Power Electronics – Useful Resources The following resources contain additional information on Power Electronics. Please use them to get more in-depth knowledge on this. Useful Video Courses 252 Lectures 35.5 hours 10 Lectures 41 mins 315 Lectures 8.5 hours 14 Lectures 29 mins 132 Lectures 13 hours 128 Lectures 10.5 hours Learning working make money
Power Electronics – Pulse Width Modulation PWM is a technique that is used to reduce the overall harmonic distortion (THD) in a load current. It uses a pulse wave in rectangular/square form that results in a variable average waveform value f(t), after its pulse width has been modulated. The time period for modulation is given by T. Therefore, waveform average value is given by $$bar{y}=frac{1}{T}int_{0}^{T}fleft ( t right )dt$$ Sinusoidal Pulse Width Modulation In a simple source voltage inverter, the switches can be turned ON and OFF as needed. During each cycle, the switch is turned on or off once. This results in a square waveform. However, if the switch is turned on for a number of times, a harmonic profile that is improved waveform is obtained. The sinusoidal PWM waveform is obtained by comparing the desired modulated waveform with a triangular waveform of high frequency. Regardless of whether the voltage of the signal is smaller or larger than that of the carrier waveform, the resulting output voltage of the DC bus is either negative or positive. The sinusoidal amplitude is given as Am and that of the carrier triangle is give as Ac. For sinusoidal PWM, the modulating index m is given by Am/Ac. Modified Sinusoidal Waveform PWM A modified sinusoidal PWM waveform is used for power control and optimization of the power factor. The main concept is to shift current delayed on the grid to the voltage grid by modifying the PWM converter. Consequently, there is an improvement in the efficiency of power as well as optimization in power factor. Multiple PWM The multiple PWM has numerous outputs that are not the same in value but the time period over which they are produced is constant for all outputs. Inverters with PWM are able to operate at high voltage output. The waveform below is a sinusoidal wave produced by a multiple PWM Voltage and Harmonic Control A periodic waveform that has frequency, which is a multiple integral of the fundamental power with frequency of 60Hz is known as a harmonic. Total harmonic distortion (THD) on the other hand refers to the total contribution of all the harmonic current frequencies. Harmonics are characterized by the pulse that represent the number of rectifiers used in a given circuit. It is calculated as follows − $$h=left ( ntimes P right )+1 quad or quad -1$$ Where n − is an integer 1, 2, 3, 4….n P − Number of rectifiers It is summarized in the table below − Harmonic Frequency 1st 60 Hz 2nd 120 Hz 3rd 180Hz 4th 240Hz 5th . . 49th 300Hz . . 2940Hz Harmonics have an impact on the voltage and current output and can be reduced using isolation transformers, line reactors, redesign of power systems and harmonic filters. Series Resonant Inverter A resonant inverter is an electrical inverter whose operation is based on oscillation of resonant current. Here, the switching device and the resonanting component are connected in series to each other. As a result of the natural features of the circuit, the current passing through the switching device drops to zero. This type of inverter yields a sinusoidal waveform at very high frequencies in the range of 20kHz-100kHz. It is therefore, most suitable for applications that demand a fixed output such as induction heating and flourescent lighting. It is usually small in size because its switching frequency is high. A resonant inverter has numerous configurations and thus it is categorized into two groups − Those with unidirectional switches Those with bidirectional switches Learning working make money
AC to DC Converters Solved Example A single-phase AC voltage converter has the following details − ON time = 6 min, OFF time = 4 min, frequency = 50Hz, and Voltage source Vo = 110V Calculate the following. Triggering angle α Solution − $T=2times left ( T_{ON}+T_{OFF} right )$ but $f=50Hz,$ $T=2times left ( 6+4 right )=20mins$ $360^{circ}=20min,$ $1min=18^{circ}$ Therefore, $T_{OFF}=4min$ Then, $$alpha =frac{4}{0.1}times 1.8=72^{circ}$$ Voltage Output Solution − $$V_{0}=left ( V_{S}times D right ),quad where quad D=frac{T_{ON}}{T_{ON}+T_{OFF}}=frac{6}{10}=0.6$$ $$T_{ON}=6min,quad T_{OFF}=4 min,quad V_{S}=110V$$ $$V_{0}left ( Voltage Output right )=V_{S}times D=110times 0.6=66Volts$$ Learning working make money
Power Electronics – Choppers A chopper uses high speed to connect and disconnect from a source load. A fixed DC voltage is applied intermittently to the source load by continuously triggering the power switch ON/OFF. The period of time for which the power switch stays ON or OFF is referred to as the chopper’s ON and OFF state times, respectively. Choppers are mostly applied in electric cars, conversion of wind and solar energy, and DC motor regulators. Symbol of a Chopper Classification of Choppers Depending on the voltage output, choppers are classified as − Step Up chopper (boost converter) Step Down Chopper(Buck converter) Step Up/Down Chopper (Buck-boost converter) Step Up Chopper The average voltage output (Vo) in a step up chopper is greater than the voltage input (Vs). The figure below shows a configuration of a step up chopper. Current and Voltage Waveforms V0 (average voltage output) is positive when chopper is switched ON and negative when the chopper is OFF as shown in the waveform below. Where TON – time interval when chopper is ON TOFF – time interval when chopper is OFF VL – Load voltage Vs – Source voltage T – Chopping time period = TON + TOFF Vo is given by − $$V_{0}=frac{1}{T}int_{0}^{T_{ON}}V_{S}dt$$ When the chopper (CH) is switched ON, the load is short circuited and, therefore, the voltage output for the period TON is zero. In addition, the inductor is charged during this time. This gives VS = VL $Lfrac{di}{dt}=V_{S},$ $frac{Delta i}{T_{ON}}=frac{V_{S}}{L}$ Hence,$Delta i=frac{V_{S}}{L}T_{ON}$ Δi = is the inductor peak to peak current. When the chopper (CH) is OFF, discharge occurs through the inductor L. Therefore, the summation of the Vs and VL is given as follows − $V_{0}=V_{S}+V_{L},quad V_{L}=V_{0}-V_{S}$ But $Lfrac{di}{dt}=V_{0}-V_{S}$ Thus,$Lfrac{Delta i}{T_{OFF}}=V_{0}-V_{S}$ This gives,$Delta i=frac{V_{0}-V_{S}}{L}T_{OFF}$ Equating Δi from ON state to Δi from OFF state gives − $frac{V_{S}}{L}T_{ON}=frac{V_{0}-V_{S}}{L}T_{OFF}$, $V_{S}left ( T_{ON}+T_{OFF} right )=V_{0}T_{OFF}$ $V_{0}=frac{TV_{S}}{T_{OFF}}=frac{V_{S}}{frac{left ( T+T_{ON} right )}{T}}$ This give the average voltage output as, $$V_{0}=frac{V_{S}}{1-D}$$ The above equation shows that Vo can be varied from VS to infinity. It proves that the output voltage will always be more than the voltage input and hence, it boosts up or increases the voltage level. Step Down Chopper This is also known as a buck converter. In this chopper, the average voltage output VO is less than the input voltage VS. When the chopper is ON, VO = VS and when the chopper is off, VO = 0 When the chopper is ON − $V_{S}=left ( V_{L}+V_{0} right ),quad V_{L}=V_{S}-V_{0},quad Lfrac{di}{dt}=V_{S}-V_{0},quad Lfrac{Delta i}{T_{ON}}=V_{s}+V_{0}$ Thus, peak-to-peak current load is given by, $Delta i=frac{V_{s}-V_{0}}{L}T_{ON}$ Circuit Diagram Where FD is free-wheel diode. When the chopper is OFF, polarity reversal and discharging occurs at the inductor. The current passes through the free-wheel diode and the inductor to the load. This gives, $$Lfrac{di}{dt}=V_{0}………………………………….left ( i right )$$ Rewritten as −$quad Lfrac{Delta i}{T_{OFF}}=V_{0}$ $$Delta i=V_{0}frac{T_{OFF}}{L}……………………………..left ( ii right )$$ Equating equations (i) and (ii) gives; $frac{V_{S}-V_{0}}{L}T_{ON}=frac{V_{0}}{L}T_{OFF}$ $frac{V_{S}-V_{0}}{V_{0}}=frac{T_{OFF}}{T_{ON}}$ $frac{V_{S}}{V_{0}}=frac{T_{ON}-T_{OFF}}{T_{ON}}$ The above equation gives; $$V_{0}=frac{T_{ON}}{T}V_{S}=DV_{S}$$ Equation (i) gives − $Delta i=frac{V_{S}-DV_{S}}{L}DT$, from $D=frac{T_{ON}}{T}$ $=frac{V_{S}-left ( 1-D right )D}{Lf}$ $f=frac{1}{T}=$chopping frequency Current and Voltage Waveforms The current and voltage waveforms are given below − For a step down chopper the voltage output is always less than the voltage input. This is shown by the waveform below. Step Up/ Step Down Chopper This is also known as a buck-boost converter. It makes it possible to increase or reduce the voltage input level. The diagram below shows a buck-boost chopper. When the chopper is switched ON, the inductor L becomes charged by the source voltage Vs. Therefore, Vs = VL. $$Lfrac{di}{dt}=V_{S}$$ $$Delta i=frac{V_{S}}{L}T_{ON}=frac{V_{S}}{L}Tfrac{T_{ON}}{T}=frac{DV_{S}}{Lf}$$ Because − $D=frac{T_{ON}}{T}$ and $f=frac{1}{T} ………………………………………. left ( iii right )$ When the chopper is switched OFF, the inductor’s polarity reverses and this causes it to discharge through the diode and the load. Hence, $$V_{0}=-V_{L}$$ $$Lfrac{di}{dt}=-V_{0}$$ $Lfrac{Delta i}{T_{OFF}}=-V_{0}$, thus $Delta i=-frac{V_{0}}{L}T_{OFF}…………………………..left ( iv right )$ Evaluating equation (iii) and (iv) gives − $frac{DV_{S}}{Lf}=-frac{V_{0}}{L}T_{OFF}$, $DV_{S}=-DV_{S}=-V_{0}T_{OFF}f$ $DV_{S}=-V_{0}frac{T-T_{ON}}{T}=-V_{0}left ( 1-frac{T_{ON}}{T} right )$, $V_{0}=-frac{DV_{S}}{1-D}$ Because $D=frac{T_{ON}}{T}=frac{T-T_{OFF}}{1-D}$ This gives, $V_{0}=frac{DV_{S}}{1-D}$ D can be varied from 0 to 1. When, D = 0; VO = 0 When D = 0.5, VO = VS When, D = 1, VO = ∞. Hence, in the interval 0 ≤ D ≤ 0.5, output voltage varies in the range 0 ≤ VO < VS and we get step down or Buck operation. Whereas, in the interval 0.5 ≤ D ≤ 1, output voltage varies in the range VS ≤ VO ≤ ∞ and we get step up or Boost operation. Learning working make money
Power Electronics – Integral Cycle Control Integral cycle controllers are converters with the ability to perform direct switching without losses. The process directly converts AC to AC without having to perform the intermediate processes of AC to DC then DC to AC. The basic integral control cycle is sinusoidal in nature. It operates by combining and eliminating higher frequency half cycles from AC input. The controllers are normally, turned ON of OFF during half cycles where the voltage input is at zero since only the full or half cycles are utilized. Therefore, integral cycle circuits achieve switching at zero voltage without requiring a resonant circuit. The diagram below shows a simple integral cycle controller. It contains a load and a power switch, which performs the direct conversion. This diagram shows the conversion of source frequency from a factor of three to one. Power Factor Control Power factor control, also known as correction of power factor, is the process of reducing the amount of reactive power. The power electronic device used in this case is called a power factor controller (PFC). From the power triangle (which comprises reactive, true and apparent power), the reactive power is at right angle (90°) to the true power and is used to energize the magnetic field. Although reactive power does not have a real value in electronic equipment, the bill for electricity comprises real and reactive power costs. This makes it necessary to have power factor controllers in electronic devices. Power factor (k) is defined as the ratio of the real power (in kW) to the reactive power (in kVAr). Its value ranges from 0 to 1. If a device has a power factor of 0.8 and above, it is said to be using power efficiently. Incorporating a PFC ensures the power factor ranges from 0.95 to 0.99. Power factor controllers are mainly in industrial equipment to minimize reactive power generated by fluorescent lighting and electric motors. To ensure power factor is improved without causing harmonic distortion, the conventional capacitors should not be used. Instead, filters (combination of capacitors and reactors) for harmonic suppression are used. The figure below shows a harmonic filter. The above type of harmonic filter is referred to as a single tuned filter. A quality factor Q of this filter is defined as quality factor of its reactance (XL) at Q (tuning frequency) where Q is given by (nXL/R). Learning working make money
Power Electronics – Control Methods In a converter, there are two basic methods of control used to vary the output voltage. These are − Time ratio control Current limit control Time Ratio Control In time ratio control, a constant k given by $frac{T_{ON}}{T}$ is varied. The constant k is called duty ratio. Time ratio control can be achieved in two ways − Constant Frequency In this control method, the frequency (f = 1/T0N) is kept constant while the ON time T is varied. This is referred to as pulse width modulation (PWM). Variable Frequency In variable frequency technique, the frequency (f = 1/T) is varied while the ON time T is kept constant. This is referred to as the frequency modulation control. Current Limit Control In a DC to DC converter, the value of the current varies between the maximum as well as the minimum level for continuous voltage. In this technique, the chopper (switch in a DC to DC converter) is switched ON and then OFF to ensure that current is kept constant between the upper and lower limits. When the current goes beyond the maximum point, the chopper goes OFF. While the switch is at its OFF state, current freewheels via the diode and drops in an exponential manner. The chopper is switched ON when the current reaches the minimum level. This method can be used either when the ON time T is constant or when the frequency (f=1/T). Learning working make money
Phase Controlled Converters Solved Example A separately excited DC motor has the following parameters: 220V, 100A and 1450 rpm. Its armature has a resistance of 0.1 Ω. In addition, it is supplied from a 3 phase fullycontrolled converter connected to a 3-phase AC source with a frequency of 50 Hz and inductive reactance of 0.5 Ω and 50Hz. At α = 0, the motor operation is at rated torque and speed. Assume the motor brakes re-generatively using the reverse direction at its rated speed. Calculate the maximum current under which commutation is not affected. Solution − We know that, $$V_{db}=3sqrt{frac{2}{pi }}times V_{L}-frac{3}{pi }times R_{b}times I_{db}$$ Substituting the values, we get, $220=3sqrt{frac{2}{pi }}times V_{L}-frac{3}{pi }times 0.5times 100$ Therefore, $V_{L}=198V$ Voltage at rated speed = $220-left ( 100times 0.1 right )=210V$ At the rated speed, the regenerative braking in the reverse direction, $=3sqrt{frac{2}{pi }}times 198cos alpha -left ( frac{3}{pi }times 0.5+0.1right )times I_{db}=-210V$ But $cos alpha -cos left ( mu +alpha right )=frac{sqrt{2}}{198}times 0.5I_{db}$ For commutation to fail, the following limiting condition should be satisfied. $mu +alpha approx 180^{circ}$ Therefore, $quad cos alpha =frac{I_{db}}{198sqrt{2}}-1$ Also, $frac{3}{pi }I_{db}-frac{3sqrt{2}}{pi }times 198-left ( frac{3}{pi }times 0.5+0.1 right )I_{db}=-210$ This gives, $quad 0.3771I_{db}=57.4$ Therefore, $quad I_{db}=152.2A$ Learning working make money
Power Electronics – Resonant Switching Resonant switch converters refers to converters that have inductor and capacitor (L-C) networks and whose current and voltage waveforms vary in a sinusoidal manner during each period of switching. There are various resonant switch converters − Resonant DC to DC converters DC to AC inverters Resonant AC inverters to DC converters In this tutorial, we will focus on Resonant DC to DC converters Resonant DC to DC Converters The concept of switch mode power supply (SMPS) is explained below using a DC to DC converter. The load is given a constant voltage supply (VOUT) that is obtained from a primary source of voltage supply VIN. The value of VOUT is regulated by varying resistor in series (RS) or the current source connected in shunt (IS). By controlling VOUT through varying IS and ensuring RS is kept constant, a considerable amount of power is lost in the converter. Switched Mode Power Supply (SMPS) An SMPS (switched mode power supply) refers to an electronic device that uses a switching regulator for the purpose of converting electrical power in an efficient manner. SMPS takes power from the main power lines and transfers it to a load. For example, a computer while ensuring the voltage and current characteristics are converted. The difference between an SMPS and a linear supply of power is that the former keeps switching ON and OFF during low dissipation and uses less time during high dissipation regions. This ensures less energy is wasted. Actually, an SMPS does not dissipate any power. The size of an SMPS is smaller and very light, compared to a normal linear supply power device of the same size and shape. The figure below shows the circuit diagram for an SMPS. When the switching frequency is varied, the stored energy can be varied for each cycle and hence the voltage output is varied. The waveforms below are for a half bridge converter also known as a push-pull. It is used in applications utilizing high power. The input voltage is halved as indicated in the waveform. Learning working make money
Power Electronics – BJT A Bipolar Junction Transistor (BJT) is a transistor whose operation depends on the contact made by two semicondutors. It can act as a switch, amplifier or oscillator. It is known as a bipolar transistor since its operation requires two types of charge carriers (holes and electrons). Holes constitute the dominant charge carriers in P-type semiconductors while electrons are the main charge bearers in N-type semiconductors. Symbols of a BJT Structure of a BJT A BJT has two P-N junctions connected back to back and sharing a common region B (base). This ensures contacts are made in all the regions that are base, collector and emitter. The structure of a PNP bipolar transistor is shown below. The BJT shown above consists of two diodes connected back to back, resulting in the depletion of the regions called quasi-neutral. The width of quasi-neutral of the emitter, base and collector are indicated above as WE’, WB’ and WC’. They are obtained as follows − $$W_{E}^{”}=W_{E}-X_{n,BE}$$ $$W_{B}^{”}=W_{B}-X_{p,BE}-X_{p,BC}$$ $$W_{C}^{”}=W_{C}-X_{n,BC}$$ The conventional signs of the currents for the emitter, base and collector are denoted by IE, IB and IC respectively. Therefore, the collector and base current are positive when a positive current meets the collector or base contact. In addition, the emitter current is positive when current leaves the emitter contact. Thus, $$I_{E}=I_{B}+I_{C}$$ When a positive voltage is applied to the base contact relative to the collector and emitter, the base-collector voltage as well as base-emitter voltage becomes positive. For simplicity, VCE is assumed to be zero. Diffusion of electrons occur from the emitter to the base while diffusion of holes originates from the base to the emitter. Once the electrons reach the base-collector depleted region, they are swept through the region by an electric field. These electrons form the collector current. When a BJT is biased in the forward active mode, the total emitter current is obtained by adding the electron diffusion current (IE,n), the hole diffusion current (IE, p) and the baseemitter current. $$I_{E}=I_{E,n}+I_{E,p}+I_{r,d}$$ The total collector current is given by the electron diffusion current (IE,n), less the base recombination current(Ir,B). $$I_{C}=I_{E,n}-I_{r,B}$$ The sum of the base current IB is obtained by adding the hole diffusion current (IE, p), base recombination current (Ir,B) and the base-emitter recombination current of the depletion layer (Ir,d). $$I_{B}=I_{E,p}+I_{r,B}+I_{r,d}$$ Transport Factor This is given by the ratio of the collector current and the emitter current. $$alpha =frac{I_{C}}{I_{E}}$$ Applying the Kirchhoff’s current law, it is found that the base current is given by the difference between the emitter current and the collector current. Current Gain This is given by the ratio of the collector current to the base current. $$beta =frac{I_{C}}{I_{B}}=frac{alpha }{1-alpha }$$ The above explains how a BJT can produce current amplification. The transport factor (α) approaches one if the collector current is almost equivalent to the emitter current. The current gain (β) thus becomes greater than one. For further analysis, the transport factor (α) is rewritten as a product of the emitter efficiency (γE) the base transport factor (αT) and the recombination factor of the depletion layer (δr). It is rewritten as follows − $$alpha =gamma _{E}times alpha _{T}times delta _{r}$$ The following is a summary of the discussed emitter efficiency, base transport factor and depletion layer recombination factor. Emitter Efficiency $$gamma _{E}=frac{I_{E,n}}{I_{E,p}+I_{E,P}}$$ Base Transport Factor $$alpha _{T}=frac{I_{E,n}-I_{r,b}}{I_{E,n}}$$ Depletion Layer Recombination Factor $$delta _{r}=frac{I_{E}-I_{r,d}}{I_{E,n}}$$ Learning working make money
Effect of Source Inductance The analysis of most converters is usually simplified under ideal conditions (no source impedance). However, this assumption is not justified since source impedance is normally inductive with a negligible resistive element. Source inductance has a significant impact on the converter performance because its presence alters the output voltage of the converter. As a result, the output voltage reduces as the load current reduces. In addition, the input current and output voltage waveforms change significantly. Source inductance effect on a converter is analyzed in the following two ways. Effect on Single Phase Assuming that the converter operates in conduction mode and the ripple from the load current is negligible, the open circuit voltage becomes equal to average DC output at a firing angle of α.The diagram below shows a fully controlled converter with source in single phase. The thyristors T3 and T4 are assumed to be in conduction mode when t = 0. On the other hand, T1 and T2 fire when ωt = α Where − Vi = input voltage Ii = input current Vo = output voltage Io = output voltage When there is no source inductance, commutation will occur at T3 and T4. Immediately thyristors T1 and T2 are switched ON. This will lead the input polarity to change instantaneously. In the presence of source inductance, change of polarity and commutation does not occur instantaneously. Thus, T3 and T4 do not commutate as soon as T1 and T2 are switched ON. At some interval, all the four thyristors will be conducting. This conducting interval is called the overlap interval (μ). The overlap during commutation reduces the DC output voltage and the angle of extinction γ resulting in failed commutation when αis close to 180°. This is shown by the waveform below. Effect on Three Phase Just like the single-phase converter, there are no instantaneous commutations due to the presence of the source inductances. Taking the source inductances into consideration, the effects (qualitative) on the converter performance is the same as in a single phase converter. This is shown in the diagram below. Learning working make money