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The value of an a.c. voltage is defined by v(t) = 5.sin(314.t).
Determine the frequency and peak amplitude of this voltage waveform
and hence sketch the voltage waveform to scale.
A sinusoidal voltage is described mathematically by an expression of
the form
v(t) = VPEAK.sin(2.π.f.t)
where:
VPEAK represents the peak amplitude of the sinusoid
f represents the frequency of the sinusoid
Thus by comparison, VPEAK = 5 V and 314  2. . f  f  50 Hz
The voltage waveform looks like –
5
4
3
2
1
0
0
10
20
30
40
50
-1
-2
-3
-4
-5
The vertical axis is calibrated in units of Volts and the horizontal axis
is calibrated in units of mS.
The voltage defined in section (a) is applied across a 10 Ω resistor in series
with a 1.6 μF capacitor - Calculate the impedance of this series R-C
circuit
Impedance
Z = R + XC
1
j.2. . f .C
1
Z  10 
 10  j.20
j.2. .50.1.6 106
Z  R
In polar form:
  20 

Z  10 2  20 2  tan 1 
  22.4  63.4 
 10 
Calculate the peak magnitude of the current flowing in the series R-C
circuit
V
5314.t

 0.22(314.t  63.4 ) A

Z 22.4  63.4
Current magnitude = 0.22 A
I
Calculate the phase shift between the applied voltage and the current
From above, the phase shift has current leading voltage by 63.4˚.
A circuit comprises an inductor of 10 H in series with a 10 μF capacitor.
The wire used to wind the inductor has resistance of 2 Ω.
Explain the circumstances under which resonance occurs in the circuit
described above.
Resonance occurs at the frequency which gives capacitive and
inductive reactances of equal magnitude. These two components of the
overall impedance have opposite polarity and thus cancel out at
resonance. This leaves the impedance equal to the resistance in the
circuit when operating at the resonant frequency.
Calculate the resonant frequency of the circuit.
The resonant frequency is determined from the equation –
1
1
fo 

 16 Hz
2. . L.C 2. . 10.10  10 6
Calculate the Q-Factor for the inductor used in this circuit
The Q factor for an inductor is defined as –
X
2. .16.10
Q L 
 500
R
2
State how the inductor should be modified in order to improve the QFactor and explain why the modification leads to an improvement.
The inductor could be wound from a thicker gauge wire in order to
reduce the self-resistance of the coil. This might require extra turns and
the reduction in resistance might be compromised by the extra length –
but some improvement may be obtained.
T1
D1
D2
D4
D3
Vmains
RLoad
Figure 1
Figure 1 shows the circuit diagram for a simple d.c. power supply.
Identify the type of rectifier circuit represented in figure 1 and explain the
operation of the circuit with reference to the function of each component
within the circuit.
This is a bridge rectifier circuit.
The mains voltage is applied to the primary winding of the transformer
T1. This typically produces a reduced amplitude voltage on the
secondary winding – closer to the desired d.c. output voltage than the
original mains voltage. The transformer also provides electrical
isolation between the mains supply and the load.
The diodes D1 to D4 perform rectification.
On positive half cycles of the secondary voltage, D2 and D4 are
forward biased and conduct – connecting positive half cycles across
RLoad, with the top of RLoad positive.
On negative half cycles of the secondary voltage, D1 and D3 are
forward biased and conduct – connecting negative half cycles across
RLoad, with the top of RLoad still positive.
RLoad represents the circuit to which it is desired to supply d.c.
electricity.
Sketch the voltage across RLoad as a function of time showing its
relationship to the secondary voltage from the transformer.
Secondary Voltage
Load Voltage
The load voltage reaches a peak value that is approximately 1.2 V
lower than the peak value of the secondary voltage. This is due to the
forward voltage drop across two silicon diodes used in the bridge
rectifier at any time.
The rectifier circuit shown in figure 1 requires the addition of a filter to
produce a near constant d.c. voltage across RLoad. Redraw figure 1
showing where a smoothing capacitor should be connected.
T1
D1
D2
D4
D3
Vmains
RLoad
+
C
The smoothing capacitor, C, should be included us shown above.
Explain how the smoothing capacitor sustains a d.c. voltage across RLoad,
despite the pulsating nature of the rectifier output.
The capacitor charges rapidly from the transformer secondary voltage
via the diodes in the bridge when the a.c. rises towards its peak
voltage. When the a.c. has reached its peak and starts to drop again, the
capacitor holds on to its voltage and the diodes in the bridge become
reverse biased. Current can only now be delivered to the load by
discharge of the capacitor. The capacitor voltage will drop gradually
under these circumstances at a rate dependent on the capacitor value
and the load resistance value. After a short time, the a.c. starts to rise
back towards its peak and will forward bias the bridge diodes when it
exceeds the voltage to which the capacitor has dropped. The capacitor
charge will be ‘topped-up’ ready for the next half cycle of a.c.
Provided that the capacitor is large enough, its terminal voltage will
not drop substantially between peaks of the a.c. Thus, the load voltage
is held almost constant.
Calculate a value for the smoothing capacitor in order to keep the
percentage ripple voltage across RLoad below 5%. Assume a value of 500
Ω for RLoad and a mains frequency of 50 Hz.
Peak-peak ripple voltage is given by –
VLOAD NOM
VRIPPLE 
2  RLOAD  C  f
Where:
VLOAD NOM = nominal output voltage from PSU
C = the value of the smoothing capacitor
f = the frequency of the a.c. supply
VRIPPLE
1
 0.05 
VLOADNOM
2  RLOAD  C  f
1
C
 400F
2  500  0.05  50
for 5% ripple, =>
=>
The power supply shown in figure 1 is said to be unregulated. Explain the
meaning of this term and show how a three terminal regulator chip may
be used to provide a regulated output voltage.
An unregulated power supply is one where the output voltage may
change substantially from the stated nominal output voltage under
specific operating conditions. In particular, change in output voltage
may arise from –
 Fluctuations in the mains supply voltage
 Change in the amount of current drawn by the load
A three terminal regulator chip is an integrated circuit designed to
prevent changes in the output voltage occurring – within design limits.
It is used as shown in the circuit below
T1
D1
D2
U1
In
Vmains
D3
+
D4
C
Out
Gnd
RLoad
D1
S1
M
R2
Q1
V1
5.6k
Vcc
24V
5V
Figure 2
Figure 2 shows a schematic diagram of a bipolar junction transistor
(BJT) used to switch a motor on and off. The BJT is specified with βDC =
100 and BVCEO = 40 V. The motor presents a load resistance of 300 Ω
when running.
Calculate the base current in the BJT when S1 is closed
The base current may be calculated from –
V
V
5V  0.6V
I B  Control BE 
 786A
RB
5.6k
Calculate the base current in the BJT when S1 is opened
The base current will be zero
Sketch to scale an approximation to the collector characteristic curves for
the BJT used in this circuit when the switch is open and closed and
identify on the characteristic the active region, the saturation region and
the breakdown region.
Collector Characteristics
100
80
IC
60
40
20
0
-5
5
15
25
35
45
VCE
Char S1 Closed
Char S1 Opened
Load Line
Set out the load-line equation for the circuit and plot this on the same
graph as the collector curve. Hence or otherwise determine the load
current flowing when the BJT is switched on.
The load line equation is –
VCC  VCE
RLoad
This may be plotted by finding two points and joining up – see above.
IC 
The load current is got by finding the intersection of the load line with
the characteristic curve – i.e. approximately 80 mA.
With reference to typical values where necessary, estimate the power
dissipation in the BJT when switched on.
Typically we might expect about 0.3 volts between collector and
emitter when turned fully on. Given 80 mA of load current, power is –
P  0.3V  80mA  24mW
State the circumstances under which the power dissipation in the BJT is
at a maximum and hence calculate the maximum instantaneous power
dissipation in the BJT.
Maximum power dissipation occurs in the BJT as it switches from
fully on to fully off. Specifically, there will be an instant during the
switching action where VCE = VCC/2 and load current is one half of the
fully on current. The value of this maximum instantaneous power is
V
I
PPEAK  CC  MAX  12V  40mA  0.48W
2
2
Explain the consequences for the BJT if a large amount of power is
dissipated and describe a technique that may be used to minimise the
effect.
Excessive power dissipation in a BJT will lead to the device heating up
to a point where the temperature of the device exceeds the maximum
rated operating temperature. In this case the device may fail or at least
the operating lifetime of the device may be shortened.
To reduce the heating, the device may be fitted with a heatsink to
conduct and radiate heat away from the device.
Explain the reason for including the diode D1 in the circuit.
The motor is an inductive load that the transistor is required to switch
on and off. Switching the current in an inductor will cause it to
generate a large voltage in an attempt to prevent the current changing.
This inductive voltage could potentially be large enough to damage the
BJT. D1 is connected so as to present a low impedance to the induced
voltage and hence protecting the BJT. The diode is reverse biased
under static operating conditions and does not affect normal operation
of the circuit. D1 is often called a flywheel diode.
A Darlington circuit is sometimes used in place of the BJT in switching
applications. Sketch a Darlington switching circuit and explain the
advantages associated with the circuit compared to a single BJT.
M
S1
R2
Vcc
V1
The Darlington circuit uses two BJTs. The first one provides base
current to the second one from its emitter. Thus the collector current of
the second BJT is related to the base current of the first BJT by –
I C 2  1  2  I B1
I.e. the effective current gain of the combined BJTs is equal to the
product of the individual current gains of the BJTs. The result is a
device with a much higher current gain than is possible with a single
BJT. This allows even smaller control currents to control larger load
currents.