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Ras Al Khaimah

Presentation On Electronic Devices

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Published in: Electronics
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Nithya P / Ajman

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Qualification: B Tech in Electronics And Communication Engineering, MBA in E Business

Teaches: Electronics, Physics, Engineering, Electrical Technology, Chemistry, Computer Science, Maths, Mathematics

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  1. Electronic Devices Ninth Edition Floyd LO-4 ELE-3413: Electronics-Il
  2. Sum"/ Basic filter Responses A filter is a circuit that passes certain frequencies and rejects all others. The passband is the range of frequencies allowed through the filter. The critical frequency defines the end (or ends) of the passband. Basic filter responses are: Gain Low-pass Gain Gain Band-pass Gain High-pass Band-stop
  3. Sum"/ The Basic Low-Pass Filter The low-pass filter allows frequencies below the critical frequency to pass and rejects other. The simplest low-pass filter is a passive RC circuit with the output taken across C. Gain (nonnalimd to l) O dB -3dB - P as sban d -20 dB -40 dB -60 dB 001 fe 0.1 fr Actual response of a single-poleRC filter Transition on Stopband region IOC 100K 1000fc
  4. Sum"/ The Basic High-Pass Filter The high-pass filter passes all frequencies above a critical frequency and rejects all others. The simplest high-pass filter is a passive RC circuit with the output taken across R. Gain (nonmlized to l) -3dB-0-d-B- Actual response of a single-pole Passband — -20 dB _ 40 dB -60 dB RC filter 0001fc 001 C 0.1K Vs 100K
  5. Sum"/ The Band-Pass Filter A band-pass filter passes all frequencies between two critical frequencies. The bandwidth is defined as the difference between the two critical frequencies. The simplest band-pass filter is an RLC circuit. Vow (nornulized to I) 0.707 Vs c L
  6. Sum"/ The Band-Stop Filter A band-stop filter rejects frequencies between two critical frequencies; the bandwidth is measured between the critical frequencies. The simplest band-stop filter is an RLC circuit. Gain (dB) fo fez
  7. Sum"/ Active Filters Active filters include one or more op-amps in the design. These filters can provide much better responses than the passive filters illustrated. Active filter designs optimize various parameters such as amplitude response, roll-off rate, or phase response. Chebyshev: rapid roll-off characteristic Butterworth: flat amplitude response Bessel: linear phase response
  8. Sum"/ The Damping Factor The damping factor primarily determines if the filter will have a Butterworth, Chebyshev, or Bessel response. DF = 2 — _L The term pole has mathematical significance with the higher level math used to develop the DF values. For our purposes, a pole is the number of non-redundant reactive elements in a filter. For example, a one-pole filter has one resistor and one capacitor. uency- selective RC ciltuit Anvlifier Negati ve feedback circuit
  9. Basically, the damping factor affects the filter response by negative feedback action. Any attempted increase or decrease in the output voltage is offset by the opposing effect of the negative feedback. This tends to make the response curve flat in the passband of the filter if the value for the damping factor is precisely set. By advanced mathematics, which we will not cover, values for the damping factor have been derived for various orders of filters to achieve the maximally flat response of the Butterworth characteristic. The value of the damping factor required to produce a desired response characteristic depends on the order (number of poles) of the filter. A pole, for our purposes, is simply a circuit with one resistor and one capacitor. The more poles a filter has, the faster its roll-off rate is. To achieve a second-order Butterworth response, for example, the damping factor must be 1.414. To implement this damping factor, the feedback resistor ratio must be 1.414 — 0.586 This ratio gives the closed-loop gain of the non-inverting amplifier portion of the filter Aci(N1), a value of 1.586, derived as follows: Ac/(NI) 0.586 + I I .586
  10. The critical frequency is determined by the values of the resistors and capacitors in the frequency-selective RC circuit shown in the previous figure. For a single- pole (first-order) filter, the critical frequency is: 1 27TRC The same formula is used for thefc of a single pole high-pass filter. The number of poles determines the roll-off rate of the filter. So, a first-order (one-pole) filter has a roll-off of -20 dB/decade; a second-order (two-pole) filter has a roll-off rate of -40 dB/decade; and a third-order (three-pole) filter has a roll-off rate of -60 dB/decade and so on. Single-pole low-pass circuit c FIGURE 15-7 First-order (one-pole) low-pass filter.
  11. Generally, to obtain a filter with three poles or more, one-pole or two-pole filters are cascaded, as shown in Figure 15—8. To obtain a third-order filter, for example, cascade a second-order and a first-order filter; to obtain a fourth-order filter, cascade two second-order filters; and so on. Each filter in a cascaded arrangement is called a stage or section. RC circuit RC circuit circuit A FIGURE 15-8 The number of filter poles can be increased by cascading.
  12. Sum"/ The Damping Factor Parameters for Butterworth filters up to four poles are given in the following table. (See text for larger order filters). Butterworth filter values Order dB/decadé 1 2 3 4 -20 -60 -80 Poles 1 2 2 2 Optional .414 1.00 1.848 R,/R2 0.586 1.00 0.152 Poles 2 1.00 0.765 RI/R2 1.00 1.235 Notice that the gain is I more than this resistor ratio. For example, the gain implied by this ratio is 1.586 (4.0 dB).
  13. Single-pole Filter A single-pole filter is an active filter with a single low-pass RC frequency- selective circuit that provides a roll-off of -20dB/decade above the critical frequency, as indicated by the response curve in Figure 15—9(b). The critical frequency of the single-pole filter isfc = 1/(27tRC). The op-amp in this filter is connected as a non-inverting amplifier with the closed-loop voltage gain in the passband set by the values of RI and R2. One pole A FIGURE 15-9 c/(NI) — R 2 c Gain (dB) 1 0 -3 20 (b) -20 dB/decade Single-pole active lowpass filter and response curve.
  14. Sum"/ Two-pole Low-Pass Butterworth Design As an example, a two-pole VCVS (voltage-controlled voltage source) Butterworth filter is designed in this and the next two slides. Assume thefc desired is 1.5 kHz. A basic two-pole low-pass filter is shown. Step I : Choose R and C for the desired cutoff frequency based -1 on the equation f 2ÆRC By choosing R = 22 kQ, then C = 4.8 nF, which is close to a standard value of 4.7 nF. 22 4.7 nF 22 4.7 nF
  15. Sum"/ Two-pole Low-Pass Butterworth Design Step 2: Using the table for the Butterworth filter, note the resistor ratios required. Butterworth filter values dB/decade Order 2 3 4 -20 -60 -80 Poles 2 2 2 Optional .414 1.00 1.848 0.586 1.00 0.152 Poles 1 2 1.00 0.765 RI/R2 1.00 1.235 Step 3: Choose resistors that are as close as practical to the desired ratio. Through trial and error, if RI = 33 kQ, then R2 = 56 kQ.
  16. Sum"/ Two-pole Low-Pass Butterworth Design The design is complete and the filter can now be tested. You can check the design using Multisim. The Multisim Bode plotter is shown with the simulated response from Multisim. I Set„, I I $47 22 4.7 nF 22 4.7 nF 33 kQ 56 kQ To read the critical frequency, set the cursor for a gain of I dB, which is — 3 dB from the midband gain of 4.0 dB. The critical frequency is found by Multisim to be 1.547 kHz.
  17. Sum"/ Four-pole Low-Pass Butterworth Design What changes need to be made to change the two-pole low-pass design to a four-pole design? Add an identical section except for the gain setting resistors. Choose RI-R4 based on the table for a 4-pole design. 4.7 nF 22kQ 22kQ cm 22 kQ 3.3kQ The resistor ratio for the 1st section needs to be 0.152 (gain = 1.152); the 22 2nd section needs to be 1.235 (gain = 2.235). Use standard values if possible. 4. nF 22 kQ 15 kQ 12kQ
  18. EXAMPLE 15-3 FIGURE 15-11 Solution Determine the critical frequency of the Sallen-Key low-pass filter in Figure 15—11, and set the value of RI for an approximate Butterworth response. 1.0 kn 1.0 kQ 0.022 "F CD 0.022"F 1.0kO Since RA = = R = 1.0kO andCA = C = 0.022 - 7.23kHz 27RC For a Butterworth response, RI/R2 = 0586. RI = 0.586R2 = 0.586(1.OkQ) 5860 Select a standard value as near as possible to this calculated value.
  19. Cascaded Low-Pass Filters A three-pole filter is required to get a third-order low-pass response (-60 dB/decade). This is done by cascading a two-pole Sallen-Key low-pass filter and a single-pole low-pass filter, as shown in Figure 15—12(a). Figure 15—12(b) shows a four-pole configuration obtained by cascading two Sallen-Key (2-pole) low-pass filters. In general, a four-pole filter is preferred because it uses the same number of op-amps to achieve a faster roll-off.
  20. Cascaded Low-Pass Filters FIGURE 15-12 Cascaded lowpass filters. (a) cmfigwalön
  21. EXAMPLE 15-4 Solution For the four-pole filter in Figure 15—12(b), determine the capacitance values required to produce a critical frequency of 2680 Hz if all the resistors in the RC low-pass circuits are I .8 kQ. Also select values for the feedback resistors to get a Butterworth response. Both stages must have the same L. Assuming equal-value capacitors, 27Rfc - 0.033 27(1.8 Hz) CAI = cm = cm = cm = 0.033 Also selectR2 = R4 = 1.8 kO for simplicity. Refer to Table 15—1. For a Butterworth response in the first stage, DF = 1.848 and RI/R2 = 0.152. Therefore, RI = 0.152R2 = 0.152(18000) = 2740 Choose RI = 270 0. In the second stage, DF= 0.765 and R3/R4 = 1.235. Therefore, R3 = 1.235R4 = 1235(18000) = 2.22 kn Choose R3 2.2 kn. 2 poles
  22. Sum"/ High-Pass Active Filter Design The low-pass filter can be changed to a high-pass filter by simply reversing the R's and C's in the frequency-selective circuit. For the four-pole design, the gain setting resistors are unchanged. 4.7 nF 22 kO 4.7 nF 22kO 4.7 nF 3.3kQ 22 kQ 4.7 nF High-pass 22 22 12kQ
  23. EXAMPLE 15-5 Solution Choose values for the Sallen-Key high-pass filter in Figure 15—15 to implement an equal-value second-order Butterworth response with a critical frequency of approxi- mately 10 kHz. Start by selecting a value for RA and RB (RI or R2 can also be the same value as RA and RB for simplicity). R = RA = RB = R2 = 3ßkQ (an arbitrary selection) Next, calculate the capacitance value fromfc 1/(27RC). = 0.0048 For a Butterworth response, the damping factor must be 1.414 and RI/R2 = 0.586. RI = 0.586R2 = 0.586(3.3 = 1.93kQ If you had chosen RI = 3.3 kO, then 0.586 3.3kO 563 kO 0.586 Either way, an approximate Butterworth response is realized by choosing the nearest standard values. high-p»s circuit „O_IA < FIGURE IS-IS Basic Sallen•Key high-pass filter.
  24. Cascading High-Pass Filters As with the low-pass configuration, first- and second-order high-pass filters can be cascaded to provide three or more poles and thereby create faster roll-off rates. Figure 15—16 shows a six-pole high-pass filter consisting of three Sallen-Key two-pole stages. With this configuration optimized for a Butterworth response, a roll-off of -120dB/decade is achieved. cm RA2 cm A FIGURE 15-16 Sixth-order high-pass filter. cm
  25. Cascaded Low-Pass and High-Pass Filters One way to implement a band-pass filter is a cascaded arrangement of a high-pass filter and a low-pass filter, as shown in Figure as long as the critical frequencies are sufficiently separated. Each of the filters shown is a Sallen-Key Butterworth configuration so that the roll-off rates are -40dB/decade, indicated in the composite response curve of Figure 15—17(b). The critical frequency of each filter is chosen so that the response curves overlap sufficiently, as indicated. The critical frequency of the high-pass filter must be sufficiently lower than that of the low-pass stage. This filter is generally limited to wide bandwidth applications. The lower frequency fcl of the passband is the critical frequency of the high-pass filter. The upper frequency fc2 is the critical frequency of the low-pass filter. Ideally, as discussed earlier, the center frequency fo of the passband is the geometric mean of andfc2. The following formulas express the three frequencies of the band-pass filter in Figure 15—17: fcl fc2 — 27 CBI 27 RA2RB2CA2CB2
  26. Cascaded Low-Pass and High-Pass Filters FIGURE 15-17 Band-pass filter formed by cascading a two-pole high-pass and a ngo-pole low-pass filter (it does not matter in which order the filters are cascaded). •my-pole high-pass Low-pass response •IWo-pole low-pass High-pass response fcl fc2 — RA2RB2CA2CB2 fclfc2
  27. Sum"/ Active Band-Pass Filters One implementation of a band-pass filter is to cascade high-pass and low-pass filters with overlapping responses. These filters are simple to design, but are not good for high Q designs. Low-luss response OdB -3 dB High-pass response
  28. Sum"/ Active Band-Pass Filters The multiple-feedback band-pass filter is also more suited to low-Q designs ( <10) because the gain is a function of Q2 and may overload the op-amp if Q is too high. Resistors RI and RS form an input attenuator network that affect Q and are an integral part c, of the design. Key equations are: 1 Rl+R3 21tc RIR2R3 Q=Æf0CR2 2R RI BW
  29. Sum"/ Filter Measurements Filter responses can be observed in practical circuits with a swept frequency measurement. The test setup for this measurement is shown here. The sawtooth waveform synchronizes the oscilloscope with the sweep generator. Oscill oscope filter g enerator Sawtooth output
  30. OJI!J. OA!JOE ssed ANOEI (0BJ?(!l/åp@QA&€f:' .111
  31. •Interactive:NiaeeVIGV.g 2nd order Butterworth filter LPF and HPF Butterworth Filter Design Vout
  32. Selected Pole A circuit containing one resistor and one capacitor that contributes —20 dB/decade to a filter's roll-off. Roll-off The rate of decrease in gain below or above the critical frequencies of a filter. Damping factor A filter characteristic that determines the type of response.
  33. Qui///' 1. The green line represents the response for a a. Butterworth filter b. Chebychev filter c. Bessel filter
  34. Qui///' 2. The blue line represents the response for a a. Butterworth filter b. Chebychev filter c. Bessel filter
  35. Qui///' 3. The filter that is superior for its pulse response is the a. Butterworth filter b. Chebychev filter c. Bessel filter
  36. Qui///' 4. From the table for a 4-pole Butterworth filter, the gain required for the second stage is a. 0.765 c. 1.765 Butterworth filter values •oil-o Order 1 2 3 4 -20 -40 -80 Poles 2 2 2 Optional 414 1.00 1.848 b. 1.235 d. 2.235 0.586 1.00 0.152 Poles 2 1.00 0.765 RI/R2 1.00 1.235
  37. Qui///' 5. For a 2-pole Butterworth filter, assume that RI = 39 kQ From the choices given, the best value for R2 is a. 22 ICQ c. 56 kQ Butterworth filter values •oil-o Order 1 2 3 4 -20 -40 -80 Poles 2 2 2 Optional 414 1.00 1.848 b. 27 d. 68 ICC) 0.586 1.00 0.152 Poles 2 1.00 0.765 RI/R2 1.00 1.235
  38. Qui///' 6. The type of active filter shown is a a. two-pole, low-pass b. two-pole, high-pass c. four-pole, low-pass d. four-pole, high-pass
  39. Qui///' 7. The approximately roll-off for the filter shown is 22 a. -20 dB/decade c. -60 dB/decade 4.7 nF REI 22 Q cm 4.7 nF b. -40 dB/decade d. -80 dB/decade 4. nF 22 kQ 22kQ 3.3kQ 22 15kQ 12kQ
  40. Qui///' 8. A good choice for a high-Q active band-pass filter is a. cascaded high-pass and low-pass filters b. a multiple-feedback band-pass filter c. a state-variable band-pass filter d. an inverting amplifier with a resonant filter
  41. Qui///' 9. The filter shown forms a a. band-stop filter b. band-pass filter State-van able HP filter c. low-pass filter d. high-pass filter
  42. Qui///' 10. For the swept-frequency measurement, the signal on the X-channel of the oscilloscope is a a. sine wave that changes frequency b. sawtooth wave Osci Il c. square wave d. dc level Filter
  43. Qui#h Answers: 10. b
  44. Homework Solve the following problems from the textbook. Chapter 15, Pages 801-805, Problems: 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

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