oer.uam.edu.ng · Web viewTo increase this factor necessities a reduction in T m at the design stage. This can be done in several ways-by increasing the system’s MTBF, which will

Engr. Dr. Jonathan A. Enokela

EEE382 MAINTENANCE AND REPAIRS

SYSTEM MAINTENANCE, AVAILABILITY, and DEPENDABILITY

All recoverable systems which are used for continuous or intermittent service for some period of time are subject, at one time or another, to maintenance.

Maintenance actions can be classified in two categories. First, there is off-schedule maintenance, necessitated by system in-service failure or malfunction. Its purpose is to restore system operation as soon as possible by replacing, repairing, or adjusting the component or components which cause interruption of service. Second, there is scheduled maintenance at regular intervals; its purpose is to keep the system in a condition consistent with its built-in levels of performance, reliability, and , where applicable safety.

Scheduled maintenance fulfils this purpose by servicing, inspections and minor or major overhauls during which

(i) regular care is provided to normally operating subsystems and components which require such attention (lubrication, refuelling, cleaning, adjustment, alignment, etc).

(ii) failed or redundant components are checked, replaced, or repaired if the system contains redundancy, and

(iii) components which are nearing a wear out condition are replaced or overhauled.

These actions are performed to prevent component and system failure rates from increasing over and above the design levels. Therefore, scheduled maintenance is also called preventive maintenance.

The frequency at which maintenance actions of type (i) must be performed to prevent system reliability degradation depends on the physical characteristics of the components involved; for type (iii) items it depends on the wear out statistics of the components and their number in a system.

The frequency of type (ii) maintenance actions is of a different character because it is governed by probabilities. It depends on the failure rate of the components in a redundant system and on the reliability requirements at which such a system must operate. If failed components are detected, they must be replaced or repaired to restore the system to its design reliability level. The time required for such inspection and maintenance actions can be estimated in man hours and in clock hours and allocated in advance as an additional time to be added to the fixed times provided for types (i) and (iii).

As to the frequency of the off-schedule maintenance, this is strictly a function of the failure rates of those components or units which cause in-service failures of the system, and therefore it is a function of the reciprocal of the system’s mean time between failures, (MTBF) m.

For every t operating hours there will be on the average t/m in-service system failures, and therefore on the average t/m off-schedule maintenance actions will have to be performed. Off-schedule maintenance is also called Corrective maintenance. The man hours required for these repairs will vary with the components which cause the failure. The t/m off-schedule maintenance actions can be broken down

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into t /m1 , t /m2 , t /m3 etc, i.e. the number of maintenance actions to be performed on the individual components and units which have MTBF m1, m2, m3, etc. The total average number of off-schedule maintenance actions for t system operating hours is thus given by:

tm

= tm1

+ tm2

+ tm3

+…=∑i=1

n tmi

=∑i=1

n

λi t…………………………….(1)

Now if the time in man-hours required for repairing a system failure caused by component 1 (which has a MTBF of m1) is known to be T 1 and for component 2 it is T 2, etc, then the total average man-hours H0 involved in off-schedule maintenance for every t system operating hours become

H 0=tm1

T 1+tm2

T 2+tm3

T 3+…=∑i=1

n T i

mit=∑

i=1

n

λ iT i t ………………….(2)

The individual times T i which represent the times needed to restore system operation in each particular case depend on the physical location and design characteristics of the components in the system, including their accessibility, their mountings, and the provisions made to detect the component which has failed. The times T i may become longer if unskilled personnel is used, in which case T i has to be multiplied by a factor larger than unity. This factor can be defined as the reciprocal of personnel efficiency or skill.

Assuming that H 0 in eqn (2) is the total average time in manhours required for a system’s off-schedule maintenance with personnel of a given skill, and that it includes fault location before repair and check out after repair, it can be converted into a total clock time T 0 according to the available manpower. When this time T 0 is added to the total time allocated to system preventive maintenance, we obtain the total maintenance clock time T m, or time to repair, which has to be spent on the average for each t system operating hours:

T m=T p+T 0…………….(3)

Where T p is the time for preventive maintenance

and T 0 is the time for corrective maintenance.

This also represents the system’s functional down time. Thus with a provision made for down time other than functional designated T r, which may be the scheduled idle periods for t system operating hours or other scheduled activities such as administrative time, etc, we obtain the system utilization factor:

U= totaloperating timetotal operating time+total downtime

¿ tT p+T 0+T r+t

………………….(4)

The time t is the time the system will be expected to operate for the calendar time, T p+T0+T r+t . It is customary to choose one year or 8760 hours as calendar time.

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If the total down time is made up of only the maintenance time Tm=T p+T 0, then we obtain the maximum possible utilization factor and this is given by

Um=t

T m+t……………… ..(5)

To increase this factor necessities a reduction in Tm at the design stage. This can be done in several ways-by increasing the system’s MTBF, which will reduce the number of in-service failures, and, in addition, by reducing T i in equation (2). These measures will result in a reduction of man hours H 0 and the time T 0 required for off-schedule maintenance. The most effective design procedure is to reduce the times T i of those components which have the lowest mi, which means to locate these components in the system so that they will be easily accessible and to see that their mounting arrangements will allow fast replacement.

When several maintenance actions are being performed simultaneously on a system by a crew, the maintenance clock time required for restoring system to operation is determined by the longest maintenance action. In this case the shorter maintenance actions do not enter into the clock time, or down time, computation at all. But the number of man hours required by each individual action is important because it is a measure of the total maintenance effort which has to be expended and allows us to plan for the optimum manpower and to establish optimum maintenance charts, so that the maintenance clock time, or down time, can be minimized.

In equation (5) we defined the maximum possible system utilization factor, Um=t /(T m+ t). This factor is a measure of the system’s availability because it gives the percentage of time the system will be available for operation. We have seen that to operate for t hours out of a total time of Tm+t hours, the system requires T m hours for scheduled and off-schedule maintenance. If for the system operating time we select its MTBF m, we can then derive the average maintenance time or mean time to repair MTTR, Tm ' which is required for every m system operating hours. Because it is customary to choose Tm+t=8760 hours so that T m is the clock time in hours, the average maintenance time or MTTR T m ' per m system operating hours will be

MTTR=Tm' =T m.

mt=

mTm

8760−Tm…………………(6)

Where m is the system’s MTBF for a given maintenance policy, and T m the system’s down time per year. Using m and T m ' instead of t and T m in the expression for utilization factor [equation (5)] we obtain

Um=mTm /T m '

Tm+mTm

T m '

=mTm( 1Tm ' )

mT m( 1m+ 1Tm ' )

¿ 1Tm 'm

+1= mm+T m '

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The above equation is, by definition, called the system availability A:

A= mm+T m '

…………………..(7)

Or

A= MTBFMTBF+MTTR

……………….(8)

This equation gives the percentage of time the system will be available for service. The highest value of A is unity or 100% and occurs only if a system never requires maintenance and therefore if MTTR is zero. When analyzing A we find that it is a ratio of system operating hours to the sum of system operating plus maintenance hours. By its nature, therefore, this ratio called system availability A, is a probability. Conversely, we can also define a complementary probability B so that A+B = 1, and B is the probability that the system will not be available for operation. This probability, called system unavailability, is given by

B=T m '

m+Tm '= MTTR

MTBF+MTTR………………………..(9)

By substituting for T m ' the system’s functional down time T m per year in equations (7) and (9) can be obtained in a much simpler way and these are:

A=1−T m

8760=1− Downtime

total ( calendar ) time…………………….(10)

B=T m

8760=Downtime

total time……………………… ..(11)

Sometimes it is required to know the probability that a system will be available for operation between any two scheduled maintenances, i.e. when assigned to continuous duty so that it must be available at a moment’s notice or is required to operate without interruption. We call this probability system Dependability D. For the value of D we must leave out T p (which stands for average scheduled maintenance) from the term T m so that T m=T 0, the average off-scheduled maintenance time and take this on a per year basis. Thus

D= mm+T m '

= m

m+T 0mt

[using equation (6 )]

¿ 1

1+T 0t

= 1

1+T 0

8760−T 0¿8760−T0

8760−T 0+T 0=8760−T 08760 ∴D=1−

T 08760

…………………. (12)

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OVERHAULS AND PART REPLACEMENTS

The repair time required for each component varies according to its physical characteristics and location in the system but variations of the repair time also exist for identical components in identical locations. There is a certain minimum time required to get to the faulty component and to remove it as well as to put it back again or to replace it by another. This time varies with skill, as does the actual repair time, but it is safe to assume that for a given component in a given location, the over-all repair time is approximately normally distributed, with some mean and standard deviation. From the normal curve the probability that the repair will be performed in a time equal to the mean time M is 0.5 or 50%, and the probability that the repair will be completed in a time of, say, M−3σ is only 0.0014 or 0.14 percent, etc. This probability of repairing a particular component in a given system in a given time is sometimes referred to as the maintainability.

The functional down time between two regular or scheduled overhauls depends on system reliability which determines the average number of system failures and on the time it takes to repair these failures. When the MTBF of a system is m, conclusions can be drawn as to how often the system will fail in a time T 0 between two regular overhauls, where T 0 is the system’s operating time and amounts to the calendar time T less the total down time T D.

If between two regular overhauls the system is not affected by wear out failures so that it behaves exponentially, its reliability of operating for T 0 hours between the two overhauls is

R (T 0 )=e−T 0/m……………………… ..(13)

where T 0 is the required operating time.

For instance, if the overhaul time of an equipment or system is fixed for every T 0=800 hours of operation and MTBF of this equipment is m = 4000 hours, the probability that the equipment will operate without failing for 800 hours is

R=e−T0 /m=e−800/4000=e−0.2=0.81873

This means that we would expect about 82% of these equipments to reach the regular overhaul time without failing, and about 18% would fail before the time T 0. Therefore about 18% of the equipment would have to be overhauled before the regular 800 hours. Thus the expected average overhaul time for all these equipments would be less than 800 hours.

The average time between overhaul T avg is computed as a mean time for the operating period, T 0, similar to the way the MTBF is computed from R as the integral from zero to infinity. The time T avg is actually a mean time between both scheduled and unscheduled overhauls.

T avg=∫0

T0

Rdt……………… (14)

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In the exponential case this equals

T avg=∫0

T0

e−t /mdt=−m [e−t /m ]T00

i .e .T avg=−m(e¿¿−T0 /m−1)=m (1−e−T0 /m )=mQ (T 0 )…………………… ..(15)¿

Thus in our case, with Q (T 0 )=0.18 ,m=4000

T avg=4000×0.18=720 hours.

Thus, if we had 100 of these equipments with a scheduled time between regular overhauls of T 0=800hrs, about 82 would operate without failure for the full 800 hours, and averaging over the 100 equipments, the average time between overhauls would be 720 hours. The knowledge of this average time allows us to plan for a total number of overhauls over a long period of time.

The main function of overhaul is to prevent wear out. For single components the replacement or overhaul time must be kept at M−3σ∨M−4σ or in between to prevent wear out from appreciably increasing the failure rate. If large numbers of components are in a system, this replacement or overhaul time must be further reduced to M−5σ to M−6 σ, according to reliability requirements.

BASICS OF FAULT DIAGNOSIS

An electronic circuit is a collection of components connected together to perform a particular electronic function. Each component has its part to play in the operation of the circuit. If any component should fail, then the operation will be drastically changed. A faulty component produces a certain set of symptoms, which can be used to indicate the component and its type of fault. Such symptoms are, for example, the voltage levels at various points in the circuit.

Skilful fault diagnosis requires both theoretical knowledge and practical experience. Before attempting the diagnosis of faulty components the engineer or technician will need to understand the purpose of the circuit and its operation. This clearly presupposes that he understands the principle of operation of the various electronic components used.

COMPONENTS AND COMMON FAULTS

A component can be said to have failed if any one of its constants is out of its specified limit. For example, if a 5.6k Ω±5% resistor actually has a resistance of 6k Ω, or if the leakage current of a 64 μF ,12V electrolytic capacitor is 150 μA when it is specified as a maximum of 10 μA, then both components have failed.

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Both these examples can be described as partial failures, since they do not necessarily lead to a complete loss of performance, rather to a small change. Partial failures are especially important when the component is used in a critical circuit position.

The faults we are more concerned with are called catastrophic failures, when the failure of the component is both sudden and complete. For example, a resistor goes very high in value, or becomes open circuited, or a diode develops a short circuit between anode and cathode. Such failures lead to a complete loss of performance and are usually accompanied by a drastic change in dc bias levels.

As a general rule certain types of component fail in a particular way. When resistors, especially the film type, fail they often go open circuit, since a small break in the resistance spiral is a much more likely event than a short circuit across the whole resistor. Electrolytic capacitors on the other hand are more prone to fail short circuit. Table 1 indicates the more probable types of failure for various types of electronic component.

It is perhaps easy to understand failures caused by defects and overloads, but why should a component fail in normal use? Basically the component is ageing because of the stresses that are acting continuously upon it. These stresses are of two kinds, operating and environmental. The operating stress is due to the design conditions and the life of a component can be prolonged by operating it well within its rated maximum value of current, voltage and power. This is called derating. Environmental stresses are those caused by the surrounding conditions. High temperature, high humidity, mechanical shock and vibration, high or low pressure, and corrosive chemicals or dust in the air, are the major adverse conditions. All of these stresses affect the component and cause some deviation from the specification, and finally the component will fail.

Table 1

Components Common Type of FaultResistors High in Value or Open CircuitVariable Resistors Open Circuit or Intermittent Contact resulting

From mechanical wearCapacitors Open or Short CircuitInductors(Including Transformers)

Open Circuit; Short-Circuited turns; Short-CircuitCoil to Frame (Iron-Cored Type)

Semiconductor Devices:Diodes, Transistors, FETs, SCR, etc.

Open or Short-Circuit at any junction

For example, consider a component subjected to continual cycles of heating and cooling; this may cause the material from which the component is made to become brittle, and any mechanical shock may then cause the component to fail open circuit.

The effects of adverse environmental conditions can usually be minimized by careful design, and this is increasingly important when an electronic instrument forms an integral part of some industrial manufacturing process where high temperatures, vibration and other hazards are present.

7

Anode Cathode

Diode Symbol

Fig. 1


Another cause of component failure is high voltage pulses or spikes, generated from switched inductive loads, being transmitted along the mains and appearing on internal supply wires. These spikes can easily lead to the breakdown of junctions in semiconductor devices.

OPERATING PRINCIPLES OF COMMON ACTIVE COMPONENTS

Semiconductor Diodes

These devices have a low slope resistance when the anode is positive with respect to the cathode, a typical value being 25Ω at a forward current of 1mA. When the anode is negative with respect to the cathode, the resistance is very high, greater than 100MΩ for a silicon diode. In order to pass current in the forward direction, a small forward bias is necessary, about 200mV for a germanium device and 600mV for a silicon diode.

P N

Bipolar Transistors

The bipolar transistor is a three-layer device. Depending on the arrangement of the layers, the BJT is either NPN or PNP.

B

E

C C

E

B

BB

C CE

EN N NP P P

The transistor is best considered as a device in which a current flowing between the collector and emitter is controlled by a much smaller current flowing between the base and emitter. For correct operation the silicon transistor requires a small forward bias voltage of approximately 600mV between base and emitter to overcome the junction potential set up by the fixed charges in the depletion region.

8

Fig. 2

P+

P+

Fig. 3


The transistor can be operated in three possible configurations, called common base, common emitter, and common collector. The base, the emitter or the collector is made the common terminal for the input and output signals. The relationship between the three currents flowing in a bipolar transistor, neglecting any leakage current, can be written as

I E=IC+ I B . . .(16)

The base current I B is much smaller than both I E and IC.

The Field Effect Transistor

The operation of FETs differs from the bipolar transistor in that the current flowing through the FET is controlled by a voltage input. The terminals are called drain, source, and gate and the simplified construction of an n-channel junction FET is shown in figure 3:

G

G

D

D

S

S n-channel

Figure 3 shows that the FET is made from a bar of n-type material to which contacts called drain and source are made at each end. Two P-regions formed into the bar directly opposite each other are connected together and are called the gate.

A current will flow between source and drain when the voltage between drain and source is positive. However, this current will fall if the gate voltage is made negative with respect to the source. When the gate is negative, depletion regions are formed, and this reduces the channel width between source and drain, thus the current falls. When the gate is made sufficiently negative, say -3V, these depletion regions meet and the drain current is cut-off.

MEASURING INSTRUMENTS AND TESTING METHODS

To get information about the symptoms of a particular fault, a set of voltage readings at critical points in the circuits must be taken. This information, together with additional information on waveforms, is usually all that is necessary for correct fault diagnosis. So the most essential pieces of test equipments

9


for fault finding are: a good, general purpose multi range meter, and an oscilloscope. The meter should have sensitivity on dc range of at least 20k Ω /V in order to prevent the loading of circuit from which measurement is being made.

The meter could be a portable moving-coil meter or a digital multimeter. The latter displays the measured voltage, current or resistance on a three or more in-line digital display. The input resistance of these instruments is typically10MΩ, which means that the unit takes only a small current from the circuit being measured.

THE CATHODE RAY OSCILLOSCOPE

This is perhaps the most versatile measuring instrument available. With it, it is possible to measure dc and ac voltage, current, phase angle, and many other quantities. The accuracy depends to a great extent on the care paid to the instrument’s calibration. The typical input impedance of a CRO is 1MΩ which has a capacitance of about 20 ρF in parallel with it.

FAULT FINDING ON ELECTRONIC INSRUMENTS AND SYSTEMS

When a complete instrument is returned for repair, the service engineer must first locate which block of the instrument is faulty before he can locate the actual component that has failed. There are various methods used to narrow down the fault to one block, but before these are discussed, it is useful to consider some fairly obvious but often over looked points:

1. The service engineer must have a maintenance manual with up-to-date circuit diagrams. This manual should also give the figures of the performance specification.

2. The engineer must have all the necessary TEST equipments3. The engineer then has to define the fault accurately. This point is very important. The symptoms

must be accurately noted and this means that a functional test must be made on the instrument.

For example consider that a signal generator has been returned for repair with a suspect power supply failure. Before taking off the cover and checking power supply line, the service engineer would

(a) Check the mains fuses and if not blown(b) Check for sine wave output on all ranges(c) Then make notes of the fault symptoms.

The complete circuits of most electronic instruments can be broken down into a series of functional blocks; for example in a general purpose sine wave generator these would be power supply, variable sine wave oscillator, buffer amplifier, and output attenuator. By treating the instrument in blocks, rather than as a whole, it is possible to narrow down the search for a faulty component first of all to one block, then by measurements within that block to locate the actual faulty component. The methods used to decide which block is faulty are

(a) Input to output (or beginning to end)

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(b) Output to input(c) Random(d) Half-split

All of these have their particular advantages and uses. The random method, which implies a totally non-systematic approach, is rarely used. A method based on the reliability of components can also be used when there is a wealth of service knowledge and experience concerning a particular instrument. For example, a service engineer might make the reasonable assumption that, because a particular electrolytic capacitor has been at fault in 60% of the instruments recently returned, it is a strong possibility that the next faulty instrument also has a faulty electrolytic capacitor. He would naturally check this first, and in most cases save valuable service time. It must be stressed, however, that this method depends upon the availability of a large amount of data on the reliability of the various components within an instrument. Most service engineers would use a logical systematic approach to system fault location.

The input to output and output to input methods are examples of this systematic approach. The method is fairly obvious. A suitable input signal (if required) is injected into the input block and then measurements are made sequentially at the output of each block in turn, working either from the input towards the output or from the output back to the input, until the faulty block is located. This logical method is the one most service engineers use on equipment containing a limited number of blocks.

The half-split method is very powerful in locating faults in instruments made up of large number of blocks in series. Take for example a superheat radio receiver shown in block diagram in figure 4. Since there are eight blocks it is possible to divide the circuit in half, test that half, decide which half is working correctly, then split the non functioning section into half again to locate the fault. Assume that a fault exists in the demodulator of the receiver. The sequence of tests would be as follows:

2 3 4 5 6 7 8

1Aerial

RF Amp Freq. Changer IF1 IF2 Demodulator AF Amp Loud Speaker

(a) Split in half, inject signal into the input of (1) (aerial circuit) and check output at (4)(IF). Output is correct. Therefore the fault is somewhere in block (5) to (8).

(b) Split blocks (5) to (8) in half by checking output of (6). Input signal can be left at (1). No output.(c) Leaving signal at (1), check output from (5). Output should be correct, indicating that the faulty

block is (6), the demodulator.

11

Fig. 4


The half-split method is most useful when the number of components or blocks in series is very large, for example where several series plug and socket connections are used. There are, however, several assumptions made for the half-split:

(a) That all components are equally reliable(b) That it is possible and practical to make measurements at the desired point and(c) That all checks are similar and take the same amount of time. These assumptions will not always

be valid and it is up to the service engineer to then decide the best method of approach.

The half-split method can also be easily complicated by

(a) An odd number of series units(b) Divergence: an output from one block feeding two or more units(c) Convergence: two or more inputs being necessary for the correct operation of one unit.(d) Feedback: which may be used to modify the characteristics of the unit or in fact be a sustaining

network for an oscillator.

When using any of the methods as described, try and use the method, or a combination of them, that will locate the faulty block in a system in the shortest possible time.

SINGLE STAGE TRANSISTOR AMPLIFIER

Before considering any fault conditions, the operation of the circuit must be understood. Consider the single stage amplifier circuit of figure 5. The bias circuit achieves stabilization by fixing the value of the base voltage V B and by keeping it constant irrespective of changes in the base current. To do this the values of R1 and R2 must be chosen so that the current flowing through them is much greater than the transistor base current. These resistors form a potential divider and if we neglect base current, the dc base voltage can be calculated from

V B≈V CC

R2R1+R2

And the emitter voltage

V E=V B−V BE

And

I E=V E

R4

SinceIE≈ IC, the dc voltage at the collector V C is given by

V C=V CC−ICR3

12

VCC = 12V

Fig. 5

R1

R2

R3

R4

C1

C2

C3

R1 = 47k, R2 = 12k, R3 = 2.2k, R4 = 0.560k, C1 = 10µF, C2 = 10µF, C3 = 100µF

Input

Output

1

2

3


Now, since V B is fixed, the dc current through the transistor will be fixed, and this gives the operating point V C.

+

+

+

-

-

-

For the circuit shown in figure 5 the calculations of the dc bias voltages would be as follows:

V B=V CC

R2R1+R2

=12×1247+12

=2.4VV E=V B−V BE=2.4−0.7=1.7V

IC=IE=V E /R4=1.70.56

=3.05mA

V C=12−3.05×2.2=5.3V

These values are tabulated as:

TP 1 2 3V 2.

45.3 1.7

When the circuit is built, the actual voltages measured with a 20k Ω /V meter will be slightly different due to resistor tolerances and meter loading. The actual readings are:

TP 1 2 3V 2.

35.5 1.7

13


Now let us consider the effect of component failures, taking each in turn.

Resistor Faults

R1 open circuit

TP 1 2 3 No OutputSignalMeter Reading (MR) 0 12 0

When R1 goes open circuit, the current flowing in R2 and the base is zero. The transistor is cut off so both the emitter and base voltages are zero. Since no collector current is flowing the voltage dropped across the collector load R3 is zero and the collector voltage itself is the same as the supply voltage V CC.

R2 open circuit

TP 1 2 3 Grossly distorted output; negative-goingSignals clippedMR 3.

22.6 2.5

Without R2 in the circuit the current that was flowing through R2 now tries to flow into the base of the transistor. This increased base current causes the transistor to be saturated so that V CE becomes very small.

R3 Open Circuit

TP 1 2 3 No outputsignalMR 0.75 0.

10.1

Without R3 in circuit the collector current is zero, so any current flowing in the emitter must now be supplied from the base. The base-emitter junction acts like a forward biased diode placing R4 in parallel with R2. Since R4 is a low value resistor the emitter voltage falls to a very low value. The base voltage is about 650mV greater than the emitter voltage.

It might be reasonable to assume that the voltage reading at the collector would be zero since the resistance is open circuit. However, when the meter is connected it presents a high resistance path from the collector to ground and the base-collector junction acts like a forward biased diode passing a small current through the meter.

R4 Open Circuit

TP 1 2 3 No outputsignalMR 2.3 12 2

14


With an open circuit between emitter and ground, no current flows through the transistor. The collector voltage therefore rises to V CC. The voltage at the base is fixed by the potential divider R1 and R2 and since the base current is small in comparison to the current through R2 this voltage hardly changes at all.

As with the previous example when the meter is connected between emitter and ground, a small emitter current flows so the voltage indicated at the emitter is slightly higher than normal.

CAPACITOR FAULTS

C1∨C2 Open Circuit

TP 1 2 3 No outputsignalM

R2.3 5.5 1.7

With this type of fault the bias conditions of the circuits are unchanged. The fault can only be an open circuit coupling capacitor. A check with an oscilloscope is necessary to determine which one is actually faulty.

C3 Open Circuit

TP 1 2 3 Low GainMR

2.3 5.5

1.7

Again the bias conditions are unchanged. The symptom that identifies this fault is the fact that the amplifier voltage gain has fallen. With C3 open circuit, R4 introduces negative feedback thus causing a reduction in gain.

C3 short circuit

TP 1 2 3 No outputsignalMR 0.7 0.15 0

The emitter resistor R4 is shorted out, so the emitter voltage reads 0V . The transistor, heavily forward biased, saturates and therefore attempts to pass a large current. However, the transistor current is

15


limited to a value given by V CC /R3 which prevents the transistor from being damaged. The base voltage must be 0.7V higher than the emitter.

TRANSISTOR FAULTS

Collector-Base junction open circuit

TP 1 2 3 No outputsignalMR 0.75 12 0.1

Since the collector is open there can be no collector current flowing, so the voltage at TP2 rises to +12V. The base-emitter junction now acts as a forward biased diode in a similar way as for the fault of R3 open circuit.

Collector-Base junction short circuit

TP 1 2 3 No outputsignalM

R3 3 2.3

As with any short circuit a clue to the fault is given by the fact that the voltages on the base and collector are equal. With this fault the circuit effectively reduces to R3 in series with the base-emitter diode and R4. The resistance of this path is much lower than R1∧R2, so the effect of the latter resistors can be neglected. The current flowing in R4 is given by

I=V CC−V BE

R3+R4=12−0.7

2.76=4mA

The voltage at the emitter will then be I ×R4=2.3V . The voltages at TP1 and TP2 will be 0.7V higher than this, sufficient to forward bias the base-emitter diode.

Emitter-Base junction open circuit

TP 1 2 3 No output signalMR 2.3 12 0

With this fault there can be no current flowing in the transistors. The voltage drops across R3∧R4 are zero, so the collector voltage rises to V CC and the emitter voltage is 0V . The voltage on the base is

16


determined by the potential divider R1∧R2 and, therefore, remains at 2.3V . There is no difference in the symptoms if the base or the emitter connection to the junction is open circuit.

Emitter-Base junction short circuit

TP 1 2 3 No output signalM

R0.13 12 0.13

The voltage at TP1 and TP3 will be equal and at a low value since R4, a low resistance, is placed directly in parallel with R2. With a shorted base-emitter junction all transistor action ceases, so the collector voltage rises to V CC.

Collector-Emitter short circuit

TP 1 2 3 No output signalM

R2.3 2.5 2.5

The voltage at the emitter is equal to that on the collector, indicating a short. The value of the voltage will be determined by R3∧R4 which now form a potential divider. The base voltage remains unchanged at 2.3V since the emitter voltage has risen, thus cutting off the base-emitter diode.

POWER SUPPLY CIRCUITS

Practically all electronic instruments require a source of dc power before they will operate. Sometimes the source is a battery, but more usually the power is obtained from a unit that converts the normal single phase ac mains supply (220V ,50Hz) to some different value of dc voltage. The function of the power supply is to provide the necessary dc voltage and current, with low levels of ac ripple (mains hums) and with good stability and regulation. In other words it must provide a stable dc output voltage irrespective of changes in the mains input voltage and of changes in the load current. A further important requirement of a modern unit is that it should be able to limit the available output current in the event of an overload (current limiting) and also limit the maximum output voltage. Damage to sensitive components, such as integrated circuits, in the instrument can easily occur if excessive voltages appear on the power supply lines.

There are various methods of achieving a stable dc voltage from the ac mains, but only two methods are commonly used. These are

(i) Using a linear stabilizer(ii) Using a switching mode stabilizer.

17

Vo

Vo

Vs

Fig. 7a Half-wave Rectifier

Input

Vo

t

Input

Vs

VsVo

RL

RL

D1

D2 Fig. 7b Full-wave Rectifier


Both have their advantages and disadvantages. The switching mode power supply unit (SMPU) finds its main use in high-power applications (100W upwards).

THE LINEAR STABILIZED POWER SUPPLY

The block diagram of a conventional power unit is shown in figure 6.

TransformerRectifier

Filter

Regulator LoadMains Input220V, 50Hz

+

-

Stable dc Output

Fig. 6

The transformer servers two main purposes; it isolates the equipment dc power lines from the mains supply, and it changes the level of the ac mains voltage to some lower or higher value. The ratio of the secondary voltage to primary voltage is determined by the number of turns on each winding. The rectifier unit converts the ac voltage from the transformer secondary winding into pulses of unidirectional current. Three types of rectifier circuit are used for single phase: the half-wave, the full-wave and the bridge. These, together with their output waveforms, are shown in figure 7.

The half-wave rectifier, although being a simple circuit, has the main disadvantage of low efficiency. The diode conducts only on one half of the cycle, so the efficiency cannot be greater than 50%. The full-wave rectifier uses two diodes each conducting on alternate half cycles to give much higher efficiency. However, to achieve this, a transformer with a centre tapped secondary winding is necessary. This means that twice the number of turns is required on the secondary winding. The bridge rectifier, which is the most common circuit, uses four diodes to achieve rectification over the whole cycle, and no centre tap is required. The four diodes can now be supplied in one encapsulated unit, which is more convenient and somewhat cheaper than wiring in four separate diodes.

++

+

-

-

-

18

Vd

t


++

+

-

-

-

+

-

Following the rectifier is the filter which serves to smoothen out the pulses received from the rectifier. The circuit can have either a capacitive or an inductive input as shown in figure 8, or it can simply be a single capacitor.

The inductive filter, or choke input filter, is more commonly used when the power unit has to supply a large load current. On low power equipment a capacitive input filter is more typical. The input capacitor, called the “reservoir”, is used as a storage device for electric charge. Let us suppose that a reservoir

++ +- --

(a) Capacitor Input (b) Choke InputFig. 8

capacitor is connected to the output of a half-wave rectifier as shown in figure 9. When the diode conducts on the positive half cycle, the capacitor is charged and a large pulse of current is taken. The voltage across the capacitor then rises to nearly the peak value of the ac secondary voltage. When the secondary voltage begins to fall, the diode becomes reverse biased, and the capacitor now discharges through the load resistor. The voltage across the load now more represents a dc level, but superimposed upon it is an alternating waveform, called the ripple. The value of the ripple amplitude depends upon

19

VoRL

Vo

C

D

t

t

I

Current throughCapacitor

Fig. 9


the size of the capacitor and the load resistance. To achieve low values of ripple a high-value electrolytic capacitor has to be used. A number of points should be noted concerning the reservoir capacitor.

(a) Since it is an electrolytic, it is polarized and must be connected correctly in the circuit.(b) Its dc working voltage must be greater than the peak of the transformer secondary voltage.(c) It must be physically large since it has to absorb large pulses of current when it is charging, the

peak values of which may be several amperes. If too small a capacitor is fitted it may overheat and possibly explode.

The other components of the circuit form a low pass filter, which reduce still further the output ripple voltage. Typical values for the iron-cored inductor are 1-5 Henries and for C2, 500 μF. The inductor is often replaced by a wire-wound resistor of low value.

+

-

+

-

The last block is the regulator, which is used to keep the output voltage constant irrespective of changes in the mains input voltage and of changes in the load current. These two functions are called line stabilization and load regulation, respectively. As shown in figure 10, all linear regulators comprise

(a) Control unit(b) A reference element (usually a Zener diode)(c) An error amplifier

20

R1

R2

Fig. 10

Potential divider


+

-

+

-

ControlElement

ReferenceElement

ErrorAmp

UnstabilzedInputFrom Filter

StabilizedDC Output

In operation, the circuit compares a portion of the dc output voltage with the reference voltage. Any difference between the two levels is amplified by the error amplifier whose output is fed to the control unit. The stability and regulation of the output voltage depends on the stability of the reference element and the gain of the error amplifier. High gain operational amplifiers in IC form are now commonly used as the error amplifier to give power supplies of excellent performance. The main advantage of the linear regulator is that the output is continuously controlled to give good stabilization against mains input changes. A typical specification for an output voltage of +15V @100mA load current is:

Line Stability 10,000:1(a 10V change in mains supply giving a 1 mV in DC Output)

Output Ripple 0.1 mV peak-to-peak at full loadDC Output Impedance 0.05 OhmsTemperature Coefficient 200 µV per degree CentigradeLoad Regulation 0.033% from zero to full load

(i.e. an output change of 5 mV)

The limitation in the linear regulator circuit is that good performance is achieved at the expense of inefficiency. Power is dissipated and lost in the series control transistor and this power loss increases with load current. A large heat sink is required to ensure that the junction temperature of the series transistor is kept within its rated value. For power units supplying above about 100W, the switching mode regulator becomes a preferred alternative.

SWITCHING MODE POWER SUPPLIES (SMPU)

21


There are two main variations of this type. In one, a fast switching transistor is used as the control element in the regulator (figure 11). This transistor is switched on and off at a frequency above audio (usually 20kHz). The dc output voltage, after being smoothened by a low pass filter, is controlled by varying the mark-to-space ratio of the switching signal. Such techniques are known as secondary switching. The error signal generated by comparing the dc output with a reference level, is used to control the duty cycle of a free-running oscillator. The advantage of this type of circuit is that the series transistor heat dissipation is greatly reduced, hence greater regulator efficiency.

UnstabilizedSupply

Square WaveOscillator

Duty CycleControl

SwitchFilter

Comparator

ReferenceSupply

LoadL

N

+

-

Fig. 11

+

-

Another form of the SMPU is shown in figure 12 and uses a principle called primary switching. The mains supply itself, after rectification and smoothening, is switched at high frequency by high voltage switching transistors. With this method the transformer following the switching transistors can be much smaller than the bulky 50Hz transformer required in conventional supplies.

Regulation is achieved by again varying the switching duty cycle of the transistors. Naturally RF suppression circuits must be included to reduce the switching spikes that would otherwise be fed back into the mains supply. This SMPU offers considerable advantages in terms of efficiency, reduction in heat loss, and reduction in overall volume. However it does not possess the regulation performance that can be achieved in the linear circuit. Switching mode supplies are now commonly used where large currents at low voltage are required, as in equipment using many digital ICs.

22

Fig. 12

C


Bridge Rectifier

Push-PullDrive

Pulse WidthModulator

20 kHz SquareWave Oscillator

Comparator

Reference Voltage

StabilizedDC Output

+

-

L

N

POWER SUPPLY PROTECTION CIRCUITS

Some form of protection must be incorporated in even the simplest power supply. A common form is the standard fuse which serves to disconnect the unit from the main’s supply when an overload or short circuit occurs. A power unit may have fuses in the line and neutral mains wires, and also a fuse in the dc unstabilized line. Fuses usually do not blow soon enough to protect the series transistor in the regulator if the output is short-circuited, and so some form of current limiting is used. A simple circuit for achieving this is shown in figure 13 where the load current flows through a low value current monitoring resistor. If the load current increases beyond a pre-determined value, the voltage developed across this resistor turns on Q2 which in turn tends to turn off Q1.

Over voltage protections can be provided by a circuit which senses the dc output voltage, and compares it with a reference level as in figure 14. If the dc output voltage rises above V Z a signal is generated which triggers the thyristor and this short circuits the output, either blowing a dc line fuse or operating the current limit. Such circuits are called “crow bars”. Naturally the fault must be cleared before the circuit can be reset.

TESTING POWER SUPPLY CIRCUITS

The main parameters which ought to be measured either in a test department or by the service technician after he has repaired a power unit are the following:

(a) Dc output voltage(b) Available dc output current(c) Output ripple voltage at full load

23

CurrentLimitSeries

Transistor

ErrorAmplifier

ReferenceZener

Q1

Q2

Q3R1

R2R3

R4

Z

rm

SCRR

Z


(d) stabilization against mains supply changes(e) regulation from zero to full load.

StabilizedDC Output

+

-

+

-

Fig. 13

+

-

Power Supplywith currentlimit or fusein dc line

Fig. 14

L

N

24

Vdc Vo

Io+

-

Variable AutoTransformer


These can be measured using a standard test set-up as shown in figure 15.

L

N

Fuse PowerSupplyUnderTest

Load

CRO

Fig. 15

The dc output voltage should be measured, and if necessary adjusted, when the unit is fully loaded. However, it is sometimes advisable to measure the output on a low load and then gradually increase the load current to maximum. There should, of course, be little change in the output voltage.

The peak-to-peak ripple amplitude can be checked best my measuring at the output with an oscilloscope. A sensitive ac range must be selected because the ripple should be quite low, typically less than 20mV .

Measurement of stabilization and regulation requires that any small change in dc output be carefully noted, and therefore a digital voltmeter is often necessary. For stabilization measurement, the unit should be fully loaded and the change in dc output voltage noted for say a ±10% change in the a.c input. The mains input can be varied using an adjustable auto-transformer as shown. Then, if for example, the dc output changed by 50mV from 10V , i.e. an output change of 0.5%, then the line stabilization would be 40:1.

Load regulation is measured, keeping the ac input constant, by noting the change in output when the load is varied from zero to full load.

load regulation= change∈dcoutputDcoutput onno load

×100%

For example suppose the output changed by 20mV from 10V . The load regulation is

20×10−3

10×100%=0.2%

25

Vo

IoNo load Full load

Current limit

Fig. 16


To obtain fuller information on a power supply’s performance it is often necessary to plot the load regulation curve. This is a plot of output voltage against load current. A typical result for a unit with current limiting is shown in figure 16.

FAULT FINDING TECHNIQUES AND TYPICAL FAULT CONDITIONS

When a faulty power unit is returned for repair, the fault has to be isolated to some particular portion of the unit. The fault may lie in the transformer, the rectifier, the filter section, or the regulator, and measurement with a voltmeter will be necessary to locate the fault.

However it’s probably best to start diagnosis with a few rather obvious but often overlooked checks. First measure the dc output voltage. If this is zero, the next check should be on the mains input. Is the mains supply reaching the transformer primary? If it is not, there is the possibility of a faulty plug, open circuit mains, wires, or a blown fuse. If the fuse is suspected, always test its continuity with an Ohm meter, never rely on just a visual inspection.

If the fuse is blown it has done so because of some fault condition and the fault must be cleared before a new fuse is fitted. Resistance checks must be used to locate such a fault. Use an Ohm meter to measure the resistance of the transformer primary, the secondary, the rectifiers, and so on. The winding resistance depends on the size of the transformer, should be low, typically about 50Ω. The secondary, usually supplying a lower voltage, may have a resistance of only a few ohms. Detecting shorted turns on a winding can therefore be quite difficult. Another useful check is to run the transformer off load and test for overheating.

Suppose, however, that the fuse is intact, and that the mains is reaching the primary. The next step is to measure the secondary ac voltage, the unstabilized dc voltage, then the dc voltage in the regulator and so on, until the fault is located.

Table II lists some typical faults together with the associated symptoms. The faults are only a sample of those which may occur. Locating a faulty component from a given set of symptoms will come with practice.

26


Table II

S/N

Fault Symptoms

1. Mains transformer, open circuit primary or secondary

Dc output zero, secondary ac zero, high resistance primary or secondary

2. Mains transformer, shorted turns on primary or secondary

Two possibilities: (a) mains fuses blown or (b) low dc output and transformer overheating because of excessive current being drawn.

3. Mains transformer, windings shorting to frame or screen

Fuses blown. Low resistance between windings and earth.

4. One diode in bridge open-circuit

Circuit behaves as a half wave rectifier. Lower dc output with poor regulation. Increased ripple at 50Hz not 100Hz as should be the case.

5. One diode in bridge short circuit

Mains fuse blown, since secondary winding will be practically shorted every other half cycle. A resistance check across each arm of bridge is required, measuring the resistance of each diode in the forward and reverse direction.

6. Reservoir capacitor open circuit

Low dc output with very high values of ac ripple on output.

7. Reservoir capacitor short circuit

Fuses blown. DC resistance of unstabilized line low in both directions

8. Error amplifier in regulator open circuit

High dc output that is unregulated. No control signal for the series element.

9. Series transistor open circuit base- emitter

Zero dc output. The unstabilized dc will be slightly higher than normal since no current is being drawn.

10. Reference Zener short circuit

Low dc output. Possibility of series transistor overheating.

EXERCISE: POWER UNIT WITH A SIMPLE LINEAR REGULATOR

In figure 17, R1=470Ω ,R2=1.2k ,R3=820Ω ,R4=1 k ,C1=3300μF ,C 2=0.22 μF ,Q1=BC 108 ,Q2=BFY 51.

The unit incorporates most of the features discussed earlier and is designed to give an output of 12V at 100mA. The output resistance is less than 0.5Ω, the load regulation better than 0.5%, and the ripple less than 5mV pp on full load. The unstabilized dc is obtained from a bridge rectifier circuit and a reservoir capacitor of 3300 μF. The transformer has a secondary voltage of 12V rms so the unstabilized voltage across C1 will be approximately 12√2 i.e. about16V .

The reference voltage is provided by a 5.6V Zener diode; Q1 is the dc error amplifier which compares a portion of the dc output voltage, the voltage across R4, with the reference. Any difference between the two voltages is amplified by Q1 and the amplified signal is fed to the base of Q2. Consider the case when

27

C1 C2

R1 R2

R3

R4

Q1

Q2

Z1

T1


the dc output falls when more load current is taken; the base voltage of Q1 decreases and it conducts less current. Therefore Q1 collector voltage rises, and this rise in voltage is coupled through Q2, which acts as an emitter follower to counteract the original fall in output. Thus, the circuit operates to maintain the output as nearly constant as possible.

1

2

4

3

12

3

4+

-

+ +- -

L

NSW1

Fuse

Fig. 17

220V rms50 Hz

The normal dc voltages measured with a standard multi range meter are as follows:

Test Point

1 2 3 4

Voltage 16 13 5.8 12.2

First let us consider the following fault condition:

TP

1 2 3 4

V 17.5 17.5 0 0

The dc output is zero, but the unstabilized input to the regulator has risen, indicating that little current is being drawn. Also TP2 is at the same voltage as TP1. This, further, shows that no current at all is flowing through R1 into the base of Q2. The only possible fault is that Q2 has an open circuit base-emitter junction. Note that if R1 were open circuit TP2 would be at zero Volts.

28


Consider a fault condition when all test points are at zero Volts. Further inspection shows that the fuse has blown. Resistance checks give the primary resistance as 43Ω, the secondary resistance as 4Ω, but TP1 to ground is zero Ohm. The fault in this case can only be C1 short circuit.

PROBLEMS

The following table lists a series of fault conditions; in each case state which component or components could cause the fault and give a supporting reason. The solutions to the problems have been provided. The reader is required to compare his own independent reasoning with the suggested solutions.

Voltages at indicated Test Points

Fault 1 2 3 4 Other symptoms. SolutionA 16 15 14.5 14.5 Z1openB 11 6 4.8 5 Increased ripple. C1openC 16 15 5.8 14.5 Poor regulation. R3 open or BE of Q1

openD 17.5 0 0 0 R1 openE 16.5 2.1 0 1.5 Z1 shortF 17.5 17.5 0 0 BE of Q2 openG 16 7.5 5.8 7 R4 openH 16 5.9 5.9 5.2 Poor regulation. CEof Q1 short

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