DESIGN+SIMULATE LAB 2022
SCR CURRENTCONTROLLED BATTERY CHARGER
 BACKGROUND
Battery chargers can be constructed from a variety of circuits, and the complexity versus price is always an important tradeoff. The recharging ability of batteries is fundamentally rooted in the bidirectional chemical process in the battery. The slower a chemical reaction takes place, the more reversible it will be. Therefore, the slower a battery is discharged and charged, the longer the lifetime of the battery. However, for practical reasons, charging time cannot be indefinite, and is typically required to be overnight (12 hours) or even 24 hours. Therefore, charging tempo and the control thereof is a very important part of the battery charging field.
Before attempting to understand the theory behind a controlled (SCR) battery charger, insight into an uncontrolled (diode) battery charger is helpful. Refer to p.6 of “Chapter 3A Lecture Extra HandWritten Notes” for background and also to the attached pages of the Power Electronics textbook by Barry Williams. Start by working through the following background tasks which will be reused for your report:
 Review the circuit diagram and waveforms of the uncontrolled singlediodebattery charger.
 Derive the dc and rms equivalent output voltages.
 Now derive the dc equivalent current (can’t use Ohm’s law directly – which is only valid for the resistor).
 Work through example 11.1.
Then write a professional individual report to accurately document the tasks below. This should include some background theoryandreferences too.
 TASK A – SET UP THE SIMULATION AND DERIVE EQUATIONS
 Construct a single phase, onepulse SCR battery charger in Multisim. The same basic circuit ^{as in section 1 above can be used, but replace the diode with a SCR and include a 340V}P ^{: ≈}^{ }^{200V}P ^{(depending on your student number) stepdown transformer between the ac supply}^{ }and the SCR. Implement your circuit in Multisim using ‘virtual’/ideal components. Also add a 10V firing/triggering pulse via a 10Ω resistor to Gate and Cathode. You must have some form of delay in the pulse to fire at a certain angle. Use component values per student number as specified below.
 Simulate typical output waveforms for the circuit (choose a firing angle ). Show voltage as well as current – and use a current probe to send the signal to the oscilloscope. Briefly discuss the waveforms to summarise the operation of the circuit.
 ^{Now, from the waveforms, derive the dc equivalent voltage as a function of V}M^{, E, and }^{}^{,}^{ }^{where V}M ^{is}^{ }^{the}^{ }^{peak}^{ }^{secondary}^{ }^{voltage. The}^{ }^{onvoltage}^{ }^{across}^{ }^{the}^{ }^{SCR}^{ }^{may}^{ }^{be}^{ }^{neglected.}
 Next, derive the dc equivalent current (can’t use Ohm’s law directly, first get the voltage across the resistor).
 For the given specs below, calculate the expected minimum and maximum firing angle, as well as expected maximum and minimum output dc current.
The following values must be used for the analysis and iterative design of your battery charger:
 ^{Rectified}^{ }^{Peak}^{ }^{Voltage:}^{ }^{V}M ^{=}^{ }^{200V}^{ }^{plus}^{ }^{sum of}^{ }^{last}^{ }^{three}^{ }^{digits}^{ }^{of}^{ }^{your}^{ }^{student}^{ }^{number.}
 Battery Voltage: E = 100 V minus sum of last 2 digits of your student number (SN).
 ^{Internal}^{ }^{Resistance: R}INT ^{=}^{ }^{1}^{ }^{}^{ }^{plus one}^{ }^{tenth}^{ }^{of}^{ }^{the}^{ }^{last digit}^{ }^{of}^{ }^{your}^{ }^{SN.}
 ^{Target}^{ }^{Current: I}DC ^{=}^{ }^{2}^{ }^{A}^{ }^{minus}^{ }^{one}^{ }^{tenth}^{ }^{of}^{ }^{the}^{ }^{2nd}^{ }^{last digit}^{ }^{of}^{ }^{your}^{ }^{SN.}
 TASK B – SET UP A SPREADSHEET TO IMPLEMENT EQUATIONS
 For a certain wanted safe dc charging current, cannot be solved symbolically, as the ^{equation for I}DC ^{contains }^{}^{ as well as a trigonometrical function of }^{}^{. This is called a}^{ }transcendental equation. One way of solving this problem, is to follow a trialerror approach. ^{I}DC ^{can be calculated for a certain hypothetical }^{}^{, the result evaluated, and a better }^{}^{ can be}^{ }guessed. By repeating this iterative procedure, a good answer for can be obtained. The ^{number of iteration steps required, depends on the accuracy of I}DC ^{required. Your task is to}^{ }^{implement such a trialerror approach in Excel. First implement the formula for I}DC ^{in one}^{ }row of your spreadsheet. Confirm that the answers obtained are correct by comparing with hand calculations and with your simulated value. Then copy and paste this row a number of ^{times, changing }^{}^{ in each row to try and find the correct I}DC ^{to an accuracy of 0.01 A.}^{ }Include screenshots of your spreadsheet in the report.
 ^{Also}^{ }^{draw}^{ }^{a}^{ }^{graph}^{ }^{in}^{ }^{Excel,}^{ }^{showing}^{ }^{I}DC ^{vs.}^{ }^{}^{.}
 Finally, go back and simulate your circuit in Multisim with the correct to confirm the waveforms, and measure if the average dc current flowing in the circuit is as expected.
 EVALUATION
Submit your .pdf report through Moodle, as well as your Excel spreadsheet and your Multisim simulation file. Each student must start a fresh, NEW word document, a fresh, NEW Excel spreadsheet, and a fresh, NEW simulation file, else penalties will apply.
The report will contribute 20% to your total grade, and will be marked according to the following criteria:
 Report style, language, references and tidiness 15%
 Graphics quality and usefulness 25%
 Technical content, accuracy and correctness 60%
~ ~ ~
11
NaturallyCommutatingACtoDCConverters
–UncontrolledRectifiers
The rectifier converter circuits considered in this chapter have in common an ac voltage supply input and a dc load output. The function of the converter circuit is to convert the ac source energy into fix dc load voltage. Turnoff of converter semiconductor devices is brought about by the ac supply voltage reversal, a process called linecommutationor naturalcommutation.
Converter circuits employing only diodes are termed uncontrolled (or rectifiers) while the incorporation of only thyristors results in a (fully) controlled converter. The functional difference is that the diode conducts when forwardbiased whereas the turnon of the forwardbiased thyristor can be controlled from its gate. An uncontrolled converter provides a fixed output voltage for a given ac supply and load.
Thyristor converters allow an adjustable output voltage by controlling the phase angle at which the forward biased thyristors are turned on. With diodes, converters can only transfer power from the ac source to the dc load, termed rectification and can therefore be described as unidirectionalconverters.
Although rectifiers provide a dc output, they differ in characteristics such as output ripple and mean voltage as well as efficiency and ac supply harmonics. An important rectifier characteristic is that of pulse number, which is defined as the repetition rate in the direct output voltage during one complete cycle of the input ac supply.
A useful way to judge the quality of the required dc output, is by the contribution of its superimposed ac harmonics. The harmonic or ripple factor RFis defined by
RF
_{v} = = =
where FF is termed the form factor. RF_{v}is a measure of the voltage harmonics in the output voltage while if currents are used in the equation, RF_{i}gives a measure of the current harmonics in the output current. Both FFand RFare applicable to the input and output, and are defined in section 11.4.
The general analysis in this chapter is concerned with single and three phase ac rectifier supplies feeding inductive and resistive dc loads. Purely resistive load equations generally can be derived by setting inductance Lto zero in the LR load equations.
 Singlephaseuncontrolledconvertercircuits–acrectifiers
 Halfwave circuit with a resistive load, R
The simplest meaningful singlephase halfwave load to analyse is the resistive load. The supply is impressed across the load every second ac cycle half period, when load current flows.
The load voltage and current shown in figure 11.1a are defined by
^{v}^{ }^{}(^{ω}^{t}^{ }) ^{=}^{ }^{i}^{ }^{R}^{ }^{=}^{ }^{{}^{}
2V sin ωt
0 ≤ ωt ≤ π
(11.1)
o o ^{{}
_{¯}^{0}
π ≤ ωt ≤ 2π
The circuit voltage and current equations can be found by substituting L= 0, β = π and φ = 0 in the
generalised equations (11.18) to (11.20) in section 11.1.3. The average dc output current and voltage
are given by
_{1}_{ }π
0 V
_{o} = I_{o}_{ }R = _{2}_{π}_{ }_{∫}
2V sin ωt d ωt =
V = 0.45V
(11.2)
BWW
NaturallyCommutatingACtoDCConvertersUncontrolledRectifiers^{236}
The rms voltage across the load, and rms load current, are
½^{「}^{ }1 ^{π}^{º}^{ }1
^{V}o rms ^{=}^{ }^{}
_{∫}_{ }2V ^{2}^{ }sin^{2} ωt d ωt _{}
= I _{o}_{ }_{rms}_{ }R = V
(11.3)
¬ ^{2}^{π}^{ }0 ]
and the power dissipated in the load, specifically the load resistor, is
P = I ^{2}
R = ½ ^{V}^{ }^{2}
(11.4)
The ac current in the load is
o o rms
R
¬^{V}^{ }「 ^{2}^{ }º^{½}
I _{ac}_{ }=
= _{R}_{ }_{}½ −
_{π}_{ }2 _{]}
(11.5)
The load voltage harmonics are
^{v}^{ }(^{ω}^{t}^{ }) ^{= }^{+}^{ }^{}^{2}^{V}^{ }^{sin}^{ }^{ω}^{t}^{ }^{−}
^{「}^{ }^{1}^{ }cos 2ωt + ^{1}^{ }cos 4ωt .. +
^{1 }cos nωt ^{º}
(11.6)
^{o }π 2
π ^{}_{¬}1 × 3 3 × 5
n ^{2}^{ }− 1 ^{}_{]}
for n= 2, 4, 6, …
 Halfwave circuit with a resistive and back emf RE load
With an opposing emf Ein series with the resistive load, the load current and voltage waveforms are as shown in figure 11.1b. Load current commences when
ωt = α = sin^{−}^{1} ^{E}
(11.7)
and ceases when
ωt = π − α = π − sin^{−}^{1} ^{E}
(11.8)
The diode conducts for a period θ = π – 2α, during which energy is delivered to both the load resistor R
and load back emf E.
(a)
(b)
Figure 11.1. Singlephasehalfwaverectifiers:
 purelyresistiveload,Rand(b)resistiveloadRwithbackemf,E.
237
The load average and rms voltages are
PowerElectronics
_{V }( ^{α}^{ }⎞
_{1}_{ }π −α
α^{ }
o _{π} = _{}½ + _{}_{ }E
⎝ ⎠
^{+}^{ }2π ^{∫}
2V sin ωt d ωt
(11.9)
= ^{(}½ + ^{α}^{ }^{⎞}^{ }E + ^{1}
2V cos α
 _{π}_{ } _{π}
⎝ ⎠
^{V}o rms
「
^{=}^{ }^{E}^{ }^{2}
(_{½}_{ }_{+}_{ }^{α}^{ }⎞ _{+}_{V}_{ }_{2}_{ }(
   π
½ − ^{α}^{ }+ ^{1}
π 2π
sin 2α
_{⎞}_{2}_{ }_{º}½
_{}_{ }
(11.10)
_{¬} ^{⎝}^{ }^{⎠}^{ }^{⎝}^{ }^{⎠}^{ }^{}_{]}
The load average and rms currents are
I = ^{1}^{ }^{「}
cos α − E ^{(}½ − ^{α}^{ }^{⎞}^{º}^{ }= ^{1}^{ }^{「}
sin½θ − E ^{θ}^{ }^{}^{º}
(11.11)
^{o }R ^{}^{ }π
 _{π}_{ }^{}
R ^{}^{ }π
2π ^{}
_{¬}
1 ^{「}V ^{2}
^{⎝}^{ }^{⎠}_{]}_{ }_{¬}
_{]}
_{θ}_{ }_{º}½
I = _{}
sinθ −
V E sin½θ + _{(}V ^{2}^{ }^{}+ E ^{2}^{ }_{)}_{ }_{}
(11.12)
o rms
R _{}_{¬}_{ }2π π
2π _{}_{]}
The total power delivered to the REload is
P = P + P = I ^{2}^{ }^{}R + E I
(11.13)
o R E o rms o
Example11.1: Halfwaverectifierwithresistiveandbackemfload
A dc motor has series armature resistance of 10Ω and is fed via a halfwave rectifier, from the single phase 230V 50Hz ac mains. Calculate
 The rectifier diode peak current
 The motor average starting current
If at full speed, the motor back emf is 100V dc, calculate
 The average and rms motor voltages and currents
 The motor electrical losses
 The power converted to rotational energy
 The supply power factor and motor efficiency
 Diode approximate loss if modelled by v_{D}= 0.8 + 0.025× i_{D}.
Solution
Worst case conditions are at standstill when the motor back emf is zero and the circuit and waveforms in figure 11.1a are applicable.
 The peak supply and peak load voltage is √2×V = √2×230 = 325.3V. The peak diode and load current is
i^{ˆ}^{ }^{}= i^{ˆ}
_{=}_{ }_{v}ˆ
= ^{325.3V}^{ }= 32.5A
o D o _{R}
10Ω
 The motor average current, at starting, is given by equation (11.2)
V_{o}_{ }= I_{o}_{ }R = 0.45 × 230V=103.5V
I = ^{V}^{o}
o _{R}
= ^{103.5V}^{ }= 10.35A 10Ω
With a 100V back emf, the circuit and waveforms in figure 11.1b are applicable.
The current starts conducting when
ωt = α = sin^{−}^{1} ^{E}
= sin^{−}^{1}
= 17.9°
The current conducts for a period θ = π – 2α = 180° – 2×17.9 = 144.2°, ceasing at ωt= π – α = 162.1°.
 The average and rms load currents and voltages are given by equations (11.9) to (11.12).
V = ^{(}½ + ^{α}^{ }^{⎞}^{ }E + ^{1}
2V cos α
^{o } _{π}_{ } _{π}
⎝ ⎠
= ^{(}½ + ^{17.9}^{°}^{ }^{⎞}^{ }× 100V + ^{1}
2 × 230V × cos17.9° = 158.5V
^{ }180° ^{ }π
⎝ ⎠
NaturallyCommutatingACtoDCConvertersUncontrolledRectifiers^{238}
¬
] I = ^{1}^{ }^{「}^{ }sin½θ − E ^{θ}^{ }^{}^{º}
^{o }R _{}_{ }π 2π _{}
= ^{1}^{ }^{「}^{ }^{2}^{ }^{× }^{230}^{ }^{V}^{ }sin½ × 144.2° − 100V × ^{144.2}^{°}^{º}^{ }= 5.85A
¬
] 10Ω _{ }π 360° _{}
^{V}o rms
「
^{=}^{ }^{E}^{ }^{2}
(_{½}_{ }_{+}_{ }^{α}^{ }⎞ _{+}_{V}_{ }_{2}_{ }(
π
½ − ^{α}^{ }+ ^{1}
π 2π
sin 2α
_{⎞}_{2 }_{º}½
_{}_{ }
 
_{¬} ^{⎝}^{ }^{⎠}^{ }^{⎝}^{ }^{⎠}^{ }^{}_{]}
^{「} ( ^{17.9}^{°}^{ }⎞ ( ^{17.9}^{° }^{1}
_{⎞}_{2}_{ }_{º}½
^{=}^{ }^{1002}^{ }_{}^{½}^{ }^{+ }_{}_{ }^{+}^{ }^{2302}^{ }_{}^{½}^{ }^{− }^{+}
^{sin}^{ }^{2}^{ }^{×}^{ }^{17.9}^{°}_{}_{ }
= 179.2V
_{¬}_{ }^{⎝}
1 ^{「}V ^{2}
180° _{⎠ }_{⎝}
180°
2π
_{θ}_{ }_{º}½
^{⎠}^{ }_{]}
I = _{}
sinθ −
V E sin½θ + _{(}V ^{2 }^{}+ E ^{2}^{ }_{)}_{ }_{}
o rms
R _{}_{¬}_{ }2π π
2π _{}_{]}
1 ^{「}230^{2}
= _{}
sin144.2° − × 230V × 100V × sin½ × 144.2° + _{(}230^{2} + 100^{2} _{)}
144.2°^{º}^{½}

10Ω _{}_{¬}
= 10.2A
2π π
360° _{}_{]}
 The motor loss is the loss in the 10Ω resistance in the dc motor equivalent circuit
P = I ^{2}^{ }^{}R = 10.2^{2} × 10Ω = 1041.5W
R o rms
 The back emf represents the source of electrical energy converted to mechanical energy
P_{E}_{ }= E × I _{o}_{ }= 100V × 5.85A = 585W
 The supply power factor is defined as the ratio of the supply power delivered, P, to apparent supply power, S
pf = ^{P}
_{=}_{ }P_{R}_{ }+ P_{E}
_{=}_{ }1041.5W + 585W _{=}_{ }_{0.69}
^{S }^{V}^{ }^{×}^{ }^{I}o rms
The motor efficiency is
230V × 10.2A
η = ^{P}^{E}
P_{R}_{ }+ P_{E}
_{= }585W 1041.5W + 585W
× 100 = 40.0%
 By assuming the diode voltage drop is insignificant in magnitude compared to the 230V ac supply, then the currents and voltages previously calculated involve minimal error. The rectifying diode power loss is
P = 0.8 × I + 0.025Ω × I ^{2}
D o o rms
= 0.8 × 5.85A + 0.025Ω × 10.2^{2} = 7.3W
♣
 Singlephase halfwave rectifier circuit with an RL load
A singlephase halfwave diode rectifying circuit with an RL load is shown in figure 11.2a, while various
circuit electrical waveforms are shown in figure 11.2b. Load current commences when the supply voltage goes positive at ωt = 0. It will be seen that load current flows not only during the positive half of the ac supply voltage, 0 ≤ ωt≤ π,but also during a portion of the negative supply voltage, π ≤ ωt≤ β.
The load inductor stored energy maintains the load current and the inductor’s terminal voltage reverses
and is able to overcome the negative supply and keep the diode forwardbiased and conducting. This current continues until all the inductor energy, ½Li^{2}, is released (i= 0) at the currentextinctionangle(or
cutoffangle), ωt=β.
During diode conduction the circuit is defined by the Kirchhoff voltage equation
_{L }didt
+ Ri
= v_{R}_{}+ v_{L}
= v=
Vsin ωt
(V)
(11.14)
where Vis the rms ac supply voltage. Solving equation (11.14) yields the load (and diode) current
i(ωt) =
^{2}^{ }^{V}^{}_{{}sin (ωtφ) +
Z
sin φ
_{e}ωt/ tan φ _{}}
(A)
(11.15)
0 ≤ ωt
≤ β ≥ π
(rad)