1. No category

ENGN3225
Tutorial 1
2002
Power
Power in a three terminal device: Choose a reference terminal, then sum over the other two:
say ref. is E on a transistor, then
P = VCEIC+VBEIB.
P = V1I1+V2I2
Extension of Power Factor to non-Sinusoidal Waveforms: PF = IS1/IS ´ DPF (Keith should cover)
Why is complex power S = V ´ I* ?
We need to use a formalism which is independent of phase angle – two of the three phase voltage sources are
displaced in phase (Ð120°,Ð240°). Assume we want the power when Z = 1W (i.e. real), and V = 1+j.
S = V´ I*
=(1+j)(1-j) = 1 – –1 + j (1-1) = 2Watts (Real only – reactive power = 0).
This is correct as V and I are in phase. If we had (incorrectly) used V´ I, (complex quantities, but omitting
the *) we would have calculated 2VAr (2 units of reactive power).
Three phase Rotation Conventions
Phasors are assumed to rotate anti-clockwise in time. So if we draw a phasor for Va at three consecutive time
intervals, it looks like fig a. Va (phase a) is taken to be the reference phase, so that its phase angle is 0 by
definition. phase b lags by 120°, so at that instant, it will be drawn at –120°, or “7o’clock” (fig. b), and phase c.
Note that this requires that we order the phases CW, so that it is consistent with rotation CCW.
Va
Va
VC –e.g. 240Ð-240°
(240° lagging º
120° leading)
Va
t=0
rotation
t =T/3
t =2T/3
fig 1a/ Va phasor at 3 consecutive times
Vb e.g. 240Ð-120°
(120° lagging)
Va 0°–e.g.
240Ð0°
fig 1b/ three phase
phasors all at t=0
Why Distribute Power as AC, three phase?
Why AC?
· Transformers can change voltage/current level to match current magnitude with mechanical design
considerations – and in transmission, high voltage low current can be used for long-distance, reduced I2R
loss, c.f. low voltage for distribution inside buildings.
· Transformers can isolate for safety.
· Fuses and Circuit Breakers interrupt AC much more readily than DC.
· sliding contacts are much simpler (or not required at all) in motors – so higher currents and powers are
possible. (DC motors must have a commutator )
Why three phase?
· More efficient transmission (less copper) – regard as three separate circuits, (Vo, VoÐ120°, VoÐ240°),
whose return currents cancel (if the system is balanced) so no return conductor required in theory (in
practice smaller gauge “neutral” conductors are used at the user end of the network. (i.e. ~half the mass of
conductor compared to a single phase system)
· Provides a natural source of rotation for machinery (in principle a two phase, quadrature (Vo, VoÐ90°),
system would do too, but the neutral is not balanced, so still need three wires)
· Interfaces better with rectifiers – load current has less ripple, line current drawn is closer to a sinusoid than
single or two phase.
· Naturally extends to 6 phase by simple inversion using a 3f transformer.
Terminology
Confusion arises over 240V vs. 415V. They are really the same thing, just measured Phase-Neutral (240) and
Line-Line(415V). When reading texts, note that American voltages are different. The nominal voltages there
are 120 and 208(l-l) respectively, but there are many alternatives e.g. “domestic” 115V/230V two-phase, and
220V/380V 3 phase – a real mess, spawning a sizeable industry in power/phase conversion.
Best advice is to be unambiguous. Never say Vphase (strict meaning is Vl-l) or Vline (strict meaning is Vl-n) –
say Vline-neutral or Vline-line, similarly for I.
B.D. Blackwell
p1/12
ENGN3225
Tutorial 1
2002
Comments on Nodal admittance matrix method:
Admittance analysis is naturally suited to power systems because they are voltage sources, with loads connected
in parallel. Nodal Admittance Matrix method is the basis for load-flow, fault level, and general power systems
analysis.
· Can be applied to one 3 phase circuit, or to transmission networks (many interconnected circuits), either
“per-phase” (if the network and load are symmetric, or if they can be transformed to be symmetric using the
SCT below) or explicitly for all phases.
· Can write by inspection. (see two simple results for delta and star configurations in Ass 1 qu’s 1&2). Often
simple, may have many zeros (zero for each element Yij for which there is no direct connection between
nodes i and j. If there are “internal” nodes, you probably want to consider these as explicit nodes to restore
the simplicity (but at the expense of increasing the size of your matrix). These can then be eliminated first if
they are simple – e.g. no generator connected to them. (Internal nodes occur for example when connections
between explicit nodes are “T” networks rather than “p” networks.)
· Nodal impedance matrix Z ºY-1is usually not as simple in mains circuits (which often contain more
paralleled elements than elements in series) à zeros are rare, can’t easily write by inspection. Invertability
is not guaranteed, such as when there is no connection to the reference node e.g. 3 terminal delta config.
(hence Keith’s use of y44 = 1e12 in Ass 1)
· Nodal impedance matrix is different to the mesh impedance matrix: The nodal method naturally applies to
power systems, where the all three phases are referred to the one node, the common node
· The NAMS of two circuits in parallel simply add. (e.g. loads connected to a mains network).
· For a passive circuit, the NAM is symmetric about the leading diagonal. (why?)
· The number of equations is easily found – just one less than the number of nodes less the neutral node,
which is usually the reference. Mesh analysis may be harder, but the number of equations is not clear.
See KAW’s circuit maths p9-10, also Bergen Example 6.1 (but in a different context: power flow analysis)
[Symm. Cpt transform was here in 2002 – enlarged in Tut 3]
B.D. Blackwell
p2/12
ENGN3225
Tutorial 2
2002
TUTORIAL 2
RMS Phasors
In power engineering we usually use RMS phasors by default, so 240Ð120° means a cosine with amplitude
339.4V (240*sqrt(2)). Note that Spice uses amplitudes for AC, so you would need to enter 339.4 there. (It only
matters if you plot voltages/currents versus time.)
Note: In the lecture notes, Vphase, Vphase-neutral, Vline-Neutral all mean the phase-neutral voltage (e.g. VAN ,
typically 240V), whereas Vline-line refers to the voltage between phases (e.g. VAB, typically 415V).
Star-Delta transformation
What is the equivalent “Delta” circuit to a general “Wye” or “Star” arrangement of admittances YA, YB,
Y C?
Call the Delta equivalents YAB, YBC, YCA, as they connect two nodes, in contrast to the Wye with a single
subscript, as admittances here go from one node to neutral.
You can show, by writing the Nodal Admittance Matrix for both circuits, that
YAB = (YA ´ YB) / (YA + YB + YC) this is referred to “straddle” / “sum”. (meaning product of the straddling
nodes over the sum of all nodes). For the special case YA = YB = YC = Y, YAB = Y/3. etc. In other words, a star
arrangement of 10W loads would be indistinguishable from a delta arrangement of 30W loads (in terms of any
measured external currents or voltages).
Conversely the delta to star transform looks the same in terms of Z’s e.g. ZA = (ZAB ´ ZAC)/( ZAB + ZBC + ZCA).
If expressed in terms of Y’s, this looks a bit different – 7 Y terms, and is harder to remember.
Using this transformation makes the delta connected 10kVA load in example 1 trivial to incorporate: transform
to Wye (Z=Z/3), then add in the admittance to the first three loads.
References for Power Network Theory.
Johnson, Irwin, Bergen, Schaum – details in bibliography
B.D. Blackwell
p3/12
ENGN3225
Tutorial 3
2002
Lab 1 report due beginning lecture 29th Aug
Assignment I – Due beginning lecture 22nd Aug
a/ Derive the Star-Delta Transform from consideration of the nodal admittance matrix
b/ Repeat the problem in the notes (Assg 1) with loads (1 and 2) on phases a and b swapped. (Both open and
grounded neutral).
c/ Show how a two phase wattmeter (measures real power with two V sensors, two I sensors, output =
Real(V1I1)+Real(V2I2)) can measure real power in an unbalanced three-phase star-connected system.
Hints/Notes about Pspice:
·
We mainly use “Transient Analysis” because power electronics circuits are usually non-linear (rectifiers),
the signals are not “small”, or the start-up transients are significant (capacitor charge-up transient). The
usual “AC” or “small-signal” analysis although very fast, is not valid in these cases.
· This means we should use sources (VSIN, VPULSE) which are set up for transient analysis, not VAC.
· No spaces allowed in values: (e.g. 10 Ohm gives an error “Missing value”: instead use 10.Ohm - decimal
point separates the Oh from the zero clearly (or 10Ohm or 10R or 10ohm) ) Also need to spell MEG
10Mohm means 10 milliohm.
· You need an earth somewhere .e.g GND_EARTH
· Saving setup time in “Probe”:
1. Use voltage and Current markers in the schematic editor
2. If you set up complicated analysis, and wish to repeat that for a number of circuit changes, you can
check “Restore Last Probe session” under Analysis/Probe Setup in the Schematic editor. Downside can
be error messages about “missing” traces when changing files.
· RMS: Can use the RMS() function in add/Trace, or if there is time dependence (transients), 1 cycle average
is better – sqrt(avgx(I(R1)*I(R1),20ms)) is a 50Hz 1 cycle calculation of the RMS current in R1.
· Similary, average power can be calculated as sqrt(avgx(I(R1)*(V(R1:2)-V(R1:1)), 20ms)). You can define
a macro (probe/Trace/Macros) for these (don’t forget to save, or just keep the macro in a file and cut and
paste when needed) example:
avgpwr(a) = avgx((v(a:1)-v(a:2))*i(a),20ms).
rms20(a) = sqrt(avgx(a*a,20ms))
Note: the default file type is “probe utility” file .prb, which I tend to delete often……careful.
· Can see details of a model (e.g. 1N4002): MSimeV_8\lib\eval.lib (like many files) is in text format and has
part definitions and some useful comments that seem to be not otherwise visible, alternatively place the part
in the schematic editor and select Edit/Model/(Text instance) (then cancel edit) this misses comments, but
covers all library files.
http://www.cadencepcb.com/products/downloads/
Circuit simulation is limited to circuits with up to:
64 nodes, 10 transistors, 65 digital primitive devices, 10 transmission lines in total (ideal or non-ideal), 4
pairwise coupled transmission lines.
Additional limits:
The sample library includes 39 analog and 134 digital parts. Device characterization (i.e. adjusting
characteristics to make variant components) in the PSpice Model Editor is limited to diodes.
Various component characteristics:
2n2222 – 1.2W Sat R
Qbreak does not seem to have any resistance?
IRF150 goes down to 50mW at 10V VGS – this may be over the top? – yes, there is apparently no series R. (No,
it is very low, about 1mOhm – model is quite detailed)
1N4002 has the same area as Dbreak, but a different resistance turnover point (slightly) 39mW. (34 in model)
Dbreak 0.11W per unit area (model has R=0.1 and C=0.1 in it – that is all)
Area of 1 is about 1.6A at 1 volt, and about 0.1W series R.
SCR 2N1595 is too complicated – only two fit into the limits of the evaluation package.
Pspice 9.1 vs 8
Nicer interface to the “Probe” data plotting module. Darlington transistors included in (eval) library. Win 2000.
Probe in 9.1 has a distracting grid layout.
libraries are different in9.1 – can use others?
Mohan pSpice examples: 4.6-2 for buck convertor
Examples 6-2 and 6-3 in appendix to ch 6 (p158) – look like .cir files.
Problem 10-17 for forward convertor with feedback.
B.D. Blackwell
p4/12
ENGN3225
Tutorial 3
2002
Comments on Symmetric Component Transform (SCT)
The symmetric components of current and voltages in a 3 phase circuit can be used to simplify analysis, and are
the standard way [KAWCM § 4.1-2, Bergen Ch 13, Weedy Ch7] to simplify treatment of a non-symmetric
transmission fault (e.g. one phase shorted to neutral ) which we didn’t cover(faults) in detail this year.
Mathematically it is a different choice of variable for V,I in the form of a basis set
I = S aiIi where i=a,b,c
which diagonalises the circuit matrices for balanced circuits. So it is a formalism for studying unbalanced
excitation (e.g. faults) in balanced networks, and for studying unbalanced loads in balanced networks. Usually
in the context of the gross imbalance of a fault (short) the ever-present, but often small imbalances in the circuit
or network can be ignored.
The three phasors Ia, Ib, Ic are written as a sum of the positive sequence (three phasors which are the ideal
excitation (all same amplitude, exactly 120°) in the positive rotation, the negative sequence (ditto, but negative
rotation) and the zero sequence (no rotation – i.e. all in the same phase).
To obtain the Z1 (normal or positive rotation) term, we consider the network to be excited by currents [Ia, Ib, Ic]
(or voltages) of the normal positive sequence ([Ia, Ib, Ic] = [VR, VRh2, VRh].
So V1 =…
Or regarding Ia as the reference phase current, we write:
I abc
éI a ù
é1ù
é1ù
é 1 ù é1 1
ê
ú
0ê ú
1ê 2ú
2ê
= ê I b ú = I a ê1ú + I a êh ú + I a ê h úú = êê1 h 2
êë I c úû
êë1úû
êë h úû
êëh 2 úû êë1 h
1 ù é I a0 ù
ê ú
h úú ê I a1 ú = HI 012
h 2 úû êë I a2 úû
While this seems to be a complication – we now have 9 currents instead of 3, the symmetries lead to great
simplification of both the maths and understanding. Because of the simple relationship between components, we
usually consider only the a phase components (called the “lead” or “reference” phase), and obtain the b and c at
the last stage by rotation by h. Thus we often write I0 when we strictly mean Ia0. Note that H-1 = 1/3H* (*=conj.
i.e. swap “h2”s with “h1”s).
Notes:
1.
If the circuit is linear, superposition applies, and if the (symmetric) circuit only has sources just of one
sequence(+,-,0), then the responses are of the same sequence.
2. + and – sequence sets are balanced (no neutral) – so for balanced networks and loads, you can assume all
neutral currents are zero for +,- sequ. (and usually that neutral voltages are zero).
3. For the 0 sequence (all in phase), any star networks with no neutral connection can be ignored altogether.
4. Unbalanced faults can be converted to balanced Symmetric component current sources.
5. Unbalanced sources, when represented in SCT are balanced individually (i.e. for the 0,1&2 cpts) – just the
sum will be unbalanced.
The A phase (lead phase) can be reconstructed from the 0,1,2 SCT components by simple addition, but don’t
forget the phase rotation terms (h2”s and “h1”s) when reconstructing the B and C phases, or the 1/3 when going
from A,B,C, to 0,1,2.
Need some simple examples – e.g. (could be next assignment (2?, but device questions also)
1. express the unbalanced 3 phase source (240Ð0, 245Ð120,240Ð315) in terms of its nine SCT components.
2. Give the lead phase symmetric components (3) of a co-phase source with unequal voltages (V, V+D , V-D, ).
And some more complicated ones. Also, illustrate the “series equivalent connection” aspect, but point out that
solving detailed problems that way is not examinable.
More examples of this, and the DQ0 or Park transform will be dealt with in the context of motors[KAWTM],
where the relationship of electrical and mechanical rotation is particularly important (e.g. a negative sequence
component opposes motor rotation and wastes power, heating the motor.).
B.D. Blackwell
p5/12
ENGN3225
Tutorial 4
2002
Three Phase Rectifier Example
The diagram on the left shows the combination of two “halfwave” three phase rectifiers to produce one 6 pulse 3 phase
bridge rectifier. This simplifies the analysis. The usual
simplifying assumption, that the load consists of series L and
R, in which L/R >> w is made, whereby we can replace it with
a constant current source, as the current cannot change appreciably during a cycle. Consider the left rectifiers,
producing the output voltage vd1 Fig 4.9a shows the output, which is the maximum at each time instant of all
three sources. By symmetry we can see that the current Ia is equal to Id in the interval p/6 < wt < 5p/6. (30150°), and zero elsewhere – see I1 graph in the right diagram (2-21 Lander).. The other currents Ib and Ic fill up
the gaps so that Id can indeed remain constant. Looking at the right rectifier in 4.9 Kassakian, the same
waveforms obtain, but displaced by p (inverted). So the current in the a phase here is the opposite sign at the
opposite phase. We can combine the sources va on either side, so the currents add, making a symmetric 5 piece
stepwise approximation to a cosine. (Ia in Fig 2.21)
If we move the “time zero” to centre the total Ia current pulse, then the Fourier transform integral contains only
cosine terms (even symmetry about t=0) and only odd harmonics (“half wave symmetry”: f(t) = -f(t+T/2) Mohan
p41), so
FT =1/(2p) SN ( ò0p/3 cos(Nt) dt + ò2p/34p /3 cos(Nt) dt + ò5p/32p cos(Nt) dt = 4ò0p /3 cos(Nt) dt (by symmetries)
= 2/pSN [1/N sin(Nt)]0p/3 = 2/pSN1/N sin(Np/3) = Ö3/p ´ [0,1, 0, 0, 0, -1/5, 0, +1/7, 0, 0, …]
Note
· the 2nd,3rd and 4th harmonics have been eliminated, leaving just [5,7,11,13…]´w. – KAW notes show how
higher order rectifiers eliminate more harmonics
· thus although the waveform of Ia looks “rough”, it is the best approximation to a sine curve possible with
three discrete levels.
· The 3 phase circuit represents another step in the progression of improving load quality, starting with the
worst (1/2W, Cap.) then FW Cap, then FW Inductive (“square wave” current), then this.
· you should be able to see this structure in the Spice 3 phase 6 pulse rectifier in the lab report, particularly if
you put a large inductor in series with the load. (Remember – we are looking at the current drawn from the
mains, not in the load.
· If you look at the currents in a delta source (Iy fig 2.21you can see how a changing the source configuration
can make the approximation to sine even better.
There is a subtlety here: The simple waveshape of Ia is suggestive that current only flows through one diode
pair per 120° - that is true of the forward path, but the return is split between two diodes 60° each. This can be
easily seen from the comparison of vd1 and vd2 in fig 4.9, where diodes on phase b , then on phase c return the
current. during the phase a forward period. See also Mohan § 5-6.
B.D. Blackwell
p6/12
ENGN3225
Tutorial 4
2002
Hints for Phase Controlled Rectifier
I would advise you to look carefully at the VCVS. The four terminals of this device are in two distinct circuits
which can be electrically isolated, so you need to make sure that the nodes are connected to the right references.
The comparator input should be connected to VSIN and VC, and the output is the drive circuit for the SCR.
TCR problem
a=!pi/2*range(1,2,num=90) & b=!pi-a
n=1.001 & plot,(sin(n*b)*cos(a)+n*Cos(n*b)*Sin(a))/(n-n^3),col=6 , yr=[-.2,1]
for n=3.0,9,3 do oplot, (sin(n*b)*cos(a)+n*Cos(n*b)*Sin(a))/(n-n^3),col=2+n/3
B.D. Blackwell
p7/12
ENGN3225
Tutorial 5
2002
Understanding Induction Motor Theory by Comparing with DC Motors
The basic theory of an induction motor, given in Notes
KAWTM§2.3 and in more detail in Mohan § 14.2 can be
illuminated, to explain the term involving R/s, by considering
the simplest possible case – a DC motor. The equivalent
circuit for a DC motor (c.f. Mohan fig 13.2) is Lr (which we
assume is fixed), the physical winding resistance Rw, in series
V0
with the back EMF due to rotation. We use Rw here, because
st
it is the sum of rotor and stator resistances in a DC motor. (1
year text ([Cogdell]) useful introduction here – Steinmetz
transformation p805-842).
In the steady state, Ohm’s law à
Lr
Rw
EMF
I = Rw/(V0-emf).
The emf is proportional to rotation speed, reaches V0 at idealized full speed (this would be the synchronous
speed for an induction motor), and is zero when stationary, so we can obtain
emf = V0*(1-s), where s is the per-unit slip.[s = (ws –w)/ ws, where w is the rotor angular velocity and ws is the
free-running value – no slip]
We can then simplify the circuit, for the steady state, dw/dt=0, by
incorporating the emf into a new, speed dependent, equivalent resistance
given by V0/I which reduces to just Rw/s. Note that this model allows for
Rw/s
Lr
(but does not predict the value of) friction and air-resistance implicitly,
V0
because the result of these will be that s>0 even with no load, so the motor
will be modelled as consuming a finite current( at no load). (Note that most
texts treat DC motors by considering torque: my treatment shows the
relationship between synchronous motors and simpler motors.) The above
equivalent circuit is also applicable to an induction machine, although the
reasoning is more complex. A full treatment (Mohan 14-2) involves the
concept of magneto-motive force, and recognises that the frequency of the induced current in the rotor depends
on the speed of the rotor (partial explanation below).
Finally, as explained in the notes (§2.3.1-2), the induction motor model includes a transformer to model the
current in the rotor being induced rather than directly connected. (Note that if an autotransformer is used during
startup(KAWTM§2.3.5), then there is an additional transformer to consider.)
Power, Torque
For power calculations, the astute reader will notice that for the DC motor case, of the total power going into the
equivalent resistor Rw/s, the power going into mechanical rotation is I´ emf and the balance (Vo-emf)´ I is lost as
heat into Rw. So to obtain the torque (by considering work done) from the power (T=P/w), in principle, only the
power going into rotation should be calculated (i.e. P_mech = I´ emf). However for this equivalent circuit, if
this is allowed for, ( à P_mech = I2Rw(1-s)/s), the result (T=P_mech/w , ) is equivalent to I2(Rw)/(s*ws).
So the correct result for T is obtained more simply if the total power in Rw is used, provided also that the
synchronous angular velocity (ws) is used in place of the actual w.
Comparison of Equations for simple DC and Induction Motors
Quantity
Simple DC motor model
Induction motor (KAW symbols)
T = Pmech/w º PROTOR(1-s)/s
Pg(3/(2pns)
Power
Torque
where
p = number of pole pairs
ns = synchronous speed in revs/sec.
w = actual angular speed,
B.D. Blackwell
f = frequency of applied mains power
ws = 2pns = 2pf/p
ws = synchronous angular speed,
p8/12
ENGN3225
Tutorial 5
2002
Another fortuitous cancellation is the rotor inductance and frequency. The DC motor argument above would
indicate that the effective series L should also be divided by s, but the induced frequency in the rotor is less than
the applied frequency, and is proportional to the slip s. Therefore the two factors of s cancel, and we use simply
Lw in the rotor circuit. This rotating transformer changes frequency as well as V and I, so when seen from
the primary side, the (lower) frequency currents in the rotor secondary appear at the applied mains frequency.
Conclusion: The induction motor is a complex device, but is represented reasonably well by a surprisingly
simple equivalent circuit (4-6 components, 1-2 loops). A more detailed derivation is given in Mohan 14-2.
The essence of what you need to know for this course is in KAWTM.
;; Worked example for quiz Q3 ENGN4506
;;; IDL file: & is statement separator, ; is comment
;;
j = complex(0, 1)
;;;;;;;;;;;;;;R1 = 0.13 & R2 = 0.32 & X1 = 0.6 & X2 = 1.48 & Vln = 500/sqrt(3); notes 2-11
R1 = .05 & R2 = 0.2 & X1 = 0.6 & X2 = 1.48 & Vln = 415/sqrt(3); Ass2
rstr = 1/1.57
YMag = .004 -j*0.05
R2d = R2*rstr^2 & X2d = 1.48*rstr^2
; rotor locked or stationary means s=1
s = 1.0
; Using the transformer model - the 1e-6 fudge prevents overflow at s=0
Zin = ((r2d/(s+1e-6) + j*X2d)^(-1) + (YMag))^(-1) + R1 + j*X1
;;The model is usually in "star" or "Wye" form
;; This (star connection) is also stated in the problem
;; assume 415V means line to line voltage.
Istalled = VLn/Zin
help, r2d, x2d, Zin, abs(IStalled)
end
R2d = .081W, X2d = 0.6W, Zin = (
|Istalled| =
201.386 A
0.127813,
1.18287) W
This applies if the motor were directly connected. When the motor is fed from the autotransformer the voltage
and hence the current at the motor terminals is reduced to
0.65 * Istalled, and the current in the supply lines is reduced by a further factor of 0.65
So I_line_stalled is 0.65^2 * 201.4 = 85.1 amps.
Note that the value of slip at full load is not required in this problem (as s=1).
Other References for Motors and Transformers
3 Phase concepts are reviewed in Mohan Ch3, Motors in Cogdell. See also Irwin, Johnson and Schaum,
Transformers are covered in Mohan§3-3-9-1 to 4.
B.D. Blackwell
p9/12
ENGN3225
Tutorial 6
2002
Textbook Coverage:
Here is a reading guide. This does not mean that you have to know all of the material listed below; rather it
allows you to ignore sections altogether. The best indicator of the required depth is the notes. If we don’t cover
it in lectures, labs or assignments or exams handed out, or say specifically to read parts of Mohan, it is not
examinable.
Example
21 27-2-2 qualitative omit 27-2-3 on
means ch21, include up to 27-2-1, then 27-2-2 is qualitatively covered and 27-2-3 on is omitted.
Ch 1 Intro: should be straightforward except no “matrix convertors”
Ch 2 Review of Devices (all)
Ch 3 Review of basic electrical and magnetic circuits: All except magnetic circuits only in as much detail as in
the lab. Transformers only in as much detail as KAW’s notes.
Ch 4 Computer Simulation – only to the extent of the pSpice exercises given in the lab and class.
Ch 5 Line frequency rectifiers – Keith’s notes + lab. experience+pSpice exercises – i.e. not as much detail on
commutation, notching.
Ch 6 Phase Controlled Rectifiers: omit 6-3-2 to 6-3-4-1 and 6-4-3 on.
Ch 7 DC-DC switch mode convertors: basic formulae only, and know how to calculate boundary of cts
conduction, not all the detailed formulae. omit 7-6 (Cuk) on
Ch 8 DC-AC switch mode convertors – as per KAW notes (addendum) less detail than Mohan, omit 8-3-2-4 on
Ch 9 (omit all 9 in 2002 – not covered) 9-1, 9-4-5
10: Switching Power supply applications: (covered mainly in lab )10-2, 10-3, 10-6 (our IC TL494 is very similar
to 1524) , 10-8 (e.g. we estimated ESR in fuses lab (crowbar) 10-8-6 was the last part of an exam question
11: As per Keith’s recent notes. (most of chapter)
12-13-14 15– As per KAW notes: we are more at the level of Cogdell – too much detail in Mohan
16: Industrial Applications: most (16-3-1 and 16-3-2 not covered in 2002)
17: Electric Utitlity Applications: (most) 17-3 as per KAW (17-1 and 2, 4 and 5 not covered in 2002)
18: Power Quality: As covered in power quality lab. power factor, DPF, power factor correction, harmonics,
harmonic filters. less detail, omit 18-6,7
19: all
20: omit 20-5-4
21: 21-5-2,3 superficially
22: 22-3 (equn 22-2), 22-4 very brief coverage – too subtle, 22-5 qualitative (understand figs 22-8,22-9 and
corresponding text) 22-6-2 qualitative 22-6-4 qual, 22-6-5
23: All except GATT.
24: omit 24-4-3 on
25: most
26: qualitative
27: 27-2-2 qualitative omit 27-2-3 on
28: omit all but fig 28-33
29: Heatsinks, ~70%, to be determined after lectures
30: browse – mainly as per lab.
October Quiz:
11-18 not in quiz, ditto 26-30
B.D. Blackwell
p10/12
ENGN3225
Tutorial 6
2002
Bibliography
[KAWSAF] K.A.Walshe’s ENGN3225 course notes, Safety
[KAWCM] K.A.Walshe’s ENGN3225 course notes, Circuits and Mathematics
[KAWTM] K.A.Walshe’s ENGN3225 course notes, Transformers and Motors and Surge Divertors
[KAWSC] K.A.Walshe’s ENGN3225 course notes, Basic Switching Circuits
[BDBSD] B.D.Blackwell’s ENGN3225 course notes, Semiconductor Devices in Power Electronics
[BDBLAB] B.D.Blackwell’s ENGN3225 lab notes.
[Mohan] Text: N. Mohan, T.M. Undeland, W.P. Robbins “Power Electronics: Converters, Applications and
Design” 2nd Ed, John Wiley 1995.
Additional Reading:
[Cogdell] Cogdell, J.R. “Foundations of Electrical Engineering” Prentice-Hall, 2nd Ed 1995 [1st yr text: Not in
library] Very good basic text. chs 4,5,6 on AC circuits and power, ch 13-16 on motors and
transformers, but no matrices. Per-unit, per-phase on p275-86, Y-Delta but only for symmetric circuits.
Good intro to Half and FW rectifiers p 317. Magnetics, Induction motor analysis – intro p287+,
detailed(Steinmetz transformation) p805—42, DC motors ch 17. Some intro. info. on Power
Electronics Ch18, good treatment of Controlled Rectifier into inductive/motor load (FW), 1 page on
3phase version, does not cover commutation current transfer.
[Bergen] Arthur R. Bergen, a modern, comprehensive, advanced mathematical treatment “Power Systems
Analysis”, Prentice –Hall, 1986, many problems, lots of Symm. Cpt. transform (ch13), DQ0 transform
is called Park [or Blondel] transform pp 152-7, 444-6.
[Irwin] J.D. Irwin, Basic Engineering Circuit Analysis, 5th ed. Prentice Hall, simple, heaps of 3 phase problems
& answers (Ch11 polyphase, per-phase example 11.6, power, power factor (2 copies)).
[Johnson], D.E Johnson, J.R.Johnson, J.L. Hilburn, P.D.Scott, ‘Electric Circuit Analysis’, 3rd edition PrenticeHall (Also earlier editions with similar titles). Good level for this course, many 3 phase problems &
answers, some 2 port admittance matrix theory around p669. Spice (not pSpice) Appendix.
[Weedy] B.M. Weedy, “Electric Power Systems”, John Wiley, 1987. More practical, originally 1967, many
research paper references and problems, no answers.Nodal Admittance Matrix p 220+, Star-Delta
(result only) p83, symm. Cpt p258, but more applied than explanatory.
[Wildi] Theodore Wildi, “Electrical power Technology”, more hardware, less maths, problems with solutions.
John Wiley 1981.
[Schaum] Schaum Outline series J.A. Edminster /J.E.Swann (M.K.S. edition)– “Electric circuits”,
comprehensive, practical examples and problems, but polar notation only, no ejwt. Polyphase chapter for
3 phase.
Not so useful:
Bogart – Electronic Devices and Circuits – old (2nd ed.), but 32 page (old) Spice notes might be handy, no NAM,
SCT.
Greenwood – Transients in Electric Power Systems; MacGrawHill?? (can’t find, looks like it is not that relevant)
Concepts to be covered in 3225 Tuts:
Phasors – revise, rotation direction, RMS phasor
3 phase sources and loads
Star and Delta connections – transformation
Nodal Admittance Matrix – application to unbalanced loads, and to parallel connection in main network.
example in deriving Star-Delta transform
Symmetric Component transform
basic principles
application to fault analysis.
pSpice – mainly transient analysis because of inherent non-linearity
Motors – DC, Induction and Synchronous
B.D. Blackwell
p11/12
ENGN3225
Revision questions and requests
Tutorial 6
2002
The 2nd part of Keith's updated Basic Switching Circuits notes are on Switch Mode Inverters. You have told us
to omit Chapter 8 Mohan (covering switch mode converters) and switch mode inverters are not in the course
outline (Under Power Conversion).
Thus, do we have to learn about switch mode inverters i.e. is it examable? Rgrds, Andy
Dear Boyd,
Will switch mode convertors be examable? And can we classify fuses and thermal design to be under 1 topic and
power quality to be under another topic? Andy
I thought of emailing you to suggest some possible stuff to cover for this week's tutorials. I would particularly
like some more explanation of 3phase systems and ways of solving related problems eg. writing down the nodal
admittance matrix by inspection. It would also be good if you could give an overview by topic of what is going
to be examined. The notes for the course have proved somewhat difficult in providing the appropriate depth and
explanation of concepts in some places, therefore, would you could better outline this in the chapter list. Noting
however, that this is not supposed to be the knowledge of the course, but what is 'examinable', ie. directly
relating to the exam and not stuff that is extraneous. Since the course webpage has not provided us with an
adquate idea of the course outline and depth, this overview would be extremely helpful for achieving higher
marks in the exam. I hope only that I can get a higher mark which can better reflect upon the involvement in the
subject that we have had through the extenisve practical exprience through the laboratory sessions.
Also, if you take note of the assessment scheme for most unit, we usually have 2 assignments and a few labs.
Seeing as we have had to hand in a fair bit of assessment already, and we are still forced to hand stuff in the last
week (when we should ideally have a break from assessment so that we can study for exams), I would request a
fair amount of leniency in your marking of the remaining items that you request. I would almost like to say that
you give the solutions to the remaining assessment and change the marking scheme altogether to the following:
Quiz: 10% Lab1: 15% Lab2: 10% Lab3: 5% Assignment1: 15% Assignment2: 15% Exam : 30%
or a similar variation. A good source of guidance on the assessment scheme would be other units on the course
webpage which have got some laboratory involvement.
Thank you for your time and dedication. I hope that this exam period is successful for the students and not so
much of a strain on your hand and eyes. :DI was hoping that you would be able to make an outline of the
syllabus that was covered in the couse in chronological order and list the items within each topic that we
covered. Then for each item, you could put a chapter reference down that would be required to cover the material
for the course. This would reduce the level of ambiguity regarding the course material to be covered since there
is so much depth in the text book and greatly save time we have to study for the course.
Hence,
Motors - induction motors, examples, equations, basic circuit. Chapter ##
In particular, I am not sure about the depth of material we need for the part on Basic Switching circuits. There is
a fair bit of stuff in the text book on Buck/Boost/buck-boost/Cuk convertors which I don't know applies to us,
since the notes on the webpage are so small. I also wanted to ask whether just reading the material that is
provided on the webpage provides sufficient depth of knowledge for the exam or does the text book take
precedence on the depth?
Please send out an email when you would have the rest of the material ready. I am sure that students wouldn't
mind hearing from you. Also, would you be able to send out an email to confirm the tutorial on Wednesday at
1pm?
As I mentioned to you in the lab this afternoon, in the final lecture would you be able to give us a summary of
what course content you expect us to know for the exam, and to what level of detail. i.e. Could you go through
each of the topics, and let us know what you would like us to be able to describe, and what you would expect us
to be able to calculate/derive. The main reason that I would like the summary is to use it as a study guide, and to
‘tick off’ the topics as I have gone over the material.
I don’t expect as much detail as to tell us what content is in the exam, but a more in-depth coverage than the
chapter guide from the text-book, as the textbook goes into lots of detail, and I am note sure what detail you
expect us to remember.
B.D. Blackwell
p12/12