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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 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
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
L1
R2
Vcc
Q1
20V
1k
PLC
Vs
Figure 2
Figure 2 shows a bipolar junction transistor (BJT) used to switch a lamp (L1) on
and off in response to the output voltage from a programmable logic controller
(PLC). VS switches between 0 V and 5 V as the PLC output changes state. The
BJT is specified with βDC = 150 and BVCEO = 50 V. The lamp is rated at 0.5 W
when supplied from a 20 V source.
Calculate the base current in the BJT when
The PLC output is 0 V.
The PLC output is 5 V.
The base current may be calculated from –
V
V
5V  0.6V
I B  Control BE 
 4.4mA when S1 is closed
RB
1k
V
 VBE 0V
I B  Control

 0 A when S1 is opened
RB
1k
Sketch to scale an approximate collector characteristic curve for the BJT used in
this circuit when the PLC output switches to 5 V. Identify on the characteristic
the active region, the saturation region and the breakdown region.
Collector Curve
30
25
IC
20
15
10
5
0
0
5
10
15
20
25
VCE
30
35
40
45
Calculate the collector current needed to turn the lamp on fully. Hence
determine if the PLC output will be powerful enough to turn the lamp on fully
using this interface circuit.
20V 2  R  20V 2  800
0.5W 
LAMP
RLAMP
0.5W
With reference to typical values where necessary, estimate the power dissipation
in the BJT when the PLC output is at 5V.
Typical VCE when the BJT operates in the saturation region might be about 0.3
volts.
The collector current when the lamp is on should be –
20V
IC 
 25mA
800
Thus power dissipated in the BJT will be –
Pbjt  0.3V  25mA  7.5mW
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 heat
away from the device and radiate this heat to the surrounding air.
R1
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
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
R1
27 Ω
E
30 V
R2
10 Ω
Figure 3
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
I

E

R1  R 2
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