DC-DC Converters Notes

DC-DC Converters

• In many industrial applications, converting a fixed DC source into a variable-voltage DC source is required.
• A DC-DC converter directly converts DC to DC.
• A DC converter is DC equivalent to an AC transformer with a continuously variable turns ratio.
• Like a transformer, it can step down or step up a DC voltage source.
• DC converters are widely used for:
  • Traction
  • Meter control in electric automobiles
  • Trolley cars
  • Marine hoists
  • Forklift trucks
  • Mine haulers
• DC converters can be used in regenerative braking of DC motors to return energy back into the supply.
• DC converters are used in DC voltage regulators.

Step-Down Operation Principles

• The principle of operation can be explained by Figure (1-a).

• When switch SW (the chopper) is closed for time $t<em>1$, the input voltage $V</em>s$ appears across the load.

• If the switch remains off for time $t_2$, the voltage across the load is zero.

• Waveforms for output voltage and load current are shown in Figure (1-b).

• The Converter switch can be implemented using:

  • Power Bipolar Junction Transistor (BJT)
  • Power Metal Oxide Semiconductor Field-Effect Transistor (MOSFET)
  • Gate-Turn-Off Thyristor (GTO)
  • Insulated-Gate Bipolar Transistor (IGBT)

• From Figure (1), the average output voltage is given by:
  $V<em>a = \frac{1}{T} \int</em>0^{t<em>1} V</em>s dt = \frac{t<em>1}{T} V</em>s = KV_s$

• The average load current is:
  $I<em>a = \frac{V</em>a}{R} = \frac{KV_s}{R}$
  Where:

  • $T$ is the chopping period.
  • $K = \frac{t_1}{T}$ is the duty cycle of the chopper.
  • $f = \frac{1}{T}$ is the chopping frequency.

• The RMS value of output voltage is:
  $V<em>o = \sqrt{\frac{1}{T} \int</em>0^{t<em>1} V</em>s^2 dt} = \sqrt{K} V_s$

• Assuming a lossless converter, the input power is the same as the output power:

  $P<em>i = \frac{1}{T} \int</em>0^{t<em>1} V</em>o i dt = \frac{K V_s^2}{R}$

• The effective Resistance seen by the source is $R_i$:

  $R<em>i = \frac{V</em>s}{I<em>a} = \frac{V</em>s}{\frac{K V_s}{R}} = \frac{R}{K}$

• The duty cycle $K$ can be varied from 0 to 1 by varying $t<em>1$ or $f$. Therefore, the output voltage $V</em>o$ can be varied from 0 to $V_s$ by controlling $K$, and the power flow can be controlled.

Control Methods

1. Constant-Frequency Operation: The converter or switching frequency $f$ (or chopping period $T$) is kept constant, and the on-time $t_1$ is varied. This is Pulse-Width Modulation (PWM) control.
2. Variable-Frequency Operation: The chopping or switching frequency $F$ is varied. Either on-time $t<em>1$ or off-time $t</em>2$ is kept constant. This is frequency modulation. The frequency has to be varied over a wide range to obtain the full output voltage range, which can make filter design difficult.

Example (1)

A DC converter with a resistive load of $R = 10 \Omega$ and an input voltage of $V<em>s = 220V$. The converter switch has a voltage drop of $V</em>{ch} = 2V$ when on, and the chopping frequency is $f = 1 kHz$. If the duty cycle is 50% ($K = 0.5$), determine:

• (a) the average output voltage $V_a$
• (b) the RMS output voltage $V_o$
• (c) the converter efficiency$\eta$
• (d) the effective input resistance $R_i$
• (e) The RMS value of the fundamental component of output harmonic voltage

Solution

Given: $V<em>s = 220V$, $K = 0.5$, $R = 10 \Omega$, and $V</em>{ch} = 2V$

• (a) Average Output Voltage:

  $V<em>a = K(V</em>s - V_{ch}) = 0.5 (220 - 2) = 109 V$

• (b) RMS Output Voltage:

  $V<em>o = \sqrt{K} V</em>s = \sqrt{0.5} \times 220 = 155.56 V$

• (c) Output Power:
  $P<em>o = \frac{K (V</em>s - V_{ch})^2}{R} = \frac{0.5 \times (220 - 2)^2}{10} = 2376.2 W$

• Input Power:

  $P<em>i = \frac{K V</em>s (V<em>s - V</em>{ch})}{R}$
  $P_i = 0.5 \times 220 \times \frac{(220 - 2)}{10} = 2398 W$

• Converter Efficiency:

  $\eta = \frac{P<em>o}{P</em>i} = \frac{2376.2}{2398} = 0.9909 = 99.09 \%$

• (d) Effective Input Resistance:

  $R_i = \frac{R}{K} = \frac{10}{0.5} = 20 \Omega$

• (e) Fourier Series of Output Voltage:

  $V<em>o (t) = KV</em>s + V<em>s \sum</em>{n=1}^{\infty} \frac{\sin(n \pi K)}{n \pi} \cos(2n \pi f t) + V<em>s \sum</em>{n=1}^{\infty} \frac{(1 - \cos(n \pi K))}{n \pi} \sin(2n \pi f t)$

  The fundamental component (for n=1) is:

  $V<em>1(t) = V</em>s [\frac{\sin(\pi K)}{\pi} \cos(2 \pi f t) + \frac{(1 - \cos(\pi K))}{\pi} \sin(2 \pi f t)]$
  Given the values, its reet-mean square (rms) value is $V_1 = 99.04 V$

Step-Up Operation Principle

• A converter can be used to step up a DC voltage.

• Figure (2-a) shows an arrangement for step-up operation.

• When switch SW is closed for time $t_1$, the inductor current rises, and energy is stored in inductor $L$.

• If the switch is opened for time $t<em>2$, the energy stored in the inductor is transferred to the load through diode $D</em>i$, and the inductor current falls.

• Assuming a continuous current flow, the waveform of the inductor current is shown in Figure (2-b).

• When the converter is turned on, the voltage across the inductor is:

  $V_L = L \frac{di}{dt}$

  This gives the peak-to-peak ripple current in the inductor as
  $\Delta I = \frac{V<em>s t</em>1}{L}$

• The average output voltage is:

  $V<em>o - V</em>s = L \frac{di}{dt} = V<em>s (\frac{t</em>1}{t_2})$

  Which means:

  $V<em>o = V</em>s + V<em>s (\frac{t</em>1}{t<em>2}) = V</em>s (\frac{t<em>1 + t</em>2}{t<em>2}) = \frac{V</em>s T}{T - t<em>1} = \frac{V</em>s}{1 - K}$

• By controlling $K$, the output voltage can be controlled.

• If a large capacitor $C<em>L$ is connected across the load, the output voltage is continuous, and $V</em>o$ becomes the average value $V_a$.

• The voltage across the load can be stepped up by varying the duty cycle $K$.

• The minimum output voltage is $V_s$ when $K = 0$.

• The converter cannot be switched on continuously such that $K = 1$.

• For values of $K$ tending to unity, the output voltage becomes very large and is very sensitive to changes in $K$, as shown in Figure (2-c).

Energy Transfer

• This principle can be applied to transfer energy from one source to another, as shown in Fig(3-a).

• The equivalent circuits for the modes of operation are shown in figure (3-b) and (3-c), and the current waveforms in fig(3-d).

• The inductor current for mode (1) is given by:

  $V<em>s = L \frac{di</em>1}{dt}$

  And is expressed as:

  $i<em>1(t) = \frac{V</em>s}{L} t + I_1$

  Where $I_1$ is the initial current for mode (1).

• During mode 1, the current must rise; the necessary condition is that:

  \frac{di1}{dt} > 0 \implies Vs > 0

• The current for mode 2 is given by:

  $V<em>s = L \frac{di</em>2}{dt}+ E$

  And is solved as:

  $i<em>2(t) = \frac{V</em>s - E}{L} t + I_2$

  Where $I_2$ is the initial current for mode 2. For a stable system, the current must fall, and the condition is:

  \frac{di2}{dt} < 0 \implies Vs < E

  If this condition is not satisfied, the inductor current continues to rise, and an unstable situation occurs. Therefore, the conditions for controllable power transfer are:

  E>V_s

• This indicates that the source voltage $V_s$ must be less than the voltage $E$.

• In electric braking of DC motors, where the motors operate as DC generators, terminal voltage falls as the machine speed decreases.

• The converter permits transfer of power to a fixed DC source (or variable) to permit transfer of power from a fixed (or variable) to a fixed DC source or rheostat.

• When the converter is turned on, the energy is transferred from the source $V_s$ to inductor $L$.

• If the converter is turned off, a magnitude of the energy stored in the inductor is forced to battery $E$.

Step-Up Converter with a Resistive Load

• A step-up converter with a resistive load is shown in Figure (4-a).

• When switch $S_1$ is closed, the current rises through $L$ and the switch.

• The equivalent circuit during mode (1) is shown in figure (4-b), and the current is described by:

  $V_s = L \frac{di}{dt}$

  Which for an initial current of $I_1$ gives.

  $i(t) = \frac{V<em>s}{L} t + I</em>1$

  which is valid for 0 < t < KT. At the end of mode (1) at $t = KT$

  $I<em>2 = i(KT) = \frac{V</em>s}{L} KT + I_1$

• When $S<em>1$ is opened, the inductor current flows through $RL</em>oad$

• The equivalent current is shown in figure (4-c) and the current during mode (2) is described by:

  $V<em>s = R i</em>2 + L\frac{di_2}{dt} + E$

  which for an initial current of $I_2$ gives

  $i<em>2(t) = \frac{V</em>s - E}{R} (1 - e^{-\frac{R}{L}t}) + I_2e^{-\frac{R}{L}t}$

  which is valid for 0 < t < (1-K)T. At the end of mode (2) at $t = (1-K)T$

  $I<em>1 = i</em>2[t = (1 - k )T] = \frac{V<em>s - E}{R} (1 - e^{-\frac{R}{L}(1-K)T}) + I</em>2e^{-\frac{R}{L}(1-K)T}$

  Where $Z = \frac{T}{R}$

• Solving for $I<em>1$ and $I</em>2$ from equations and

  $I<em>1 = \frac{\frac{V</em>s}{R}KZ + \frac{V_s-E}{R}(1 - e^{-(1-K)Z})}{1 - e^{-Z}}$

  $I<em>2 = \frac{\frac{V</em>s}{R}KZ e^{-(1-K)Z} + \frac{V_s-E}{R}(1 - e^{-(1-K)Z})}{1 - e^{-Z}}$

• The ripple current is given by:

  $\Delta I = I<em>2 - I</em>1 = \frac{V_s KT}{L}$

• The are valid for E < Vs. If $E > V</em>s$ and the converter Switch $S_1$ is opened, the inductor transfers its stored energy through $R$ to source, and the inductor current is discontinuous.

Example

The step-up converter in figure (4-a) has $V<em>S = 10V$, $f = 1 kHz$, $R = 5 \Omega$, $L = 6.5 mH$, $E = 0$, and $K = 0.5$. Find $I</em>1, I_2$ and $\Delta I$

Solution

Given:

• $V_S = 10V$
• $f = 1 kHz$
• $R = 5 \Omega$
• $L = 6.5 mH$
• $E = 0$
• $K = 0.5$
• $\frac{V_s-E}{R} = \frac{10-0}{5} = 2$

First, calculate $Z$:

$Z = \frac{T}{R} = \frac{L}{TR} \implies Z = \frac{6.5 \times 10^{-3}}{5}=0.769$

Calculate $I_1$:

$I<em>1 = \frac{\frac{V</em>s}{R}KZ + \frac{V<em>s-E}{R}(1 - e^{-(1-K)Z})}{1 - e^{-Z}} = \frac{2*0.5 * e^{-0.5*0.769}}{1 - e^{-0.769}} + 2 = 3.64 A$ Calculate $I</em>2$:

$I<em>2 = \frac{\frac{V</em>s}{R}KZ e^{-(1-K)Z} + \frac{V_s-E}{R}(1 - e^{-(1-K)Z})}{1 - e^{-Z}} = 4.4 A$
Calculate $\Delta I$:

$\Delta I = I<em>2 - I</em>1 = 4.4 - 3.64 = 0.76 A$