Download Write out the equation for Ohms law showing the relationship

Question 1
DT002 Power Conditioning January 2007 – Answers
Page 1 of 10
R1
27 Ω
E
30 V
R2
10 Ω
Figure 1
Write out the equation for Ohms law showing the relationship between voltage, current and
resistance in an electrical circuit.
Voltage  Current  Re sis tan ce
3 marks out of 33
Given a resistance of 220 Ω connected to an e.m.f. of 10 V
Calculate the expected current flow through the resistor.

I
V
R

10V
 45.5mA
220
Explain the likely consequences of increasing the value of the resistor to 270 Ω.
Increasing R to 270 Ω will cause a reduction in current flowing in the resistor and a consequent
reduction in power dissipation
The resistor is replaced by one with a value of 22 Ω. Calculate the new value of current flow and the
required power rating for the resistor.
V
10V

 455mA
R
22
Power  I 2  R  0.455 A 2  22
I

 4.55W
6 marks out of 33
With reference to figure 1:
Calculate the current flow in resistor R1
Question 1
DT002 Power Conditioning January 2007 – Answers
I

E

R1  R 2
Page 2 of 10
30V
 0.81A
37
Calculate the voltage drop across resistor R2
V
 I  R  0.81A  10  8.1V
Calculate the power dissipation in resistor R1
Power  I 2  R1  0.81A 2  27
 17.7W
Calculate the power dissipation in resistor R2
Power  I 2  R 2  0.81A 2  10
 6.56W
10 marks out of 33
Comment on the consequences of using a 10 Ω resistor with a power rating of 5 watts as R2 in figure
1
The power dissipation for R2 has been calculated as 6.56W. A device with a rated power of 5W will
heat up excessively and experience a shortened lifetime relative to a higher wattage device.
4 marks out of 33
State what a heatsink is and explain why R2 may need one if a 10 Ω 10 watt component were used in
the construction of this circuit.
A heatsink is a mechanism designed to conduct heat away from a heat sensitive device so that the
device does not operate at an excessive temperature that would contribute to premature failure of the
device. It is usually made of metal for its high thermal conductivity. The surface area will be
maximised using fins to permit heat to escape to the surrounding air.
5 marks out of 33
Explain what the term de-rating means and describe how it may be used in order to improve the
long-term reliability of an electronic component.
De-rating is a technique used to prolong the service lifetime of a device. A device operated near to its
maximum specified operating limits will wear out faster than a device operated well below its
maximum limits. Derating calls for the use of devices with highe
limits than are strictly needed in
an application so that they will provide longer service.
5 marks out of 33
Question 2
DT002 Power Conditioning January 2007 – Answers
Page 3 of 10
T
220 Vrms
50 Hz
B
10:1
C
R
Figure 2
Figure 2 shows the circuit diagram for a simple d.c. power supply.
Identify the type of rectifier circuit represented in figure 2 and explain the operation of the circuit
with reference to the function of each component within the circuit.
This is a bridge rectifier circuit.
Transformer T provides a reduction in the amplitude of the mains voltage to a value near to that of the
required dc output voltage. It also provides electrical isolation between the mains and the load R.
The diodes in the bridge circuit B conduct on alternate half cycles to provide only positive ‘pulses’ to
the capacitor and load.
C is a capacitor included to ‘smooth’ the raw d.c. from the rectifier diodes. It accumulates electric
charge as the a.c. from the transformer approaches its peak value. After the peak, the capacitor delivers
the stored charge to the load resistor, preventing the load voltage from dropping in between peaks in
the a.c. voltage.
R represents the resistance of the load being powered from the supply.
10 marks out of 33
Sketch to scale the voltage across R as a function of time showing its relationship to the secondary
voltage from the transformer.
Unsmoothed
Smoothed
ac
Question 2
DT002 Power Conditioning January 2007 – Answers
Page 4 of 10
6 marks out of 33
Calculate a value for the capacitor, C in order to keep the percentage ripple voltage across R below
5%. Assume a value of 500 Ω for R and a mains frequency of 50 Hz.
100%
2 f  RC
100%
 5% 
2  50  500  C
100
C 
 400 F
5  2  50  500
%ripple voltage 
5 marks out of 33
The power supply unit shown in figure 2 is said to be unregulated Explain the meaning of the term unregulated as used in relation to power supplies.
An unregulated power supply is one where the output voltage is not protected from fluctuation in the
event that –
The mains input voltage changes (within permitted limits)
The current drawn by the load changes.
Either of these occurrences will result in a change in output voltage amplitude from an unregulated
power supply.
Give one example where an unregulated power supply may be acceptable and one example where a
regulated power supply is required.
An unregulated power supply example is a battery charger – many chargers provide unregulated output
voltages.
A regulated power supply is necessary for powering up digital and microprocessor circuits.
Show how a three terminal regulator chip may be used to provide a regulated output voltage and
explain how a regulated output voltage of 9 V may be obtained.
T1
7809
U1
In
Vmains
+
C1
The 9 volt output is obtained using a dedicated 9 volt regulator chip.
Out
Gnd
RLoad
Question 2
DT002 Power Conditioning January 2007 – Answers
Page 5 of 10
In relation to power supply units, explain the meaning of the terms percentage load regulation and
percentage line regulation.
Percentage line regulation is a figure of merit that describes the extent of change in the output voltage
from a power supply unit as a consequence of a change in the mains amplitude.
% Line Re gulation 
VLoad Max  VLoad Min
VLoad No min al
 100%
Where:
VLoadMax = Maximum output voltage as mains voltage increases above nominal value
VLoadMin = Minimum output voltage as mains voltage decreases below nominal value
Percentage load regulation is a figure of merit that describes the extent of change in the output voltage
from a power supply unit as a consequence of a change in the load current.
% Line Re gulation 
VLoad Max  VLoad Min
VLoad No min al
 100%
Where:
VLoadMax = Maximum output voltage as load current decreases below nominal value
VLoadMin = Minimum output voltage as load current increases above nominal value
12 marks out of 33
Question 3
DT002 Power Conditioning January 2007 – Answers
R1
Page 6 of 10
L1
220 V rms
50 Hz
T1
R2
C1
Figure 3
Identify the type of electronic component represented by each of the symbols shown in figure 3 above
and state the function of the circuit.
R1 is a resistor
R2 is a potentiometer
C1 is a capacitor
L1 is a lamp
T1 is a SCR
The function of the circuit is a Lamp Dimmer control
6 marks out of 33
Briefly describe the principle of operation of the device T1 including an explanation of how the
device is made to turn on and off.
T1 will not conduct when reverse biased – i.e. on negative half cycles of the 220 V 50 Hz supply.
When T1 is forward biased it will not commence conduction until a sufficient voltage is applied on the
gate terminal. Once triggered into conduction, the SCR will continue to conduct until it becomes
reverse biased at which point it will turn off.
6 marks out of 33
Question 3
DT002 Power Conditioning January 2007 – Answers
Page 7 of 10
Sketch the typical shape of the voltage waveform that would be measured across L1 in this circuit
given that R2 is set to approximately half of its maximum value.
400
300
200
100
AC
0
0
10
20
30
40
50
LAMP
-100
-200
-300
-400
7 marks out of 33
Explain how the circuit operates to reduce the r.m.s. voltage across L1.
The rms voltage is related to the area underneath the LAMP waveform above. Keeping the SCR turned
off until a point somewhere after zero crossing of the AC waveform has the effect of reducing this area
below the area given by a complete sine half cycle. Thus in the diagram above, each positive half cycle
is only half used giving half the rms voltage.
7 marks out of 33
State the likely effect of removing component C1 from the circuit and explain the reason why the
behaviour of the circuit is modified.
C1 is used to delay the instant at which the voltage on the SCR gate reaches the voltage necessary to
turn the SCR on. This allows triggering to occur almost anywhere between 0 and 180 degrees..
Removing C1 will remove this delay and triggering will only be possible between 0 and 90 degrees.
7 marks out of 33
Question 4
DT002 Power Conditioning January 2007 – Answers
M
D1
S1
R2
Q1
5.6k
V1
Page 8 of 10
Vcc
24V
5V
Figure 4
Figure 4 shows a bipolar junction transistor (BJT) used to switch a small motor on and off in
response to switch S1 closing and opening. The BJT is specified with β DC = 100 and BVCEO = 40 V.
The motor draws 1 Amp from a 24 volt supply when running.
Calculate the base current in the BJT when
S1 is closed.
With S1 closed, we can write Kirchoffs Voltage Law in the V1 loop as
V1  I B  R2  VBE  0
Thus:
IB 
V1  VBE 5V  0.6V

 10.7mA
R2
410
S1 is opened.
S1 opened prevents base current flowing in Q1
IB
 0
6 marks out of 33
Question 4
DT002 Power Conditioning January 2007 – Answers
Page 9 of 10
Sketch to scale approximate collector characteristic curves for the BJT used in this circuit when
S1 is open and S1 is closed
Identify on the characteristics the active region, the saturation region and the breakdown region.
10 marks out of 33
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 intersects the characteristic curve at collector current of 1 Amp approx
8 marks out of 33
With reference to typical values where necessary, estimate the power dissipation in the BJT when
switched on.
Typically, the voltage between the collector and emitter when saturated is about 0.3 volts. Thus power
dissipation is of the order of –
P  0.3V  1A  0.3W
5 marks out of 33
State the reason for including device D1 in this circuit.
Question 4
DT002 Power Conditioning January 2007 – Answers
Page 10 of 10
As the BJT turns on it attempts to change the current in the windings of the motor suddenly from zero
up to about 1 amp. The self inductance of the motor windings will resist this change in current by
creating a large reverse e.m.f. that may damage the BJT. The diode D1 is included to provide a
discharge path for this e.m.f so that it never becomes large enough to damage the BJT.
4 marks out of 33