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Malik - Formal Element Report

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Malik - Formal Element Report

  1. 1. Objective: Our objective for this laboratory was to investigate a Sallen and Key second order active low pass filter and to obtain the transient analysis using our modelled circuit and pspicesimulation. The circuit has two RC networks, R1 and C1, R2 and C2. The frequency responseof the second order low pass filter is similar to the first order type except that the roll-off will be twice that of the firstorder filters, at 40dB/decade as the operating frequency increases above the cut-off frequency. Malik: Formal Element Report Junaid Malik Formal Element, Signals and Systems 1, DT008/2 Group A
  2. 2. Circuit Design: Figure 1 Simulation settings to obtain transient analysis U1 uA741 + 3 - 2 V+ 7 V- 4 OUT 6 OS1 1 OS2 5 R1 100k R2 4.7k R3 100k R4 10k C1 820p C2 820p V2 15Vdc V3 15Vdc 0 0 neg out pos neg Junaid Malik, DT008/2, Group A, 6/11/12 V4 FREQ = 1.7k VAMPL = 1 VOFF = 0 AC = 1 pos VDB
  3. 3. Reference2 R1= 100k R2=100k C1= 820p C2 820p R4=4.7k R3=10k Vin(s)→ →Vout(s) Transfer Function G(s)= 218619869.126 s2+18658.5365854s+148720999.405 Cut-off frequency fc = 1940.91394015 (Hz) Quality factor Q = 0.653594771242
  4. 4. Figure 2: This was the layout which we followed while constructing the Sallen and Key Second- order active. Reference Paul Tobin.
  5. 5. Figure 3: This is a good layout example reference Paul Tobin.
  6. 6. Deriving the equation: An active low-pass filter is required to meet the following specification:  The maximum passband loss Amax = 0.5dB  The minimum stopband loss Amin = 12dB  The passband edge frequency ωp = 100 rs−1( f p = ωp/2π = 15.9 Hz)  The stopband edge frequency ωs = 400 rs−1 We had to obtain a Butterworth transfer function so that the Sallen and Key active filter circuit could meet the required specification. The filter order is calculated as follows => 𝑛 = log10 [ 10(0.1𝑥𝐴𝑚𝑖𝑛) − 1 10(0.1𝑥𝐴𝑚𝑎𝑥) − 1 ] 2𝑙𝑜𝑔10 ( 𝑤𝑠 𝑤𝑝 ) 𝑛 = log10 [ 10(0.1𝑥12) − 1 10(0.1𝑥0.5) − 1 ] 2𝑙𝑜𝑔10 ( 400 100 ) 𝑛 = log10[ 14.8489 0.12207 ] 2𝑙𝑜𝑔10(4) 𝑛 = 2.08529 1.204119 𝑛 = 1.731 𝑛 ≈ 2
  7. 7. 𝑆𝑖𝑛𝑐𝑒 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑅𝑏 = 10𝑘Ω 𝑎𝑛𝑑 𝑅𝑎 = 4.7𝑘Ω 𝑘 = 1 + 𝑅𝑎 𝑅𝑏 = 1.47𝑘 𝐴𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑡ℎ𝑎𝑡 𝑎 𝑣𝑎𝑙𝑢𝑒 𝑓𝑜𝑟 𝑅𝑏 = 10 𝑘 𝑎𝑛𝑑 𝑅𝑎 = 4.7 𝑘, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑔𝑎𝑖𝑛 𝑖𝑛 𝑑𝐵 𝑖𝑠 𝑘|𝑑𝐵 = 20 log(1.47) = 3.346 𝑑𝐵 𝑇ℎ𝑒 𝑙𝑜𝑤 𝑝𝑎𝑠𝑠 𝑓𝑖𝑙𝑡𝑒𝑟 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑎𝑝𝑝𝑙𝑦𝑖𝑛g 𝑡ℎ𝑒 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑖𝑣𝑖𝑑𝑒𝑟 𝑡𝑒𝑐ℎ𝑛𝑖𝑞𝑢𝑒. 𝑆𝑒𝑐𝑜𝑛𝑑 𝑜𝑟𝑑𝑒𝑟 ∶ $2 + 1.414$ + 1 𝐻($) = 1 $2 + 1.414$+ 1 𝐻( 𝑠) = 1 ( 𝑠 𝜔𝑐 ) 2 + 1.414( 𝑠 𝜔𝑐 )+ 1 = 𝜔𝑐2 𝑠2 + 1.414𝜔𝐶𝑠 + 𝜔𝑐2 𝑇𝑓 = 𝑉𝑜( 𝑠) 𝑉𝑖𝑛( 𝑠) = ( 𝑘 𝐶2 𝑅2) 𝑠2 + ( 3 − 𝑘 𝐶𝑅 ) 𝑠 + ( 1 𝐶2 𝑅2) 𝜔𝑐 = 1 𝐶𝑅 = 2𝜋𝑓𝑐 = 1 𝐶𝑅 = 𝑓𝑐 = 1 2𝜋𝐶𝑅 𝜔𝑐 𝑄 = 3 − 𝑘 𝐶𝑅 = 𝑄 = 1 3 − 𝑘
  8. 8. Results: TableofResults Frequency Vin Vout Vout/Vin 20log(vout/vin) 1Hz 1 1 0 0 dB 10Hz 1 1.390V 1.390V 2.86 dB 100Hz 1 1.405V 1.405V 2.95 dB 1000Hz 1 1.431V 1.431V 3.11 dB 1.2kHz 1 1.361V 1.361V 2.677 dB 1.6kHz 1 1.269V 1.269V 2.062 dB 1.8kHz 1 1.170V 1.170V 1.363 dB 2.0kHz 1 1.064V 1.064V 0.531 dB 2.2kHz 1 962.3mV 962.3mV -0.33 dB 2.4kHz 1 864mV 864mV -1.269 dB 2.6kHz 1 776.6mV 776.6mV -2.19 dB 2.8kHz 1 695.4mV 695.4mV -3.15 dB 3.0kHz 1 626.4mV 626.4mV -4.06 dB From this table of results we can see that as the frequency is increased, the out output voltage is reduced in effect.
  9. 9. Graph First graphoffrequencyagainstvoltage GRAPH 1: The frequency response above is the same as for a first order filter. The difference this time is the steepness of the roll-off which is at -40dB/decade. Secondgraphofvolatgeagainstfrequency GRAPH 2: Frequency 1.0Hz 3.0Hz 10Hz 30Hz 100Hz 300Hz 1.0KHz 3.0KHz 10KHz 30KHz 100KHz DB(V(OUT)) -80 -60 -40 -20 -0 20 Junaid Malik, DT008/2, Group A passband gain = 1 + ra/rb = 1.47 (23.806K,-40.2dB decade) (3.1331K,-6.01dB Decade) Frequency 1.0Hz 3.0Hz 10Hz 30Hz 100Hz 300Hz 1.0KHz 3.0KHz 10KHz 30KHz 100KHz V(OUT) 0V 0.5V 1.0V 1.5V Junaid Malik, DT008/2, Group A, 6/11/12 k = 1 +Ra/Rb=1.47k (1v,1.47k)
  10. 10. Conclusion: In conclusion there are many filter types and ways to implement them but an active low- pass filter that’s greatly simplified if R1=R2 and the op amp stage is a unity gain follower (RB=short and RA=open). The Sallen and Key filter attenuates any input signal in the frequency range above the cut-off frequency to a point, but then the response turns around and starts to increase in gain with frequency. The response in the stop band should keep decreasing as the frequency increases. Instead, the response actually will begin rising again at some high frequency. A properly functioning op amp is needed for the filters operation; op amps lose voltage gain at some frequency because of their finite bandwidth.
  11. 11. References Reference 1: Paul Tobin Reference 2: http://sim.okawa-denshi.jp/en/OPstool.php Reference 3: PSpice forFiltersandTransmissionLinesby Paul Tobin

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