
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.
 Q1

Using these values:
Gain = ( 6V − V_{ofs} ) / ( 2V − V_{ofs} )
Gain = ( 1V − V_{ofs} ) / ( 1.9V − V_{ofs} )
Thus:
( 6V − V_{ofs} ) / ( 2V − V_{ofs} ) =
( 1V − V_{ofs} ) / ( 1.9V − V_{ofs} )
( 6V − V_{ofs} ) × ( 1.9V − V_{ofs} ) =
( 1V − V_{ofs} ) × ( 2V − V_{ofs} )
11.4V² −(7.9V)V_{ofs} + V_{ofs}² =
2V² − (3V) V_{ofs} + V_{ofs}²
9.4V² = (4.9V) V_{ofs}
V_{ofs} = 9.4V²/4.9V = 1.92V
Gain = ( 6V − 1.92V ) / ( 2V − 1.92V )
= 50

(1 + R23/(R22R21) ) = Gain = 50
500k/(50−1) = 10.2k = R22R21
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

3dB point for op amp gain is f_{T}/ G_{FBN}.
Freq = 2.0MHz / (1 + 50) = = 39 kHz

Neglect V_{−} since I_{B}=0.
V_{+} of inner ideal opamp 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
 Q2
 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.
 Q3

V_{T} = 9V×20k/(20k+20k) = 4.5V

V_{+} begins at 4.5V. After V_{o4}
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×(20k20k) = 1ms

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_{+}.

At t=10ms,
transition of Vo4 to L_{−}
Calculate transition time of Vo4 to L_{+}:
T= τ ln (
(V_{t=∞} − V_{t=0}) /
(V_{t=∞} − V_{t=T}) )
T= 1ms × ln( (4.5V−( −2.5V) )
(4.5V−0V) ) = 0.44 ms
At t=10.44ms,
transition of Vo4 to L_{+}
 Q4

ε = √(10^{Amax/10}−1) = 0.35
A(ω) = 40dB = 10 log( 1 +
0.35²(ω/ω_{p})^{2N} )
(f/f_{p})^{2×3} =
(10^{40/10}−1)/(.35²) = 8.2×10^{4}
(f/f_{p})=
( 8.2×10^{4} )^{1/6}
= 6.59
f_{s} = f_{p}×6.59 = 65.9Hz

(s − p_{1})(s − p_{2})
= s² + (ω_{0}/Q)s
+ ω_{0}²
= (s + 2π (2+j9)(s + 2π (2−j9)
= s² + 25.1s + 3356
ω_{0} = √(3356) = 57.9 rad/s
Q = ω_{0}/25.1 = 2.31

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

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.
 Q5

V_{BE,Q62} = I_{L}R_{61}
R_{61} = 0.6V / 10mA = 60Ω

→Approach #1:
When Vi6 is large, then V_{E,Q61}
= 9V  V_{CE,sat} = 8.8V
8.8V = (R61 + R_{L})×10mA
R_{L}= 8.8V / 10mA − 60Ω = 820 Ω
→Approach #2:
Maximum Vi6 is from Monostable is L_{+} = 8V.
Thus V_{E,61}= 8V  0.6V = 7.4V
7.4V = (R61 + R_{L})×10mA
R_{L}= 7.4V / 10mA − 60Ω = 680 Ω
Last Updated:
$Date: 20050103 13:44:28 0500 (Mon, 03 Jan 2005) $
