Transcript
SYSC3203 Prelab 3A: Fall 2017
Question: A 2.6 kΩ
Student# 0390
25 kΩ
−
Vo +
2.6 kΩ
25 kΩ
Vi For the circuit above, Vi = 10 mV: • What is Vo if the amplifier is ideal? • What is Vo if the offset voltage, VOS = 10 µV? • What is Vo if the bias current, IB = 10 nA?
• What is Vo if the amplifier is ideal? Represent ideal as V¯o 25 kΩ Vi = (9.060 mV, V+ = 25 + 2.6 kΩ
V¯o =
25 kΩ 1+ V+ = 10.615 V+ = 96.172 mV 2.6 kΩ
• What is Vo if the offset voltage, VOS = 10 µV? Use superposition to get (Vo0 ) then add to ideal VOS : Vo0
=
25 kΩ 1+ 2.6 kΩ
VOS = 10.615 × VOS = 0.106 mV
Vo = V¯o + Vo0 = 96.278 mV • What is Vo if the bias current, IB = 10 nA? First, use superposition to get (Vo0 ) for IB into V+ . Current travels through parallel resistors. Vo0
25 kΩ (R1 kR2 )IB = −10.615 × 2.355 kΩ × IB = −0.250 mV =− 1+ 2.6 kΩ
Next, use superposition to get (Vo00 ) for IB into V− . Current through R1 , since FB keeps V− at ground. Note that this resistor configuration cancels IB . Vo00 = (25 kΩ) IB = 0.250 mV Vo = V¯o + Vo0 + Vo00 = 96.172 mV
SYSC3203 Prelab 3A: Fall 2017
Question: B
Student# 0390
The op amp is ideal, except fT (= Gain·Bandwidth) is 120 kHz. 30 kΩ 2.1 kΩ
−
Vo +
2.1 kΩ
30 kΩ
Vi For the circuit above, Vi = (20 mV) cos(2πf t): • What is the peak-to-peak amplitude of Vo if f = 3 kHz? • What is the peak-to-peak amplitude of Vo if f = 30 kHz?
First, analyse ideal gain, V¯o 30 kΩ V+ = Vi = 0.935 Vi 30 + 2.1 kΩ
V¯o =
30 kΩ 1+ V+ = 15.286 V+ = 14.292 Vi 2.1 kΩ
• What is the peak-to-peak amplitude of Vo if f = 3 kHz? Given Gain-Bandwidth, maximum possible gain is G = (G · BW )/f = 120/3 = 40. We specify a gain of 14.292 which is less than 40, so we get the specified gain. Vo = 14.292 × (20 mV) cos(2πf t), and peak-peak voltage is 2 × max(Vo ). Answer: 571.7 mV. • What is the peak-to-peak amplitude of Vo if f = 30 kHz? Given Gain-Bandwidth, maximum possible gain is G = (G · BW )/f = 120/30 = 4. We specify a gain of 14.292 which is greater than 4, so we only get a gain of 4. Vo = 4 × (20 mV) cos(2πf t), and peak-peak voltage is 2 × max(Vo ). Answer: 160.0 mV.
SYSC3203 Prelab 3A: Fall 2017
Question: C
Student# 0390 Vo 10.7 µF
29
6.7 mH
− +
Vi
8Ω
For the circuit above: • What type of filter is this? (high pass, low pass, band pass, band stop) • Sketch the amplitude of rate.
Vo Vi
as a function of frequency. Label the passband, stopband and roll-off
• What is the cut-off frequency (fc ) and damping constant (ζ)?
• What type of filter is this? (high pass, low pass, band pass, band stop) This is a high pass filter • What is the cut-off frequency (fc ) and damping constant (ζ)? ωc = √
1 1 = 3734.829 rad/s, =√ 6.7 mH · 10.7 µC LC
fc = 2πωc = 23466.132 Hz
and, R ζ= 2 • Sketch the amplitude of rate. Vo Vi
Vo Vi
r
C 298 kΩ = L 2
r
10.7 µF = 5.954 6.7 mH
as a function of frequency. Label the passband, stopband and roll-off
start low and increases at 40 dB/decade. After fc , graph is at 1.0.
