UNIVERSITI MALAYSIA PERLIS (UniMAP)
EMT 212/4 ANALOG ELECTRONIC II
TEST 2 (1½ Hours) : Tuesday, 21st October 2008
MARKSQ1 (2) / Q2 (4) / Q3 (4) / Q4 (20) / Q5 (20) / Q6 (20) / TOTAL (70)
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Programme / :
Question 1
A certain op-amp has an open-loop differential voltage gain, Aol of 150, 000 and a common mode gain, Acm of 0.4. Determine the CMRR in decibels (dB).
[2 marks]
Answer
Aol = 150, 000 and Acm = 0.4. Therefore;
Question 2
Determine the values of Upper Threshold Point (UTP) and Lower Threshold Point (LTP) for the comparator in Figure 2. Assume that maximum and minimum values of Vout are ± 5V.
[4 marks]
Answer
Question 3
A series-shunt feedback amplifier is having the open loop gain, 200k and the closed loop gain 50. The input resistance and output resistance without feedback is 2.5kΩ and 40kΩ respectively. Determine the input and output resistance of the circuit.
[4 marks]
Answer
The value given as below :
A = 200k
Af = 50
Ri =2.5kΩ
Ro = 40kΩ
Therefore ;
Question 4
a) For the relaxation oscillator shown in the Figure 4(a), sketch and label the graph of vC against t. The labeling shall include the relevant mathematical equations describing various sections of the waveform.
[5 marks]
Answer
4(a)
Question 4
b) Assume that you have sufficient quantities of the following components:
Resistors : 10 kW, 12 kW and 20 kW.
Capacitors : 1.32 nF.
Active device : Very high open-loop gain operational amplifier.
i) Show in the form of a circuit diagram, how you would connect these components to form a Wien-Bridge sinusoidal oscillator. Label your circuit diagram with the values of all relevant components.
[5 marks]
ii) Derive the expression for the oscillation frequency, fo of the oscillator in (i) and hence calculate the oscillation frequency.
[10 marks]
Answer
4(b)(i)
(ii)
Let the equivalent impedance of the parallel combination of R and C be Zp and that of series combination be Zs i.e.;
Simplifying;
Substituting for s = jw;
The gain of the non-inverting amplifier configuration is;
Hence the loop gain is;
Since the loop gain, (bA) must be real at the oscillation frequency wo, the imaginary part of the above equation must be zero. Hence;
Therefore;
Substituting values;
Question 5
a) Give the definition of the band-pass filter and sketch and label the response curve.
[6 marks]
Anwer
5(a)
Band-pass filter is a filter that passes all signals lying within a band between a lower-frequency limit and upper-frequency limit and essentially rejects all other frequencies that are outside this specified band.
Question 5
b) Draw the band-pass filter by cascading a two-pole high-pass filter and a two-pole low-pass filter.
[6 marks]
Anwer
5(b)
By cascading a two-pole high-pass filter and a two-pole low-pass filter, a band-pass filter is constructed as follows:
Question 5
c) For the single-pole low-pass filter circuit shown in Figure 5;
R = 2.5 kΩ, C = 0.04 μF, R1 = 10 Ω, R2 = 2 Ω,
i) Determine the cutoff frequency, fc;
[1 marks]
(ii) Determine the pass band voltage gain or the gain of non-inverting amplifier, Acl(NI)
[1 marks]
iii) Prove the equation below;
where vo is the output voltage, f is the operating frequency and fc is the cutoff frequency.
[6 marks]
Answer
5(c)
(i)
The cutoff frequency for a single-pole low-pass filter is;
(ii)
The pass band voltage gain or the gain of non-inverting amplifier is calculated as follows:
(iii)
Applying the voltage divider rule, the input voltage, which is the voltage across the capacitor C, is determined as follows:
At the cutoff frequency fc, the magnitude of the capacitive reactance XC equals the resistance of the resistor R.
Substituting for 2πRC in Eq. results in the following equation for va:
Where f is the operating frequency and fc is the cutoff frequency.
The output vo is the amplified version of vi.
where Acl(NI) is the pass-band voltage gain or the gain of the non-inverting amplifier.
Question 6
(a) Explain the following terms as applied to a voltage regulator;
i) line regulation
[3 marks]
ii) load regulation
[3 marks]
Anwer
6(a)(i)
Line regulation as applied to a voltage regulator is a measure of its ability to maintain a constant output voltage despite variation in the input voltage. It is the change of output voltage per unit change in the input voltage and is normally expressed in percentage form, thus;
ii)
Load regulation as applied to a voltage regulator is a measure its ability to maintain a constant output voltage despite variation in the load connected its terminal, and is normally express in percentage form, thus;
Question 6
(b) The full-load current of a voltage regulator is 1.5 A and the output voltage at this current is 14.86 V. When the load is disconnected from the regulator, its output voltage is 15.2 V. Determine the load regulation of the regulator.
[2 marks]
Answer
6(b)
Question 6
(c) For the shunt regulator shown in Figure 6, VZ = 3.6 V.
i) Find Vo;
[4 marks]
ii) If IL and Vo are constant, find the change in IC1 if Vi changes by 1.2 V;
[4 marks]
iii) If the maximum input voltage Vi = 12 V, find IL if the load resistance, RL is shorted.
[4 marks]
Answer
6(c)(i)
(ii)
If IL and Vo are constant, IR3 will also be constant. Hence;
(iii)
When RL is shorted, Vo = 0. Therefore;
15