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 bi-directional 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 Hand-Written 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 re-used for your report:

  1. Review the circuit diagram and waveforms of the uncontrolled single-diodebattery charger.
  2. Derive the dc and rms equivalent output voltages.
  3. Now derive the dc equivalent current (can’t use Ohm’s law directly – which is only valid for the resistor).
  4. Work through example 11.1.

Then write a professional individual report to accurately document the tasks below. This should include some background theoryandreferences too.

  1. Construct a single phase, one-pulse 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 340VP : ≈ 200VP (depending on your student number) step-down transformer between the ac supply and the SCR. Implement your circuit in Multisim using ‘virtual’/ideal components. Also add a 10-V firing/triggering pulse via a 10-Ω 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.
  1. 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.
  2. Now, from the waveforms, derive the dc equivalent voltage as a function of VM, E, and , where VM is the peak secondary voltage. The on-voltage across the SCR may be neglected.
  1. Next, derive the dc equivalent current (can’t use Ohm’s law directly, first get the voltage across the resistor).
  1. 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:

  1. Rectified Peak Voltage: VM = 200V plus sum of last three digits of your student number.
  2. Battery Voltage: E = 100 V minus sum of last 2 digits of your student number (SN).
  3. Internal Resistance: RINT = 1 plus one tenth of the last digit of your SN.
  4. Target Current: IDC = 2 A minus one tenth of the 2nd last digit of your SN.
  1. For a certain wanted safe dc charging current,  cannot be solved symbolically, as the equation for IDC contains as well as a trigonometrical function of . This is called a transcendental equation. One way of solving this problem, is to follow a trial-error approach. IDC 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 IDC required. Your task is to implement such a trial-error approach in Excel. First implement the formula for IDC 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 IDC to an accuracy of 0.01 A. Include screenshots of your spreadsheet in the report.
  1. Also draw a graph in Excel, showing IDC vs. .
  1. 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.

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:

  1. Report style, language, references and tidiness 15%
  2. Graphics quality and usefulness 25%
  3. Technical content, accuracy and correctness 60%

~ ~ ~




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. Turn-off 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 forward-biased whereas the turn-on of the forward-biased 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


v = = =

where FF is termed the form factor. RFvis a measure of the voltage harmonics in the output voltage while if currents are used in the equation, RFigives 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 L-R load equations.

  1. Single-phaseuncontrolledconvertercircuitsacrectifiers
  1. Half-wave circuit with a resistive load, R

The simplest meaningful single-phase half-wave 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 (ωt ) = i R = {|

2V sin ωt

0 ≤ ωt ≤ π


o o {


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 = Io R = 2π

2V sin ωt d ωt =

V = 0.45V




The rms voltage across the load, and rms load current, are

½ 1 πº 1

Vo rms = |

2V 2 sin2 ωt d ωt |

= I o rms R = V


¬ 2π 0 ]

and the power dissipated in the load, specifically the load resistor, is

P = I 2

R = ½ V 2


The ac current in the load is

o o rms


¬V 2 º½

I ac =

= R |½ −

π 2 |]


The load voltage harmonics are

v (ωt ) = + 2V sin ωt

1 cos 2ωt + 1 cos 4ωt .. +

1 cos nωt º


o π 2

π |¬1 × 3 3 × 5

n 2 − 1 |]

for n= 2, 4, 6, …

  1. Half-wave circuit with a resistive and back emf R-E 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 = sin1 E


and ceases when

ωt = π − α = π − sin1 E


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.



Figure 11.1. Single-phasehalf-waverectifiers:

  1. purelyresistiveload,Rand(b)resistiveloadRwithbackemf,E.


The load average and rms voltages are


V ( α

1 π −α


o π = |½ + | E

⎝ ⎠


2V sin ωt d ωt


= (½ + α E + 1

2V cos α

| π | π

⎝ ⎠

Vo rms

= |E 2

(½ + α +V 2 (

| | | π

½ − α + 1

π 2π

sin 2α

2 º½

| |


¬| |]

The load average and rms currents are

I = 1

cos α − E (½ − α º = 1

sin½θ − E θ º


o R | π

| π ||

R | π



1 V 2

|] |¬


θ º½

I = |

sinθ −

V E sin½θ + (V 2 + E 2 ) |


o rms

R |¬ 2π π


The total power delivered to the R-Eload is

P = P + P = I 2 R + E I


o R E o rms o

Example11.1: Half-waverectifierwithresistiveandbackemfload

A dc motor has series armature resistance of 10Ω and is fed via a half-wave rectifier, from the single- phase 230V 50Hz ac mains. Calculate

  1. The rectifier diode peak current
  2. The motor average starting current

If at full speed, the motor back emf is 100V dc, calculate

  1. The average and rms motor voltages and currents
  2. The motor electrical losses
  3. The power converted to rotational energy
  4. The supply power factor and motor efficiency
  5. Diode approximate loss if modelled by vD= 0.8 + 0.025× iD.


Worst case conditions are at standstill when the motor back emf is zero and the circuit and waveforms in figure 11.1a are applicable.

  1. The peak supply and peak load voltage is √2×V = √2×230 = 325.3V. The peak diode and load current is

iˆ = iˆ

= vˆ

= 325.3V = 32.5A

o D o R


  1. The motor average current, at starting, is given by equation (11.2)

Vo = Io R = 0.45 × 230V=103.5V

I = Vo

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 = sin1 E

= sin1

= 17.9°

The current conducts for a period θ = π – 2α = 180° – 2×17.9 = 144.2°, ceasing at ωt= π – α = 162.1°.

  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

⎝ ⎠



] 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° |

Vo rms

= |E 2

(½ + α +V 2 (


½ − α + 1

π 2π

sin 2α

2 º½

| |

| |

|¬| |]

( 17.9° ⎞ ( 17.9° 1

2 º½

= |1002 |½ + | + 2302 |½ +

sin 2 × 17.9°| |

= 179.2V


1 V 2



θ º½


I = |

sinθ −

V E sin½θ + (V 2 + E 2 ) |

o rms

R |¬ 2π π


1 2302

= |

sin144.2° − × 230V × 100V × sin½ × 144.2° + (2302 + 1002 )



10Ω |¬

= 10.2A

2π π

360° |]

  1. The motor loss is the loss in the 10Ω resistance in the dc motor equivalent circuit

P = I 2 R = 10.22 × 10Ω = 1041.5W

R o rms

  1. The back emf represents the source of electrical energy converted to mechanical energy

PE = E × I o = 100V × 5.85A = 585W

  1. The supply power factor is defined as the ratio of the supply power delivered, P, to apparent supply power, S

pf = P

= PR + PE

= 1041.5W + 585W = 0.69

S V × Io rms

The motor efficiency is

230V × 10.2A

η = PE


= 585W 1041.5W + 585W

× 100 = 40.0%

  1. 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 × 10.22 = 7.3W

  1. Single-phase half-wave rectifier circuit with an R-L load

A single-phase half-wave diode rectifying circuit with an R-L 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 forward-biased and conducting. This current continues until all the inductor energy, ½Li2, is released (i= 0) at the currentextinctionangle(or

cut-offangle), ωt=β.

During diode conduction the circuit is defined by the Kirchhoff voltage equation

L didt

+ Ri

= vR+ vL

= v=

Vsin ωt



where Vis the rms ac supply voltage. Solving equation (11.14) yields the load (and diode) current

it) =

2 V{sin (ωt-φ) +


sin φ

et/ tan φ }



0 ≤ ωt

≤ β ≥ π


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