2 V DD VDD When V o2 is zero, then V2   ( 2 )  2  1 VDD 2 and these two voltages are the thresholds at which Vo1 causes switching. The voltage Vo1 thus ramps up and down at different rates as shown: So in order to achieve a duty cycle of D:1, I 2  I1  I1 D or D I2  I1  I1 DI 2  1 D I1 DVin  1 DVin 2R 4 R 2R 3  DR 3 4 1 DELE2504 – Electronic design and analysis 7 2 Timing V V During the time t1, the voltage at Vo1, Vo1(t)  in  DD 1  2R 3C 2 V and at t 1, Vo1(t1)  DD 1  2 Hence VD D 1   Vin t1  VDD 1  2 2R 3C 2 V 2R C and so t1  DD 1  1  3 2  2V DDR3C V in V in Since t2  Dt1 t  2DV DDR3C V in 1 D 2DVDDR C and T  3 V in or f  Vin 1  D 2VDDR3C Design example Choose R5 and R6 Vo 1 range is 2 to 7V 2  VDD 1  2  4.51    0.55 7  VDD 1  2  4.51    0.55 Choose resistors to be as large as possible consistent with the assumption that the current into V 2+ is negligible, say: R 5 = 100k Ω R 6 = 82k ΩELE2504 – Electronic design and analysis 8 Frequency Choose R 3 and C as a pair by trial and error until practical values of both are found. f (min)  100 Hz V in (min)  2V 100  2 f (max)  300 Hz V in (max)  6V 300  6 4 2 9 0.55 R 3C 4 4 2 9 0.55 R 3C  Say So R 3C  5.0510 C  3.3 nF R 3 150 k R 4 112.5 k say 120 k Transistor Maximum current in transistor  Vin (max) / 2 R4  33 A Assume h FE  100, so I B  0.3 A Transistor is turned on when V o2  9V so R 7  8.3 0.3 A  25 M Note this resistor is impractically large. So allow more base overdrive and choose R 7 = 1 MΩ. (The 33pF capacitor is to aid switching speed.) Report ● Show all design calculations ● Show full circuit diagram with all component values ● Results from testing or simulation ● Comparison of results with specifications with observations and conclusions.ELE2504 – Electronic design and analysis 9 Part D: Semiconductor devices (4%) Select two devices and submit a short report for each, covering the following details: ● what the device is (name, type) ● explanation of the devices' features and how it works ● sample circuit application making use of its feature(s) with a circuit operation explanation (include any relevant calculations). Device list: 2N4871 QED223 BPV11 1N6276CA (or 1.5KEI6CA) IRF730 BUK854-500 V33ZA7 (GE-MOV) BD333ELE2504 – Electronic design and analysis 10ELE2504 – Electronic design and analysis 11 Design assignment A Description Marks out of Wtg(%) Due date Design assignment A 0 0 Week 5 This assignment does not have to be submitted for marking. However it may be the subject of a question in the examination. Part A: Logic design (1) Astable Use a single 4528 dual monostable to construct an astable circuit with a mark-space ratio of 0.5 and times of the order of 1 to 2 seconds. Put indicating LEDs on the circuit to indicate both times. Design hints 1. The final transition of the times pulse from one monostable may be used to trigger the other monostable, producing waveforms as shown: 2. Mark-space means high time to low time. 3. You will need to look up the data for the I.C. you are using since details of timing relationships are different for each (see data sheets). Test Assemble the circuit and check that your design works. Design a logic interface circuit to allow the connection of data from either of the two specific logic gates shown (one TTL and one CMOS), to either of two other specific logic gates (TTL and CMOS) as shown in the sketch.ELE2504 – Electronic design and analysis 12 (2) Logic Design & Family Compatibility Specifications 1. Five volt power supplies are used on both source and destination circuits. 2. The signal must not be inverted in passing through the interface. (It may of course be inverted within the interface, but must emerge the same as it went in.) 3. The interface is to contain an LED (with current between 5mA and 10mA) to indicate the state of the data (at point X), as well as any extra gates (of any type) needed to provide the interface function. (Use gates only, not transistors.) 4. The interface must be properly designed to ensure that it takes into account the worst case specifications of the logic gates involved. D.C. design only is required. Steps 1. Assemble the relevant worst case data on all gates and organise in a table of the following form, using data @ 25°C. Gate V OH IOH VOL IOL VIH IIH VIL IIL 74HC00 – – – – 74S02 – – – – 4001 – – – – 74LS00 – – – –ELE2504 – Electronic design and analysis 13 2. Design the circuit by carefully considering each possible position of the switches, both voltage and current and both high and low states, and choosing a combination of gates to form the interface circuit. Report Submit a report containing at least the following: 1. Data for the gates in a table as above. (Data sheets at the end of this book.) 2. For the simple case of the interface being a direct connection, give clear consideration of each possible combination of gates and logic levels, and identification of any problems. Give reasons. 3. Your solution to the problems, including a sketch of the interface circuit and justification of any resistor values. Note There is no need to assemble and test the circuit since most individual I.C.s will perform better than the worst case figures indicate, and so a simple test of operation would prove nothing.ELE2504 – Electronic design and analysis 14ELE2504 – Electronic design and analysis 15 Part B: Differential amplifier exercise The circuit Simulation Use MICROCAP (or similar) to analyse the circuit. Leave out the voltage divider network (47k, 10k, 47k) for this. 1. Determine the d.c. voltages at both collectors and at the common emitter, by grounding the base of Q1 and running a d.c. analysis. 2. Determine the d.c. response at either collector to a varying input voltage over the range –0.3v to +0.3v by adding a 1mV d.c. source as shown above and running a d.c. analysis. Use limits as shown on next page. 3. Add a 1μf capacitor to the input and so determine a.c. voltage gain, Ad (single collector to ground). 4. Connect the two bases together at the bias network and determine the a.c. common mode gain (single collector to ground). Change sine source to 1V. 5. Add the constant current source shown below in place of the common emitter resistor RE and again determine the single ended a.c. common mode gain. (i.e., repeat 4 above). Notice that it is very frequency dependent, but for frequencies less than about 1MHz, it is very much less than it was for the single resistor case. Hence common mode rejection ratio is very much improved.ELE2504 – Electronic design and analysis 16 Help using MICROCAP For determining the d.c. response at the collectors, try these analysis limits: Sample plot ....ELE2504 – Electronic design and analysis 17 Test 1. Construct the circuit shown on the previous page. 2. Vary the potentiometer over its entire range and at about nine evenly spaced intervals measure the voltage on the base and the voltage at both collectors. You will need to take additional points at the cross-over to get a good representation of the shape of the curves. The following values of base voltage are suggested: –0.3, –0.1, –0.06, –0.03, 0, +0.03, +0.06, +0.1, +0.3 volts. 3. Calculate the corresponding collector circuits. 4. Plot on the one set of axes, graphs of each collector current and the sum of the two, versus input voltage. Your graph should appear as sketched below. Report ● Submit MCap schematic (with node numbers) and DC analysis plot. ● Submit table of results from practical testing with graph.ELE2504 – Electronic design and analysis 18 Part C: Transistor Amplifier Design Design and test a common emitter amplifier using the circuit shown and the selected specifications. Specifications Get your own specifications provided on the StudyDesk Bandwidth > 100Hz → 20kHz Design (Example follows) 1. Allow about half V CC across the transistor; V CE = 0.5 VCC 2. So the voltage across RC and RE is VCC – VCE IC (RE + RC) = VCC – VCE 3. Since the voltage gain will be very close to RC / RE, R C / RE = AV Hence R C and RE can be determined, choosing appropriate preferred values.ELE2504 – Electronic design and analysis 19 E 4. Determine: VE = I CRE and hence VB = V E + VBE 5. Choose R 1 and R2. The maximum value of base current is: IB (max) = IC / hfe (min) Now let the current in R 1 and R2 (IDIV) be about 5 to 10 times IB, so ID IV = VCC / (R1 + R2) = say 10 IB (max) (1) and since I B can now be neglected relative to IDIV, VB = R 2 / (R1 + R2) × VCC (2) Hence R 1 and R2 can be chosen using the nearest preferred values. 6. Check the design using the nominal preferred values to ensure that VCE ≈ 0.5VCC. If not, modify R1 and R2 slightly, remembering that their ratio is important (equation (2) above) but their absolute magnitude is not (equation (1) above). Example Voltage gain  7 V CC  10v I C  2.5mA Assume h fe(min)  130. 1. V CE = 0.5 VCC = 5v 2. I C (RE + RC) = VCC – VCE 2.5 × 10–3 (R + RC) = 5 3. R C / RE = Av = 7 So R E + RC = 2k R C / RE = 7 ∴ R E = 250 → 270Ω R C = 1750 → say 1800Ω 4. V E = IC RE = 2.5 × 10–3 × 270 = 0.68vELE2504 – Electronic design and analysis 20 V B = VE + 0.7 ≈ 1.4v 5. I B (max) = IC / hfe (min) = 2.5 × 103/130 = 19.2µA ID IV = say 10 × IB = 192µA R 1 + R2 = VCC / IDIV = 10/192µA = 52kΩ V B = R2 / (R1 + R2) × VCC R 2 / (R1 + R2) = 1.4 / 10 = 0.14 or the voltage across R2 = 1.4v, and the voltage across R1 = 8.6v. ∴ R 2 = 0.14 × 52k = 7.3kΩ say 6.8kΩ (reducing ≈ 7%) and R 1 = 0.86 + 52k = 44.7kΩ, say 39kΩ (reducing ≈ 12.8%) 6. Check: V B = 6.8k / (39k + 6.8k) × 10v = 1.5v V E = 0.8v I C = 0.8270 = 3.0mA V C = 10 – 3.0 × 1.8k = 4.6v ∴ V CE = 4.6 – 0.8 = 3.68v This is significantly less than 5v but could be acceptable. If not, select another R1 and R2 by trial and error. Try 8.2k and 56k which gives VB = 1.3v which is about as far off in the opposite direction. This may be satisfactory if the amplifier is not required to handle large signals. If not satisfactory, repeat until some satisfactory values are found.ELE2504 – Electronic design and analysis 21 Simulation Use the circuit simulation programme MICROCAP (or similar) to analyse your design: (Help is given on the following page.) 1. Determine the d.c. voltages at all points. This will check your design – if these are correct, the circuit should amplify an a.c. signal. Check choice of C1 for specification. 2. Determine the frequency response (voltage gain and phase versus frequency) over the range 1Hz to 1MHz. 3. Bypass RE with a capacitance of 0.47 µF and again determine the frequency response over the same frequency range. For example: MICROCAP analysis of the example circuit above using transistor Q1, gives the frequency response graph on the following page, using the analysis limits on the page after that. Help using MICROCAP The following may help to get the required results using MICROCAP: ● draw the circuit ● select VIEW, show node numbers and note the node numbers of the input before the capacitor, and the output (collector) ● select RUN, Transient Analysis and set the Analysis Limits ● select AC Run and the result will be a frequency response graph. Transient response – (time base) Sample Analysis setup ... Note that the node numbers in the 'Y expression' pertain to the nodes on the schematic shown overleaf. Your numbers for these nodes may be different.ELE2504 – Electronic design and analysis 22 Transient Analysis plots of input and output ... Note use of scope cursors to determine peak-to-peak voltage of output. Circuit schematic showing node voltages (quiescent) post analysis .... Note that the Transient Analysis plot and node voltages show whether biasing is correct/optimal.ELE2504 – Electronic design and analysis 23 Frequency responses – (Bode plot) Sample setup for AC Analysis ... Sample plot of AC Analysis Note use of scope cursors to determine the –3dB break point at approximately 22 Hz.ELE2504 – Electronic design and analysis 24 Test (Optional) Note that the measurements taken here are not expected to be accurate since it is assumed they will be done with a digital voltmeter only, and waveforms may not be sinusoidal. In building the circuit you may use a BC547 in place of the 2N22222. 1. Connect your circuit and measure the d.c. voltages around the circuit. If they are not as expected, find the cause and rectify it. 2. Connect the test oscillator (shown below) to the input. It should be producing an approximately sinusoidal signal of about 100kHz frequency and you can add an attenuator to produce a small size of say 0.2v p/p. Measure, using a DVM, the magnitude of the voltage gain at 100kHz. 3. Bypass RE with a capacitance of 0.47μF and again measure the voltage gain. You will need to reduce the size of the signal input in order to keep the output sinusoidal. Reduce it to an estimated 20mV. 4. Remove the bypass capacitor. Estimate the input resistance (at 100kHz) by adding a 1kW resistor (R) in series with the oscillator to monitor base a.c. current. Measure the a.c. voltage across this resistor, v. The input current is then iin = v/R, and the input resistance R in of the amplifier is Rin = Vin/iin – R. 5. Repeat 3 and 4 with the 0.47μF capacitor bypassing RE. Note: If using 4093, parallel all 6 gates to provide maximum output current capability. Test oscillator to act as a source for the amplifier. Report ● Show all design calculations. ● Show full cct design with all component values. ● Provide either – o MCap plots for AC analysis, transient analysis and schematic showing node voltages and/or o Test results from testing built circuit. ● Comparison of results (measured or simulated) with specifications, drawing inferences and conclusions.ELE2504 – Electronic design and analysis 25 Design assignment B Description Marks out of Wtg(%) Due date Design assignment B 0 0 Week 10 This assignment does not have to be submitted for marking. However it may be the subject of a question in the examination. Part A: Switching regulator design exercise Theoretical design Note: This design is entirely theoretical. You will not be expected to assemble the circuit and test it. Collect information Data on the TL497AI switching voltage regulator is provided from the Texas Instruments 1989 Linear circuits data book, volume 3, pages 2-135 to 2-141. From this data assemble the following: ● a table showing ● allowable input voltage range ● allowable output voltage range ● V ref (typical) ● switching transistor maximum current ● diode maximum current ● A sketch of the power dissipation rating curve. Data on ferrite cores suitable for construction of inductors is also provided. Design Using this data and the typical application data (use the basic configuration even if current values fall slightly outside the specified range), design a regulator to meet the specifications below. Include component ratings in your design and choose the inductor core, wire size and number of turns. Your design should also include thermal considerations by estimating the heat loss. If necessary, specify heat sink thermal resistance.ELE2504 – Electronic design and analysis 26 Specifications: Type of regulator : Step down Input voltage Vi : 24 Output voltage Vo : +5V Output current Io : 300 mA Maximum output ripple Vr : 50 mv Ambient temperature Ta : 50 deg C Report Present a report containing the following: ● assembled data ● design calculations, including inductor core choice/details ● a circuit sketch using the same schematic layout as in the data, and showing all component values including ratings.ELE2504 – Electronic design and analysis 27ELE2504-Electronic design and analysis 40 (Source: Texas Instruments 1989, Linear circuits data book, vol. 3, p. 2-136. Reproduced with permission from Texas Instruments Ltd.)ELE2504 -Electronic design and analysis 45 (Somce:TexasInstruments1989,Linear circuitsdata book,vol. 3,p.2-140.Reproduced with permission from Texas Instruments Ltd)ELE2504 -Electronic design and analysis 46 (Somce:TexasInstruments1989,Linear circuitsdata book,vol. 3,p.2-140.Reproduced with permission from Texas Instruments Ltd)ELE2504 -Electronic design and analysis 47 (Source: Texas Instruments 1989, Linear circuits data book, vol. 3, p. 2-139. Reproduced with permission from Texas Instruments Ltd.) © University of Southern QueenslandELE2504 -Electronic design and analysis 48 (Somce:TexasInstruments1989,Linear circuitsdata book,vol. 3,p.2-140.Reproduced with permission from Texas Instruments Ltd)ELE2504 -Electronic design and analysis 49 Propert es of core assemble i s at25°C (without adjusters) Stock number RM6 RM6 RMl RM10 RMIO 228-214 228-220 228-236 228-242 221-268 Inductancefactor ""'- Tumsfactor a (turns for 1mH) Effect ve permeability JJ. Temp.coeft.of "• (+25 to50"C) ppmi"C Adjuster range Max.residualplus eddy current coreloss Tengenttand,+,111:JOkllz at lOOkHz Recommended frequency range(kHz) Energystoragecapabtkty (mJ) UJ.,., B ... mT 160 250 250 250 400 79.06 63.25 63.25 63.25 50.00 ±1% ±1% ±1% ±1% ±1% 109.5 171.1 146.0 99.67 159.5 51min. SOmin 73min. 50min. SOmin. t54max. 241max. 219max. 149max. 239max. +20% +14% +15% +17% +20% 0.34 X 10' 0.53 X 10' 0.47 X 1()-0 0.32 X 10" 0.51 X 10' 0.58 X 10" 0.91 X 1(}-' 0.82x 10"' 0.60X 1(}-' 0.96 X 10 5.5 to800 3.5to700 3to650 2to650 1.2 to500 0.383 0.245 MOo 1.731 10 . 82 250 250 250 250 250 RM DataLibrary IssuedJuly 1985 5n4 RM seriesferritecores Stock numbers228-214to 228-258 A range of 5 of the most popular pcb mounting ferrite cores covering three sizes.Of square design which 11llows moximum boord ut lisation, this series enables transformers or inductors to be constructed to meet exactcustomer requirements. The core material is equivae l nt to the commonly known grades:A13-0.3-N28.Eachcoreis supplied Inkitformandcon"slstsof the folloWing:onepairof matched half cores,one single section bobbin with Integralpins onan 0.1ingrid,onepairof retan i ing clipswithearth spikes andonecore adjuster. To determine the number of turns required for a particularinductanceuse the followingformula: No,turns= vi"" Where L =inductance innH (1 H). For frequencies in excess of 30kHz,the use of stranded wire is beneficial when maximum a is Features e 5 versions availableinthree popular sz i es e PCBmounting e Compactdesign e Mountn i g pinshave 2.64mm (0.11n)spocing • l . • required. (nH/tum3') ±2% :t2% :!:2% ±2% :t2% Magneticpropertie$of c:ores Effectivepath length Effectivepatharea Effective volume symbol RM6 I, 26.9mm A., 31.3mm' V0 840mm' RM7 29.6mm 40.3mm' 1100mm' RM JO 41.7mm 83.2mm' 3470mm' Maximumturnsacmmodatedonbobbin wiredia.(mm} RM6 RM7 RMJO wiredia.(mmJ RM6 RM7 RM10 0.2 205 306 612 0.56 25 36 87 0.224 160 250 484 0.71 19 33 59 0.25 127 209 402 0.8 13 19 44 0.315 87 131 246 1.0 9 11 25 0.4 47 76 160 1.25 4 9 19 0.5 36 50 98 1.5 3 7 11 (Source: RS Components 1985, Data sheet no.5774,July. Reproduced with permission from RS Components Pty Ltd.) UNIVERSITY !If SOUTHERN QUEE,'!'. ©University of SouthernOJeenslandELE2504 – Electronic design and analysis 48 © University of Southern QueenslandELE2504 – Electronic design and analysis 48 © University of Southern Queensland Part B: Logarithmic amplifier design exercise Design Given any of the following transistor arrays, find the collector current versus base-emitter voltage characteristic of the transistors from the data sheet which follows. Typical figures are acceptable. Note that it is logarithmic over two decades of current: LM or CA 3045, 3046 or 3086. All of these are similar electrically and are readily available from suppliers. All are in 14 pin DIL packages and contain five similar transistors. Design a logarithmic amplifier of the two-transistor type described in the Study Book based on this device and LM108 op. amps. whose characteristic is to approximate the relationship: V out = –5 log (2Vin/2Vref) as closely as possible at room temperature, over a two decade range of Vin from 0.1V to 10V. Arrange for V out to always be > 0V. Data for the LM108 is attached. Use to +/– 15 volt supplies. Simulation Using MICROCAP simulate your circuit (using LM108 and transistor 2N2222 in MICROCAP) and plot Vout versus Vin. Do this at three different temperatures, 0, 30 and 60 degree C. (Try various values for the lower limit in MICROCAP (0.2 or above until it works.) Also simulate the simple logarithmic amplifier given in the notes with the same op. amp. and transistor and R = 10k ohms. (Specify MICROCAP Vin range at 10, 0.2, 0.1 for rapid results.) Note that its variation with temperature is much greater than that of the other circuit. Test Assemble the circuit. Spurious oscillation will probably occur causing incorrect operation and will need to be suppressed with capacitors such as fairly large values (4.7 µF or more) from the output to ground. Do not forget to use compensation capacitors on the 308 according to the data sheet. If necessary, extra large values may help stop oscillation. Measure the characteristics of your amplifier using a digital multimeter over its allowed range of input voltages at room temperature. Note: If you do not succeed in getting this circuit to work, it may be because of spurious oscillations which are very common. The use of an oscilloscope would help to identify and thus avoid these oscillations. However if you do not have access to an oscilloscope, try adding capacitance in other places or ring the lecturer for advice.ELE2504 – Electronic design and analysis 49 © University of Southern QueenslandELE2504 – Electronic design and analysis 50 © University of Southern Queensland (Source: National Semiconductor 1982, Linear databook, vol. 2, p. 2-19. Reproduced with permission from National Semiconductor.)ELE2504 – Electronic design and analysis 51 © University of Southern Queensland (Source: National Semiconductor 1982, Linear databook, vol. 2, p. 2-20. Reproduced with permission from National Semiconductor.)ELE2504 – Electronic design and analysis 52 © University of Southern Queensland Assignment 2 Description Marks out of Wtg(%) Due date Assignment 2 200 20 3 October 2016 Part A: Active filter design exercise (Part 1) (8%) Design Design a 3 pole high pass filter with a Butterworth response having a cut-off frequency of 160 kHz. The filter is to use the second order circuit arrangement given below, and first order section with LM108 or equivalent op. amps. Theory From the theoretical pole positions, compute and plot the following: ● the magnitude and phase of the normalised frequency response ● the group delay (–dϕ/dω) as a function of frequency. Note that this will be a normalised plot with a cut-off frequency of 1 rad/sec. Hints: Plot ω over the range 0.01 to 10 and define i  1 , and remember that a 3-pole 1 low pass response is in the form H ()  (s  a) (s  b) (s  c) complex poles. where a, b, c are theELE2504 -Electronic design and analysis 53 The use of MathCAD or a similar programme makes this simple, but do it manually if necessary. Six or eight points on a plot should be enough. Simulation Using MICROCAP simulate your circuit and plot gain, phase and group delay versus frequency over the range 1 Hz to 10 kHz. This will also tell you if your design is correct. Build and test Using an LM324, build your filter and test it by taking about 7 readings from 20 kHz to 2kHz using an oscilloscope (if you have access to one) or multimeter. If you feel your multimeter cannot take accurate AC voltage measurements at these frequencies, go to the course website for an alternative.ELE2504 -Electronic design and analysis 54 Part B: Oscillator design exercise (8%) Measure/simulate Measure the voltage-controlled resistance characteristic of a 2N 5484 (sim: 2N 3822) N-channel JFET, by the following procedure: 1. Set up the following circuit, noting that the gate voltage is of the opposite polarity to the drain voltage. If you do not have access to both these voltages, it is easy to put a voltage divider (say a pot of value several k ohms) across almost any voltage to supply the gate because it is a low current input. 2. ● Set V GS to 0V. ● Vary RD over a range and at a variety of points, record VDS and VR in a table. ● Set V GS to –1V, –2V, –3V, –4V in turn and for each, repeat the procedure above. (simulate: –0.5, –1, –1.5, –2, –2.5). 3. Convert the V R value to ID, by dividing by the R value (120Ω). 4. Plot a series of curves, one for each value of VGS on ID versus VDS axes. 5. For each curve identify the linear region around the origin, and evaluate: R DS  V DS I D 6. Finally plot RDS versus VGS.ELE2504 -Electronic design and analysis 55 Design Design a Wein Bridge Oscillator using a 741 operational amplifier as the active element. Amplitude stabilisation is to be achieved using the JFET as a voltage controlled resistance. Follow the procedure given in the study modules. The oscillator specifications are to be individualised as supplied on the StudyDesk: Power supply voltage: ______ Frequency: ______ Amplitude: ______ Test/simulate The following test may be performed using any equipment you have access to. Results from a moving coil multimeter or a digital multimeter are satisfactory, but an oscilloscope is desirable. Check the correct operation of your oscillator by measuring operating voltages at significant nodes. These will need to be compared with design values in your report. These should include the following points: + terminal of op. amp, output of op. amp., voltage across C1, voltage on gate. NB: If you suspect the oscillator is operating but is clipping, replace R3 and R4 with a pot, say 10 k. This will act as an amplitude control, and will allow you to reduce the amplitude below clipping level. Report Submit a written report which should contain only the following: ● a sketch of the final circuit ● a comparison in a table of design and measured voltages, both a.c. and d.c. at significant nodes ● include your plots of the JFET characteristics ● any comments you may wish to make. UNIVERSITY Ill:SOUTHERN QUEE © University of Southern QueenslandELE2504 – Electronic design and analysis 48 © University of Southern Queensland (Source: Linear Technology , www.linear.com/docs/1704 , 31 July 2015)© University of Southern Queensland ELE2504- Electronic design and analysis 46ELE2504 – Electronic design and analysis 51 © University of Southern Queensland Part D: Electronic integrated circuits ICs (4%) Select two ICs and submit a short report for each, covering the following details: ● what the IC is (name, type) ● explanation of the IC's features and how it works ● sample circuit application making use of its feature(s) with a circuit operation explanation (include any relevant calculations). IC list: 4052 LM 335 NE 567 4028 LM 3915 LM 723 4N 29 LM 331ELE2504 – Electronic design and analysis 52 © University of Southern Queensland