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ELG4135 − Final 2004 / ELG4535 − final 2004

This document is the solution to the ELG4135 final 2004.
Ce document est la solution à l'examen final ELG4535 2004.
  1. Q1
    1. Using these values:
      Gain = ( 6V − Vofs ) / ( 2V − Vofs )
      Gain = ( 1V − Vofs ) / ( 1.9V − Vofs )
      Thus:
      ( 6V − Vofs ) / ( 2V − Vofs ) = ( 1V − Vofs ) / ( 1.9V − Vofs )
      ( 6V − Vofs ) × ( 1.9V − Vofs ) = ( 1V − Vofs ) × ( 2V − Vofs )
      11.4V² −(7.9V)Vofs + Vofs² = 2V² − (3V) Vofs + Vofs²
      9.4V² = (4.9V) Vofs
      Vofs = 9.4V²/4.9V = 1.92V
      Gain = ( 6V − 1.92V ) / ( 2V − 1.92V ) = 50
    2. (1 + R23/(R22||R21) ) = Gain = 50
      500k/(50−1) = 10.2k = R22||R21
      9V×R21/(R21+R22) = 1.92V
      R21 = 1.92V/9V × (R21+R22) = 0.21 × (R21+R22)
      R21 = 0.21 / (1−0.21) R22 = 0.27 R22
      Thus, 10.2k = 1/( 1/R22 + 1/(.27 R22) )
      R22/10.2k = (1 + 1/.27) = 4.69
      R22= 4.69×10.2k = 47.8k
      R21= 0.27×R22 = 12.9k
    3. 3dB point for op amp gain is fT/ GFBN.
      Freq = 2.0MHz / (1 + 50) = = 39 kHz
    4. Neglect V− since IB=0.
      V+ of inner ideal op-amp is 2V − 50n×500k = 1.995V
      →Approach #1 Vo2 = (1.995V - 1.92V)×50 + 1.92 = 5.75V
      →Approach #2 Vo2 = 6V for Vi2 = 2V. Thus Vo2 = 6V + (−0.005V)×50 = 5.75V
  2. Q2
    1. The Schmitt trigger comparator is better in a case where there is significant noise. Because it has two thresholds, it will only trigger once for a noise level below the difference in thresholds.
  3. Q3
    1. VT = 9V×20k/(20k+20k) = 4.5V
    2. V+ begins at 4.5V. After Vo4 makes a transition from 8V to 1V, V+ = 4.5V − (8V − 1V) = −2.5V; V+ then returns to 4.5V as an exponential with time constant (τ) = 100n×(20k||20k) = 1ms
    3. At t=0, V+ = 4.5V; Vi4 < V+, so Vo4 = L+.
      All signals are constant until t=10ms
      At t=10ms, Vi4 > V+; Vo4 switches to L−.
      V+ changes by L+−L−, thus immediately after t=10ms, V+= 4.5 − (8V−1V) = −2.5V.
      V+ then exponentially recovers to 4.5V.
      At that point V−=0V. When V+ > V−, Vo4 returns to L+.
    4. At t=10ms, transition of Vo4 to L−
      Calculate transition time of Vo4 to L+:
      T= τ ln ( (Vt=∞ − Vt=0) / (Vt=∞ − Vt=T) )
      T= 1ms × ln( (4.5V−( −2.5V) ) (4.5V−0V) ) = 0.44 ms
      At t=10.44ms, transition of Vo4 to L+
  4. Q4
    1. ε = √(10Amax/10−1) = 0.35
      A(ω) = 40dB = 10 log( 1 + 0.35²(ω/ωp)2N )
      (f/fp)2×3 = (1040/10−1)/(.35²) = 8.2×104
      (f/fp)= ( 8.2×104 )1/6 = 6.59
      fs = fp×6.59 = 65.9Hz
    2. (s − p1)(s − p2) = s² + (ω0/Q)s + ω0²
      = (s + 2π (2+j9)(s + 2π (2−j9)
      = s² + 25.1s + 3356
      ω0 = √(3356) = 57.9 rad/s
      Q = ω0/25.1 = 2.31
    3. Low pass RLC filter has T(s) = 1/LC / ( s² + s/CR + 1/LC )
      1/CR = 25.1
      C= 1/100Ω/25.1 = 0.4mF
      1/LC = 3356
      L= 1/0.4mF/3356 = 0.75H
    4. It would be cheaper to build an active filter. The values of 0.75H and 0.4mF are very large and expensive compared to a component such as an Antoniou inductor replacement.
  5. Q5
    1. VBE,Q62 = ILR61
      R61 = 0.6V / 10mA = 60Ω
    2. →Approach #1: When Vi6 is large, then VE,Q61 = 9V - VCE,sat = 8.8V
      8.8V = (R61 + RL)×10mA
      RL= 8.8V / 10mA − 60Ω = 820 Ω
      →Approach #2: Maximum Vi6 is from Monostable is L+ = 8V.
      Thus VE,61= 8V - 0.6V = 7.4V
      7.4V = (R61 + RL)×10mA
      RL= 7.4V / 10mA − 60Ω = 680 Ω

Last Updated: $Date: 2005-01-03 13:44:28 -0500 (Mon, 03 Jan 2005) $