Step-Down (Buck) Converter

The buck converter produces a lower average output voltage than the DC input voltage V_S.

Its main application is in regulated DC power supplies and motor speed control. The basic circuit shown in Figure 12.3 constitutes a step-down converter for a purely resistive load, instantaneous input voltage V_S, and a purely resistive load. The average output voltage can be calculated in terms of the switch-duty ratio:

(12.1)

The switch-duty ratio is defined as the ratio of the on duration to the switching time period, be expressed as  

(12.2)

where T_s is the constant switching time period, equal to t_on + t_off.

Substituting for D in Equation 12.1 gives

(12.3)

where

(12.4)

By varying the duty ratio of the switch, the output voltage can be controlled. It is important to note that the average output voltage V_O varies linearly with the control voltage, as is the case in linear amplifiers. In actual practice, the circuit in Figure 12.3 has two drawbacks:

• In practice, the load would be inductive. Even with a resistive load, there would always be certain associated stray inductance. This means that the switch has to absorb or dissipate the inductive energy and therefore it may be destroyed.
• The output voltage fluctuates between zero and V_S, which is not acceptable in most applications.

The problem of stored inductive energy is overcome by using a diode, as shown in Figure 12.3, and the output voltage fluctuations are very much diminished by using a low-pass filter, consisting of an inductor and capacitor.

During the interval when the switch is on, the diode becomes reverse biased and the input provides energy to the load as well as to the inductor. During the interval when the switch is off, the inductor current flows through the diode, transferring some of its stored energy to the load.

In steady state, the integral of the inductor voltage over one time period must be zero. In the diode-inductor-capacitor loop of Figure 12.3

(12.5)

This implies that

(12.6)

In the continuous-conduction mode, the voltage output varies linearly with the duty ratio of the switch for the given input voltage. It does not depend on any circuit parameter.

Neglecting power losses associated with all circuit elements, the input power P_S equals the output power P_O.

(12.7)

(12.8)

(12.9)

Therefore, in the continuous-conduction mode, the step-down converter is equivalent to a DC transformer, where the turns ratio of this equivalent transformer can be continuously controlled electronically in Equation 12.9.

Step-Up (Boost) Converter

Figure 12.4 [2] shows a step-up converter configuration. Its main application is in regulated DC power supplies and the regenerative braking of DC motors. As the name implies, the output voltage is always greater than the input voltage. When the switch is on, the diode is reverse biased, thus isolating the output stage. The input supplies energy to the inductor. When the switch is off, the output stage receives energy from the inductor as well as from the input. In the steady-state analysis presented here, the output filter capacitor is assumed to be very large to ensure a constant output voltage.

In steady state, the time integral of the inductor voltage over one time period must be zero, so that

(12.10)

Dividing both sides by T_S and rearranging terms yields

(12.11)

Assuming a lossless circuit,

(12.12)

(12.13)

and

(12.14)

Buck–Boost Converter

A buck–boost converter can be obtained by the cascade connection of the two basic converters: the step-down converter and step-up converter. The main application of a step-down/step-up or buck–boost converter is in regulated DC power supplies, where a negative polarity output may be desired with respect to the common terminal of the input voltage and the output voltage can either be higher or lower than the input voltage.

In steady state, the output-to-input voltage conversion ratio is the product of the conversion ratios of the two converters in cascade, assuming that switches in both converters have the same duty ratio.

(12.15)

Assuming P_S = P_O, then

(12.16)

Finally, the output voltage is higher or lower than the input voltage, based on the duty ratio D. The connection of the step-down and step-up converter can be combined into the single buck–boost converter shown in Figure 12.5 [2]. When the switch is open, the energy stored in the inductor is transferred to the output. No energy is supplied by the input during this interval.

Example 1

In a buck–boost converter operating at 20 kHz, L = 0.05 mH. The output capacitor C is sufficiently large and V_S = 15 V. The output is to be regulated at 10 V and the converter is supplying a load of 10 W. Calculate the duty ratio.

Solution:

Assuming continuous conduction, from Equation 11.15

Therefore, the initial estimate duty ratio, D = 0.4.

Now we know that

Using

Since the output current is less than I_oB, the current conduction is discontinuous, therefore

Ćuk DC–DC Converter

This Ćuk DC–DC converter is obtained by using the duality principle on the circuit of a buck–boost converter. The Ćuk converter (see Figure 12.6 [2]) provides a negative-polarity, regulated-output voltage with respect to the common terminal of the input voltage. Here the capacitor acts as the primary means of storage and transfer of energy from input to output.

  In steady state, the average inductor voltages are zero. Therefore, by inspection

When the switch is off, the inductor currents flow through the diode. Capacitor C is charged through the diode by energy from both the input and L₂. Current i_S decreases, because V_C1 is larger than V_S. Energy stored in L₁ feeds the output. Therefore, decreases.

When the switch is on, V_C reverse biases the diode. The inductor currents i_L2 and flow through the switch. Since V_C  >  V_O, C₁ discharges through the switch, transferring energy to the output and L₁. Therefore i_L2 increases. The input feeds energy to L₂ causing i_L2 to increase.

The inductor currents are assumed to be continuous. If we assume capacitor voltage V_C1 to be constant, then equating the integral of the voltages across L₂ and L₁ over one time period to zero yields

(12.17)

and

(12.18)

From Equations 12.26 and 12.27

(12.19)

Assuming P_S = P_O then

(12.20)

where

Therefore, the main benefit of the Ćuk converter is that you can control the continuous current at both in the input and output of the converter as it is based on the capacitor energy transfer. It has a low-switching loss making it more highly efficient. The downside of this converter is that it includes a high number of reactive components (L and C) and heavy current stresses on the components. And since the capacitor C provides a transfer of energy, ripples in the capacitor (C_O) current are high.

Three-Phase Inverters

Three-phase inverters are normally used for high-power applications. Three single-phase half- or full-bridge inverters can be connected in parallel to form a configuration of a three-phase inverter. The gating signals of single-phase inverters should be advanced or delayed by 120° with respect to each other to obtain three-phase, balanced fundamental voltages.

The transformer primary windings must be isolated from each other, while the secondary windings may be Y connected or delta connected. The Y connection is usually to eliminate triple harmonics (n = 3, 6, 9, …) appearing on the output voltages and the circuit arrangement. This arrangement requires three single-phase transformers, twelve transistors, and twelve diodes. If the output voltages of single-phase inverters are not perfectly balanced in magnitudes and phases, the three-phase output voltages will be unbalanced. A three-phase output can be obtained from a configuration of six transistors and six diodes.

In this setup, two control signals can be applied to the transistors: 180° conduction and 120° conduction.

180° Conduction

Each transistor conducts for 180°. Three transistors remain on at any instant of time. When transistor Q₁ is switched on, terminal a is connected to the negative terminal of the DC input voltage. When transistor Q₄ is switched on, terminal a is brought to the negative terminal of the DC source. There are six modes of operation in a cycle and the duration of each mode is 60°.

The load may be Y connected or delta connected. Figure 12.9 shows a Y-connected load. For a Y connection, the line-to-neutral voltages must be determined to find the line (or phase currents). For the delta-connected load, the phase currents can be obtained directly from the line-to-line voltages. Once phase currents are known, the line currents can be determined.

In the Y-connection configuration, there are three modes of operation in a half cycle.

1. Mode 1: 0 ≤ ωt < π/3
   
   (12.26)
2. Mode 2: π/3 ≤ ωt < 2π/3
   
   (12.27)
3. Mode 3: 2π/3 ≤ ωt < π
   
   (12.28)

The line-to-line rms voltage can be found from

(12.29)

(12.30)

The line-to-neutral voltages can be found from the line voltage

(12.31)

Voltage-Multiplying Rectifier

The simple half-wave rectifier can be built in two electrical configurations, with the diode pointing in opposite directions. One version connects the negative terminal of the output directly to the AC supply and the other connects the positive terminal of the output directly to the AC supply. By combining both of these with separate output smoothing, it is possible to get an output voltage of nearly double the peak AC-input voltage. This also provides a tap in the middle, which allows use of such a circuit as a split-rail power supply.

A variant of this is to use two capacitors in series for the output smoothing on a bridge rectifier then place a switch between the midpoint of those capacitors and one of the AC-input terminals. With the switch open, this circuit acts like a normal bridge rectifier. With the switch closed, it acts like a voltage-doubling rectifier. In other words, this makes it easy to derive a voltage of roughly 320 V ( ± 15 percent) DC from any 120 V or 230 V mains supply in the world; this can then be fed into a relatively simple switched-mode power supply. However, for a given desired ripple, the value of both capacitors must be twice the value of the single one required for a normal bridge rectifier; when the switch is closed each one must filter the output of a half-wave rectifier, and when the switch is open the two capacitors are connected in series with an equivalent value of half of one of them. A switchable full-bridge circuit is shown in Figure 12.15 [2].

Example 1

A three-phase inverter has a Y-connected resistive load of R = 5 Ω and L = 23 mH. The inverter frequency is f = 60 Hz and the DC input voltage is V_s = 220 V. Determine:

1. the rms line voltage V_L
2. rms line voltage V_L1 at the fundamental frequency
3. rms phase voltage V_p
4. total harmonic distortion
5. distortion factor

Solution:

1. 
2. The rms line voltage V_L1 at the fundamental frequency
   
3. rms phase voltage is given by
   
4. We know the fundamental value of the rms line voltage is given by
   
5.  
   

Example 2

A single-phase, half-bridge inverter has a resistive load of R = 2.4 Ω and the DC input voltage is V_S = 48 V. Determine:

1. rms output voltage at the fundamental frequency V₁
2. output power P_O
3. average and peak current of each transistor
4. total harmonic distortion (THD)
5. distortion factor (DF)

Solution:

1. The rms output voltage at the fundamental frequency is given as
   
2.  
   
   
3. The peak transistor current
   

   Since each transistor conducts for 50 percent duty cycle, the average current of each transistor is I_D = 0.5 × 10 = 5 A.

4. The rms harmonic voltage is given as
   

   Therefore, the THD is

   
5. We can find.
   

   Therefore, DF is

   

Example 3

Consider the single-phase, full-bridge inverter shown in Figure Q12.5.3 is operated in the quasi-square-wave mode at the frequency f = 100 Hz with a phase-shift of β between the half-bridge outputs v_ao and v_bo.

1. With a purely resistive load R = 10 Ω, find β so that the average power supplied to the load is P_o,av = 2 kW.
2. With a purely inductive load L = 20 mH and β = 2π/3, find the peak-to-peak value (I_pp) of the load current i_o and find the amplitude of the fundamental component (I_o1) of i_o.

Solution:

1. With a purely resistive load,

   Instantaneous power

   

   Average Power supplied to the load

   
2. With a purely inductive load, the peak-to-peak value of the load current is given by
   

   The amplitude of the fundamental component is given by

   

Example 4

The three-phase, half-bridge inverter shown in Figure Q12.5.4 feeds a balanced, Δ-connected resistive load, and is operated in the square-wave mode (pole voltages are square waves) at frequency f_s.

1. Sketch line current i_a (show all important current and time values) and find its rms value.
2. Find the total average power supplied to the load (take R = 10 Ω).

Solution:

1.  
   
   
   

   

   
2. Instantaneous power in phase A:

   Average power in phase A:

   

   
   

Example 5

A single-phase, fully controlled bridge converter supplies power to a highly inductive load with resistance R = 10 Ω from a 240 V, rms 50 Hz, AC source, as shown on Figure Q12.5.5.

1. Find the firing angle α so that the average power supplied to the load is Pav = 1,200 W.
2. For α = π/6 find the input power factor of the rectifier.

Solution:

1. The average power supplied to the load
   
2. For α = π/6 the source current is a square wave as shown:

   

The input power factor is given by

Example 6

The three-phase, half-bridge inverter, Q12.5.1 feeds a balanced, Y-connected resistive-inductive load with R = 10 Ω and L = 10 mH, and is operated in the square-wave mode (pole voltages are square waves) at frequency f s = 200 Hz.

1. Sketch phase a voltage v_an (show all important voltage and time values) and find its rms value.
2. Find the amplitude of the fundamental component of the line current i_a.

Example 7

The three-phase, half-bridge inverter shown in Figure Q12.5.2 feeds a balanced, Y-connected, purely inductive load having inductance L per phase, and is operated in the square-wave mode (pole voltages are square waves) at the frequency f_s.

1. Sketch phase-to-neutral voltage v_an (show all important voltage and time values) and find its rms value V_an,rms
2. Find the peak-to-peak value (I_pp) of the line current i_a in terms of L and f_s.

Application of DC–DC Converters

The most typical example of application of DC–DC converters is DC power supplies. However, in most cases, in commercial fields and industries, the power source is AC. In this case, DC–DC converters cannot operate stand-alone. Operating together with other circuits such as rectifiers (Figure 12.16), DC–DC converters are useful in many applications.

Examples of applications of DC–DC converters include:

• DC power supplies for DC or AC input, usually called switched-mode power supplies (SMPS)
• battery chargers
• power-factor correction converters (PFCCs)
• motor drives, such as DC drives, brushless DC drives, and switched-reluctance motor (SRM) drives
• electric brakes
• electric vehicles
• most mobile devices (such as cell phones) to maintain voltage at a fixed value whatever the voltage level of the battery
• electronic isolation.
• microgrids, where a power optimizer, is a type of DC–DC converter, maximizes the energy harvest from solar photovoltaic or wind-turbine systems

Application of AC–DC Converters

AC–DC converters (rectifiers) are used every time an electronic device is connected to the mains (computer, television, etc.). These may simply change AC to DC or can change the voltage level as part of their operation.

Rectifiers have many uses, but are often found serving as components of DC power supplies and high-voltage DC power transmission systems. Rectification may serve in roles other than to generate DC for use as a source of power. As noted, detectors of radio signals serve as rectifiers. In gas-heating systems, flame rectification is used to detect presence of a flame.

Rectifier circuits may be single phase or multi-phase (three being the most common number of phases). Most low-power rectifiers for domestic equipment are single phase, but three-phase rectification is important for industrial applications and for transmission of energy as DC (high-voltage DC or HVDC).

The primary application of rectifiers is to derive DC power from an AC supply (AC-to-DC converter). Virtually all electronic devices require DC, so rectifiers are used inside the power supplies of virtually all electronic equipment.

Rectifiers are also used for detection of amplitude-modulated radio signals. The signal may be amplified before detection. If not, a very low-voltage drop diode or a diode biased with a fixed voltage must be used. When using a rectifier for demodulation the capacitor and load resistance must be carefully matched: too low a capacitance makes the high-frequency carrier pass to the output, and too high makes the capacitor just charge and stay charged.

Rectifiers supply polarized voltage for welding. In such circuits control of the output current is required; this is sometimes achieved by replacing some of the diodes in a bridge rectifier with thyristors, effectively diodes whose voltage output can be regulated by switching on and off with phase-fired controllers.

Thyristors are used in various classes of railway rolling stock systems so that fine control of the traction motors can be achieved. Gate turn-off thyristors are used to produce AC from a DC supply, for example on Eurostar trains to power the three-phase traction motors.

Application of DC–AC Converters

DC–AC converters (inverters) are used primarily in uninterruptible power supply (UPS), renewable-energy systems, or emergency lighting systems. Mains power charges the DC battery. If the mains fail, an inverter produces AC electricity at mains voltage from the DC battery. Solar inverters, both smaller string and larger central inverters, as well as solar microinverters, are used in photovoltaics.

1. DC power source usage: An inverter converts DC electricity from sources such as batteries or fuel cells to AC electricity. The electricity can be at any required voltage; in particular, it can operate AC equipment designed for mains operation, or rectified to produce DC at any desired voltage.
2. UPS: A UPS uses batteries and an inverter to supply AC power when mains power is not available. When mains power is restored, a rectifier supplies DC power to recharge the batteries. In facilities that require energy at all times, such as hospitals and airports, UPS systems are utilized. In a standby system, an inverter is brought online when the normally supplying grid is interrupted. Power is instantaneously drawn from onsite batteries and converted into usable AC voltage by a voltage-source inverter (VSI) until grid power is restored, or until backup generators are brought online. In an online UPS system, a rectifier DC-link inverter is used to protect the load from transients and harmonic content. A battery in parallel with the DC link is kept fully charged by the output in case the grid power is interrupted, while the output of the inverter is fed through a low-pass filter to the load. High power quality and independence from disturbances is achieved [8].
3. Harmonic compensation: In power systems, it is often desirable to eliminate harmonic content found in line currents. VSIs can be used as active power filters to provide this compensation. Based on measured line currents and voltages, a control system determines reference current signals for each phase. This is fed back through an outer loop and subtracted from actual current signals to create current signals for an inner loop to the inverter. These signals then cause the inverter to generate output currents that compensate for the harmonic content. This configuration requires no real power consumption, as it is fully fed by the line; the DC link is simply a capacitor that is kept at a constant voltage by the control system [8]. In this configuration, output currents are in phase with line voltages to produce a unity power factor. Conversely, var compensation is possible in a similar configuration where output currents lead line voltages to improve the overall power factor [9].
4. Electric motor speed control: Inverter circuits designed to produce a variable output voltage range are often used within motor speed controllers. The DC power for the inverter section can be derived from a normal AC wall outlet or some other source. Control and feedback circuitry is used to adjust the final output of the inverter section, which will ultimately determine the speed of the motor operating under its mechanical load. Motor speed-control needs are numerous and include things like industrial motor-driven equipment, electric vehicles, rail transport systems, and power tools.
5. Power grid: Grid-tied inverters are designed to feed into the electric power distribution system. They transfer synchronously with the line and have as little harmonic content as possible. They also need a means of detecting the presence of utility power for safety reasons, so as not to continue to dangerously feed power to the grid during a power outage.
6. Solar: A solar inverter is a balance of system (BOS) component of a photovoltaic system and can be used for both grid-connected and off-grid systems. Solar inverters have special functions adapted for use with photovoltaic arrays, including maximum power-point tracking and anti-islanding protection. Solar microinverters differ from conventional converters, as an individual microconverter is attached to each solar panel. This can improve the overall efficiency of the system. The output from several microinverters is then combined and often fed to the electrical grid.
7. Induction heating: Inverters convert low-frequency main AC power to higher frequency for use in induction heating. To do this, AC power is first rectified to provide DC power. The inverter then changes the DC power to high-frequency AC power. Due to the reduction in the number of DC sources employed, the structure becomes more reliable and the output voltage has higher resolution due to an increase in the number of steps so that the reference sinusoidal voltage can be better achieved. This configuration has recently become very popular in AC power supply and adjustable speed drive applications. This new inverter can avoid extra clamping diodes or voltage-balancing capacitors.
8. HVDC power transmission: With HVDC power transmission, AC power is rectified and high-voltage DC power is transmitted to another location. At the receiving location, an inverter in a static inverter plant converts the power back to AC. The inverter must be synchronized with grid frequency and phase and minimize harmonic generation. The HVDC transmission method can be useful for things like solar power since solar power is natively DC.
9. Electroshock weapons: Electroshock weapons and tasers have a DC–AC inverter to generate several tens of thousands of V AC out of a small 9 V DC battery. First the 9 V DC is converted to 400–2,000 V AC with a compact, high-frequency transformer, which is then rectified and temporarily stored in a high-voltage capacitor until a preset threshold voltage is reached. When the threshold (set by way of an air gap or TRIAC) is reached, the capacitor dumps its entire load into a pulse transformer, which then steps it up to its final output voltage of 20–60 kV. A variant of the principle is also used in electronic flashes and insect zappers, though they rely on a capacitor-based voltage multiplier to achieve their high voltage.
10. Renewable-energy applications: For photovoltaic purposes, the inverter, which is usually a PWM VSI, gets fed by the DC electrical-energy output of a photovoltaic module or array. The inverter then converts this into an AC voltage to be interfaced with either a load or the utility grid. Inverters may also be employed in other renewable systems, such as wind turbines. In these applications, the turbine speed usually varies, causing changes in voltage frequency and sometimes in the magnitude. In this case, the generated voltage can be rectified and then inverted to stabilize frequency and magnitude [8]. Classes of converters are summarized in comparison in Table 12.1.

Conversion from/to	Description	Device(s)	Schematic Symbol	Applications
DC to DC	Convert from one DC power level to another DC power level	buck converter boost converter buck–boost converter, Ćuk converter flyback converter forward converter		DC power supplies battery chargers motor drives electric brakes microgrids as a power optimizer
DC to AC	Electronic device or circuitry that changes DC to AC	inverter		UPS harmonic compensation, electric motor speed control induction heating electroshock weapons renewable-energy applications
AC to DC	Converts AC to DC	rectifier		detection of amplitude-modulated radio signals supply of polarized voltage for welding
AC to AC	Converts AC waveforms of specific magnitude and specific frequency into AC waveforms with different magnitude and different frequency	cycloconverter matrix converter		exchange power between power grids of different utility frequencies

12.7 CHAPTER SUMMARY

This chapter treated the classical and new aspects of inverters and converters. It covered the definition, descriptions, configurations, and performance of the various types of converters: as DC–DC, AC–DC, DC–AC, and AC–AC. Converters function as an interface in the integration of microgrids and traditional grids. A converter interface for DG functions in such a way that it accepts power from the distributed energy source and converts it to power at the required voltage and frequency. The modeling, characteristics, configuration, and application were given. Case studies and exercises using inverters and converters in typical electronic systems were given.