SOLUTION FOR END-OF-TERM EXAM IN TERM 2 OF SCHOOL YEAR 2018 - 2019
SENTENCE 1 ( 2 M )
Full wave rectifier circuit
Average output voltage is calculated by:
2 ( 22 - 1.4 )
V0(dc) = = 18.91 V
From Ohm 's law, load current has average value as: IL = 18.91 V / 0.1 K = 189.1 mA
SENTENCE 2 ( 2 M )
Clearly, Z3 OFF and Z1, Z2 ON at the same time.
In addition, permissible maximum current flowing through Z1 and Z2 in series is calculated as:
IZmax = 120 mW / 5 = 24 mA for ensuring reliability in service.
It can be noticed that output voltage has constant amount such as: V0 = 10V
As It's known, current through resistor Rs keeps constant regardless of load resistance
In other words, it's represented as: Is = ( 50 - 10 ) / 1K = 40 mA = const
Further more, according to Kirchhoff 's current law, it can be written as:
Is = Iz + IL = const
Where,
W Iz is current through the branch including Z1 and Z2 connected in series.
h
And IL is current through load RL.
e
It's
r easy to see that, due to Is = const, we get the following expressions:
e
, Is = Izmin + ILmax = const
Is = Izmax + ILmin = const
As a result, maximum load current ILmax = 40 mA.
And minimum load current ILmin = 40 - 24 = 16 mA
Briefly, the range of load current can be depicted such as: 16 mA < IL < 40 mA
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SENTENCE 3 ( 2M )
= 5.4V
= 6K
IB = (5.4 - 0.7) / (6K + 10K) = 0.29mA
Hence, IC = IE = 29mA
From KVL, it results in:
VCE = VCC - (RC + RE)IC = - 0.4V
It can be concluded that BJT is in
saturation mode.
SENTENCE 4 ( 2M )
As it's given, Vi and Vo include ac and dc components.
In details, Vi = 0.2exp(j30) + 0.5
with Vi(ac)= 0.2exp(j30) [V] and Vi(dc)= 0.5V
Vo = 4exp(j210) + 10
where Vo(ac) = 4exp(j210) = - 4exp(j30) [V]
and Vo(dc) = 10V
By superposition,
OPAMP 1 operating as noninverting amplifier
and OPAMP 2 as inverting amplifier.
As a result,Vo1 = 6Vi1 and Vo2 = - 4Vi2
Hence, we get equation systems as following
Vi1 - Vi2 = 0.5
( dc quantities )
6Vi1 + 4Vi2 = 10
Vi1 - Vi2 = 0.2exp(j30)
6Vi1 + 4Vi2 = - 4exp(j30) ( ac quantities )
Solve with matrix of determinant, it can be described as:
Vo1 = - 1.92exp(j30) + 7.2 [V] and Vo2 = 2.08exp(j30) - 2.8 [V]
SENTENCE 5 ( 2M )
Output Boolean expression is determined by:
3 variable K map is represented as following:
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