DC chopper

A step-up (boost) chopper is a dc–dc converter whose average output voltage is greater than the input voltage. It uses an inductor to store energy while the switch is ON and releases it, added to the source voltage, to the load while the switch is OFF.

       L          D
 +---^^^^^---+---->|---+--------+
 |   iL ->   |         |        |
 |           |         |        |
(+) Vs       S        === C    Load  Vo
 |   (chopper switch)  |        |
 |           |         |        |
 +-----------+---------+--------+

Operation

Let the switch S (thyristor with commutation circuit, MOSFET, IGBT, etc.) be ON for $T_{on}$ and OFF for $T_{off}$; period $T = T_{on} + T_{off}$, duty cycle $D = T_{on}/T$.

Mode 1: S ON ($0 < t < T_{on}$)

• The inductor is connected directly across the source; diode D is reverse biased (blocked by the capacitor voltage).
• $v_L = V_s$, so $i_L$ rises linearly from $I_1$ to $I_2$; energy is stored in L.
• The capacitor supplies the load.

Mode 2: S OFF ($T_{on} < t < T$)

• The inductor current cannot change suddenly, so the inductor voltage reverses and adds to $V_s$; D becomes forward biased.
• Current flows from source + inductor to the load and capacitor: $v_L = V_s - V_o$ (negative), so $i_L$ falls from $I_2$ to $I_1$.
• Energy from both the source and the inductor is delivered to the load, so $V_o > V_s$.

 vL
 +Vs  +------+      +------+
      |      |      |      |
 0 ---+      |      |      |
             +------+      +---
 -(Vo-Vs)
 iL      /\      /\        I2 (max)
        /  \    /  \
       /    \  /    \      I1 (min)
      |<Ton>|<Toff>|

Derivation of average output voltage

In steady state the average voltage across an inductor over one period is zero (volt-second balance), i.e. the rise in current in Mode 1 equals the fall in Mode 2:

$$\begin{aligned} \Delta I &= \frac{V_s}{L}T_{on} = \frac{V_o - V_s}{L}T_{off} \\ V_sT_{on} &= (V_o - V_s)T_{off} \\ V_o T_{off} &= V_s(T_{on} + T_{off}) = V_sT \end{aligned}$$ $$V_o = V_s\frac{T}{T_{off}} = V_s\frac{T}{T - T_{on}} = \frac{V_s}{1 - D}$$

Alternative (energy method): energy taken by L in Mode 1, $V_sIT_{on}$, equals energy given in Mode 2, $(V_o - V_s)IT_{off}$, which leads to the same result (assuming ripple-free $I$).

Discussion

• As $D$ varies from 0 to 1, $V_o$ varies from $V_s$ to (theoretically) infinity; in practice losses limit the gain, and D is kept below about 0.8–0.9.
• Input current $I_s = I_o/(1-D)$ for a lossless converter, since $V_sI_s = V_oI_o$.
• Example: $V_s = 100$ V, $D = 0.6$: $V_o = 100/(1 - 0.6) = 250$ V.
• Applications: regenerative braking of dc motors (motor emf pumping energy back into a higher-voltage supply), battery-powered systems, PFC front ends and solar PV converters.

A Type-E (Class-E) chopper is a four-quadrant chopper: it can give both polarities of average load voltage and both directions of load current. It uses four switches (CH1–CH4) with anti-parallel diodes (D1–D4) in a full-bridge arrangement, and is used to drive a dc motor in forward motoring, forward braking, reverse motoring and reverse braking.

      +Vs o-------+---------------+
                  |               |
            CH1 / D1        CH3 / D3
                  |    load       |
                  A---R--L--E-----B
                  |  (+vo, +io: A->B)
            CH2 / D2        CH4 / D4
                  |               |
      -Vs o-------+---------------+
 CH: switch;  D: anti-parallel diode
 E: back emf of the dc motor

           vo
            ^
  Q2        |        Q1
 (+vo,-io)  |   (+vo,+io)
 fwd brake  |   fwd motor
 -----------+-----------> io
  Q3        |        Q4
 (-vo,-io)  |   (-vo,+io)
 rev motor  |   rev brake

Quadrant I: forward motoring ($+v_o$, $+i_o$)

• CH4 is kept ON, CH2 and CH3 OFF; CH1 is chopped.
• CH1 ON: $V_s \to$ CH1 $\to$ A $\to$ load $\to$ B $\to$ CH4; $v_o = +V_s$, $i_o$ rises.
• CH1 OFF: current freewheels through CH4 and D2; $v_o = 0$.
• $V_o = DV_s$ (positive), power flows from source to motor. Works like a Type-A chopper.

Quadrant II: forward braking / regeneration ($+v_o$, $-i_o$)

• Motor emf $E$ is positive at A (machine running forward); CH2 is chopped, others OFF.
• CH2 ON: $E$ drives current B $\to$ A (negative) through the load, CH2 and D4; energy is stored in L.
• CH2 OFF: the inductor emf adds to $E$ and current flows through D1, the source and D4, so $v_o = +V_s$ while $i_o$ is negative: energy is returned to the source. Works like a Type-B chopper.

Quadrant III: reverse motoring ($-v_o$, $-i_o$)

• CH2 is kept ON, CH1 and CH4 OFF; CH3 is chopped.
• CH3 ON: $V_s \to$ CH3 $\to$ B $\to$ load $\to$ A $\to$ CH2; $v_o = -V_s$, current negative.
• CH3 OFF: current freewheels through CH2 and D4; $v_o = 0$.
• The motor runs in the reverse direction.

Quadrant IV: reverse braking ($-v_o$, $+i_o$)

• Motor emf is now positive at B (machine running in reverse); CH4 is chopped, others OFF.
• CH4 ON: $E$ drives current A $\to$ B (positive) through the load, CH4 and D2; energy is stored in L.
• CH4 OFF: current flows through D3, the source and D2; $v_o = -V_s$ with $i_o$ positive, so power is fed back to the source.

Summary

Thus by choosing which switch is chopped and which is held ON, the Type-E chopper gives full speed and torque control of a dc motor in all four quadrants, including regenerative braking in both directions.

A type-A (first-quadrant, step-down) chopper applies $V_s$ to the RLE load during $T_{on}$; during $T_{off}$ the load current freewheels through the diode FD and $v_0 = 0$.

Given: $V_s = 220$ V, $T = 2000\ \mu$s, $T_{on} = 600\ \mu$s, $R = 1\ \Omega$, $L = 5$ mH, $E = 24$ V.

$$\begin{aligned} \alpha &= \frac{T_{on}}{T} = \frac{600}{2000} = 0.3 \\ T_a &= \frac{L}{R} = \frac{5\times10^{-3}}{1} = 5\ \text{ms} \\ \frac{T}{T_a} &= 0.4, \qquad \frac{\alpha T}{T_a} = 0.12 \end{aligned}$$

i) Continuity of load current

The current is continuous if $m = \dfrac{E}{V_s}$ is less than the critical value:

$$\begin{aligned} m &= \frac{24}{220} = 0.1091 \\ \frac{e^{\alpha T/T_a}-1}{e^{T/T_a}-1} &= \frac{e^{0.12}-1}{e^{0.4}-1} = \frac{1.12750-1}{1.49182-1} = 0.2592 \end{aligned}$$

Since $m = 0.1091 < 0.2592$, the load current is continuous.

ii) Average output current

$$\begin{aligned} V_0 &= \alpha V_s = 0.3 \times 220 = 66\ \text{V} \\ I_0 &= \frac{V_0 - E}{R} = \frac{66 - 24}{1} = 42\ \text{A} \end{aligned}$$

iii) Maximum and minimum steady-state current

$$\begin{aligned} I_{max} &= \frac{V_s}{R}\cdot\frac{1-e^{-\alpha T/T_a}}{1-e^{-T/T_a}} - \frac{E}{R} = 220\times\frac{1-0.88692}{1-0.67032} - 24 \\ &= 75.46 - 24 = 51.46\ \text{A} \\[4pt] I_{min} &= \frac{V_s}{R}\cdot\frac{e^{\alpha T/T_a}-1}{e^{T/T_a}-1} - \frac{E}{R} = 220\times 0.25923 - 24 = 33.03\ \text{A} \end{aligned}$$

$I_{min} > 0$ confirms continuous conduction. Ripple $= 51.46 - 33.03 = 18.43$ A.

Answer: continuous; $I_0 = 42$ A; $I_{max} = 51.46$ A; $I_{min} = 33.03$ A.

iv) Waveforms

v0  220|____       ____       ____
       |    |     |    |     |    |
     0 |____|_____|____|_____|____|___ t
        Ton  Toff
i0     |   /\        /\        /\     51.46 A
       |  /  \      /  \      /  \
       | /    \____/    \____/    \   33.03 A
       |______________________________ t
iT     |  /|        /|        /|      (= i0 in Ton)
       | / |       / |       / |
       |/  |______/  |______/  |____ t
vT  220|    ______     ______
       |   |      |   |      |       (Vs in Toff)
     0 |___|      |___|      |______ t
        0  0.6    2   2.6    4  (ms)

• $v_0$: 220 V during $T_{on}$ (0.6 ms), 0 during $T_{off}$ (1.4 ms).
• $i_0$: rises exponentially from 33.03 A to 51.46 A in $T_{on}$, decays back in $T_{off}$.
• $i_T$: equals $i_0$ during $T_{on}$, zero during $T_{off}$ (diode current $i_{FD} = i_0$ then).
• $v_T$: about 0 during $T_{on}$; equal to $V_s = 220$ V during $T_{off}$ (FD conducts, so $v_0 = 0$).

For a step-down chopper with RLE load in continuous conduction, the load current rises exponentially during $kT$ and decays during $(1-k)T$; the standard steady-state results (Rashid) are used.

Given: $V_s = 220$ V, $R = 10\ \Omega$, $L = 15.5$ mH, $E = 20$ V, $k = 0.5$, $f = 5$ kHz.

$$\begin{aligned} T &= \frac{1}{f} = 200\ \mu\text{s}, \quad kT = 100\ \mu\text{s} \\ \tau &= \frac{L}{R} = \frac{15.5\times10^{-3}}{10} = 1.55\ \text{ms} \\ z &= \frac{TR}{L} = \frac{200\times10^{-6}}{1.55\times10^{-3}} = 0.12903, \quad kz = 0.064516 \end{aligned}$$

Check continuity: $E/V_s = 0.0909$; critical value $\dfrac{e^{kz}-1}{e^{z}-1} = \dfrac{0.066643}{0.137727} = 0.4839$. Since $0.0909 < 0.4839$, current is continuous.

i) Minimum instantaneous load current

$$\begin{aligned} I_1 &= \frac{V_s}{R}\cdot\frac{e^{kz}-1}{e^{z}-1} - \frac{E}{R} = 22\times\frac{1.066643-1}{1.137727-1} - 2 \\ &= 22\times 0.48388 - 2 = 8.645\ \text{A} \end{aligned}$$

ii) Peak instantaneous load current

$$\begin{aligned} I_2 &= \frac{V_s}{R}\cdot\frac{1-e^{-kz}}{1-e^{-z}} - \frac{E}{R} = 22\times\frac{1-0.937521}{1-0.878946} - 2 \\ &= 11.355 - 2 = 9.355\ \text{A} \end{aligned}$$

iii) Maximum peak-to-peak load current ripple

The ripple $\Delta I = I_2 - I_1$ is maximum at $k = 0.5$:

$$\begin{aligned} \Delta I_{max} &= \frac{V_s}{R}\tanh\frac{R}{4fL} = 22\times\tanh\frac{10}{4\times5000\times0.0155} \\ &= 22\times\tanh(0.032258) = 0.709\ \text{A} \end{aligned}$$

Here $k = 0.5$ already, so the actual ripple $9.355 - 8.645 = 0.709$ A equals this maximum. (Approximation for $4fL \gg R$: $V_s/(4fL) = 0.710$ A.)

iv) Average load current

$$I_a = \frac{kV_s - E}{R} = \frac{0.5\times220 - 20}{10} = \frac{110-20}{10} = 9\ \text{A}$$

Answer: $I_{min} = 8.645$ A, $I_{peak} = 9.355$ A, $\Delta I_{max} = 0.709$ A, $I_{avg} = 9$ A.

The load current of a step-down chopper feeding an RLE load is continuous when $m = E/V_s$ is less than the critical ratio $\dfrac{e^{\alpha T/T_a}-1}{e^{T/T_a}-1}$ (equivalently, when the computed $I_{min} > 0$).

Given: $V_s = 200$ V, $R = 2\ \Omega$, $L = 10$ mH, $E = 20$ V, $T = 1000\ \mu$s, $T_{on} = 300\ \mu$s.

$$\begin{aligned} \alpha &= \frac{300}{1000} = 0.3, \qquad T_a = \frac{L}{R} = \frac{0.01}{2} = 5\ \text{ms} \\ \frac{T}{T_a} &= \frac{1}{5} = 0.2, \qquad \frac{\alpha T}{T_a} = 0.06 \end{aligned}$$

(i) Continuity of load current

$$\begin{aligned} m &= \frac{E}{V_s} = \frac{20}{200} = 0.1 \\ \frac{e^{0.06}-1}{e^{0.2}-1} &= \frac{1.061837-1}{1.221403-1} = \frac{0.061837}{0.221403} = 0.2793 \end{aligned}$$

Since $m = 0.1 < 0.2793$, the load current is continuous.

Cross-check with the current limits:

$$\begin{aligned} I_{max} &= \frac{V_s}{R}\cdot\frac{1-e^{-0.06}}{1-e^{-0.2}} - \frac{E}{R} = 100\times\frac{0.058235}{0.181269} - 10 = 22.13\ \text{A} \\ I_{min} &= \frac{V_s}{R}\cdot\frac{e^{0.06}-1}{e^{0.2}-1} - \frac{E}{R} = 100\times0.27929 - 10 = 17.93\ \text{A} \end{aligned}$$

$I_{min} = 17.93$ A $> 0$, so the current never falls to zero.

(ii) Average load current

$$\begin{aligned} V_0 &= \alpha V_s = 0.3\times200 = 60\ \text{V} \\ I_0 &= \frac{V_0 - E}{R} = \frac{60-20}{2} = 20\ \text{A} \end{aligned}$$

i0 (A)
 22.13|   /\        /\        /\
      |  /  \      /  \      /  \
 17.93| /    \____/    \____/    \__
      |__________________________________ t
       0  0.3   1   1.3   2   (ms)

Answer: (i) continuous ($m = 0.1 < 0.279$); (ii) average load current $I_0 = 20$ A (current swings between 17.93 A and 22.13 A).

A step-up (boost) chopper gives an average output voltage higher than the input. It stores energy in an inductor while the switch is ON and releases it, added to the source, into the load while the switch is OFF.

Principle and circuit

      L        D
 +---mmm---+---|>|---+------+
 |   iL    |         |      |
 Vs       CH        ===C   Load  V0
 |         |         |      |
 +---------+---------+------+

• CH ON ($0 < t < T_{on}$): D is reverse biased. $V_s$ is across L, so $i_L$ rises linearly from $I_1$ to $I_2$; energy is stored in L. The capacitor supplies the load.
• CH OFF ($T_{on} < t < T$): the inductor voltage reverses ($L\,di/dt$ adds to $V_s$), D conducts and current flows to the load at voltage $V_0 = V_s + L\,di/dt$. $i_L$ falls from $I_2$ to $I_1$.

 vCH  V0|    ____      ____
        |   |    |    |    |
       0|___|    |____|    |___ t
 iL   I2|   /\        /\
        |  /  \      /  \
      I1| /    \____/    \___  t
         Ton Toff

Output voltage

In steady state, the volt-second balance on L (average inductor voltage = 0):

$$\begin{aligned} V_s T_{on} &= (V_0 - V_s)\,T_{off} \\ V_0 &= V_s\,\frac{T_{on}+T_{off}}{T_{off}} = \frac{V_s}{1-k}, \qquad k = \frac{T_{on}}{T} \end{aligned}$$

The same result follows from energy balance: energy put into L during $T_{on}$ is $V_s I T_{on}$, energy given out during $T_{off}$ is $(V_0 - V_s) I T_{off}$. As $k$ goes from 0 to 1, $V_0$ rises from $V_s$ towards infinity (in practice limited by losses).

Controllable power transfer (energy from a low voltage to a higher voltage)

The same circuit transfers energy from a low-voltage source $V_s$ (e.g. a dc motor's back EMF in regenerative braking) into a fixed higher-voltage dc source $E$ (supply or battery) through D. With resistance $R$ of the source/inductor:

$$\begin{aligned} \text{ON: } & V_s = L\frac{di_1}{dt} + Ri_1 \\ \text{OFF: } & V_s = L\frac{di_2}{dt} + Ri_2 + E \end{aligned}$$

For the current to be controllable:

1. During ON, current must rise: $di_1/dt > 0$, which needs $V_s > 0$.
2. During OFF, current must fall: $di_2/dt < 0$, which needs $V_s < E$ (neglecting $Ri$).

So the condition for controllable power transfer is

$$0 \le V_s \le E$$

Using the boost relation, average values give $V_s = (1-k)E$; as $k$ varies from 1 to 0, $V_s$ can be anywhere from 0 to $E$. If $V_s > E$, current would rise in both intervals and could not be controlled by the switch; if $V_s \le 0$, no energy would flow to $E$.

When the chopper switch has an ON-state drop $V_{ch}$, the load gets $(V_s - V_{ch})$ during $T_{on}$, and the switch dissipates power. Efficiency is the ratio of load power to input power.

Given: $R = 20\ \Omega$, $V_s = 220$ V, $V_{ch} = 1$ V, $f = 2$ kHz, $k = 0.8$.

(i) Average output voltage

$$\begin{aligned} V_a &= k\,(V_s - V_{ch}) = 0.8\times(220-1) = 0.8\times219 \\ &= 175.2\ \text{V} \end{aligned}$$

(The frequency 2 kHz only fixes $T = 0.5$ ms and $T_{on} = 0.4$ ms; it does not change average values for a resistive load.)

(ii) Efficiency

Output power (load voltage is $219$ V for a fraction $k$ of the time):

$$\begin{aligned} P_o &= \frac{1}{T}\int_0^{kT}\frac{(V_s - V_{ch})^2}{R}dt = k\,\frac{(V_s-V_{ch})^2}{R} \\ &= 0.8\times\frac{219^2}{20} = 1918.44\ \text{W} \end{aligned}$$

Input power (source supplies current $(V_s - V_{ch})/R$ during $kT$):

$$\begin{aligned} P_i &= \frac{1}{T}\int_0^{kT}V_s\,\frac{V_s - V_{ch}}{R}dt = k\,\frac{V_s(V_s-V_{ch})}{R} \\ &= 0.8\times\frac{220\times219}{20} = 1927.2\ \text{W} \end{aligned}$$ $$\begin{aligned} \eta &= \frac{P_o}{P_i} = \frac{V_s - V_{ch}}{V_s} = \frac{1918.44}{1927.2} = \frac{219}{220} \\ &= 0.99545 = 99.55\% \end{aligned}$$

Switch loss $= 1927.2 - 1918.44 = 8.76$ W.

Answer: $V_a = 175.2$ V; efficiency $\eta = 99.55\%$.

DC choppers are classified (Type/Class A to E) by the quadrants of the load voltage–current ($v_0$–$i_0$) plane in which they can operate. Class C is a two-quadrant chopper (quadrants I and II); Class E is a four-quadrant chopper.

Class C (two-quadrant, current-reversible) chopper

  +----+------------+
  |    |            |
  |   CH1  ^D2      |
 Vs    |   |        |
  |    +---+---+  Load
  |    |   |   |  (R,L,E)
  |   D1^  CH2 |    |
  |    |   |   +----+
  +----+---+

It is a Class A chopper (CH1 + freewheeling diode D1) in parallel with a Class B chopper (CH2 + diode D2).

• Quadrant I (motoring): CH1 is chopped. When CH1 is ON, $v_0 = V_s$ and $i_0$ is positive; when OFF, $i_0$ freewheels through D1 with $v_0 = 0$. Power flows from source to load.
• Quadrant II (regenerative braking): CH2 is chopped. When CH2 is ON, the load EMF $E$ drives a negative current through L and CH2, storing energy in L ($v_0 = 0$). When CH2 is OFF, $E + L\,di/dt$ exceeds $V_s$, so negative current flows through D2 back into the source ($v_0 = V_s$). Power flows from load to source.
• Load voltage is always positive; load current can be positive or negative.
• CH1 and CH2 must never be ON together, or the source is short-circuited.
• Use: dc motor drives needing motoring and regenerative braking in one direction of rotation (e.g. traction).

Class E (four-quadrant) chopper

  +------+------------+------+
  |      |            |      |
  |     CH1 ^D1   CH3 ^D3    |
 Vs      |  |          | |    |
  |      +--+--[Load]--+-+    |
  |      |  |    R,L,E  | |   |
  |     CH4 ^D4   CH2 ^D2    |
  +------+------------+------+

Four switches with anti-parallel diodes form an H-bridge, so both $v_0$ and $i_0$ can reverse.

• Quadrant I: CH2 ON continuously, CH1 chopped; $v_0 = +V_s$ when CH1 ON; current freewheels via CH2–D4 when CH1 OFF.
• Quadrant II: with load EMF positive and current reversed, CH4 chopped; when CH4 ON, $E$ drives current through CH4–D2 storing energy in L; when OFF, current returns to the source through D1–D2.
• Quadrants III and IV are the mirror operations with the load EMF reversed.
• Use: reversible, regenerative dc motor drives (all four quadrants of speed–torque).

With an ON-state drop $V_{ch}$, the load voltage is a pulse of height $(V_s - V_{ch})$ lasting $kT$ in every period, so the average and rms values follow directly from that pulse.

Given: $V_s = 230$ V, $R = 10\ \Omega$, $V_{ch} = 1$ V, $k = 0.4$.

 v0   229|_____       _____
         |     |     |     |
       0 |_____|_____|_____|____ t
          kT    T

Average output voltage

$$\begin{aligned} V_a &= \frac{1}{T}\int_0^{kT}(V_s - V_{ch})\,dt = k\,(V_s - V_{ch}) \\ &= 0.4\times(230-1) = 0.4\times229 = 91.6\ \text{V} \end{aligned}$$

RMS output voltage

$$\begin{aligned} V_{rms} &= \left[\frac{1}{T}\int_0^{kT}(V_s - V_{ch})^2dt\right]^{1/2} = \sqrt{k}\,(V_s - V_{ch}) \\ &= \sqrt{0.4}\times229 = 0.63246\times229 = 144.83\ \text{V} \end{aligned}$$

For reference, the load currents are $I_a = 91.6/10 = 9.16$ A and $I_{rms} = 14.48$ A.

Answer: $V_a = 91.6$ V, $V_{rms} = 144.83$ V.

The output of a step-down chopper with resistive load is a rectangular pulse train (height $V_s$, width $kT$). Its Fourier series gives the dc (average) value and the harmonics; the fundamental is at the chopping frequency.

Given: $V_s = 220$ V, $R = 20\ \Omega$, $k = 0.45$, $f = 1$ kHz.

Average output voltage

$$V_a = kV_s = 0.45\times220 = 99\ \text{V}, \qquad I_a = \frac{99}{20} = 4.95\ \text{A}$$

Fourier series of the output voltage

With the pulse from $t = 0$ to $kT$:

$$v_0(t) = kV_s + \sum_{n=1}^{\infty}\frac{V_s}{n\pi}\Big[\sin(2n\pi k)\cos(n\omega t) + \big(1-\cos(2n\pi k)\big)\sin(n\omega t)\Big]$$

Peak of the $n$th harmonic:

$$V_{nm} = \frac{V_s}{n\pi}\sqrt{\sin^2(2n\pi k) + \big(1-\cos 2n\pi k\big)^2} = \frac{2V_s}{n\pi}\sin(n\pi k)$$

Fundamental component ($n = 1$)

$$\begin{aligned} V_{1m} &= \frac{2V_s}{\pi}\sin(\pi k) = \frac{2\times220}{\pi}\sin(0.45\times180^\circ) \\ &= 140.06\times\sin 81^\circ = 140.06\times0.98769 = 138.33\ \text{V} \\ V_{1,rms} &= \frac{138.33}{\sqrt2} = 97.82\ \text{V} \end{aligned}$$

Fundamental load current (resistive load):

$$\begin{aligned} I_{1m} &= \frac{V_{1m}}{R} = \frac{138.33}{20} = 6.917\ \text{A} \\ I_{1,rms} &= \frac{6.917}{\sqrt2} = 4.891\ \text{A} \end{aligned}$$

Phase: $\tan\phi = \dfrac{\sin 2\pi k}{1-\cos 2\pi k} = \tan(90^\circ - 81^\circ)$, so $\phi = 9^\circ$:

$$i_1(t) = 6.917\sin(2000\pi t + 9^\circ)\ \text{A}$$

Answer: $V_a = 99$ V; fundamental load current $= 6.917$ A peak ($4.891$ A rms) at 1 kHz.

A step-up chopper produces an average output voltage higher than the source. An inductor in series with the source stores energy while the switch is ON and pushes it into the load, in series with $V_s$, while the switch is OFF.

Circuit and operation

      L        D
 +---mmm---+---|>|---+------+
 |   iL    |  vx     |      |
 Vs       CH        ===C   Load  V0
 |         |         |      |
 +---------+---------+------+

1. CH ON ($0 \le t \le T_{on}$): node $x$ is shorted to the negative rail ($v_x = 0$). D is reverse biased by $V_0$. $V_s$ appears across L, so $i_L$ rises linearly: $L\,di/dt = V_s$. The capacitor feeds the load.
2. CH OFF ($T_{on} \le t \le T$): the inductor EMF reverses and adds to $V_s$; D conducts, $v_x = V_0$, and $i_L$ falls: $L\,di/dt = V_s - V_0$ (negative).

 vx  V0 |    ____      ____
        |   |    |    |    |
      0 |___|    |____|    |___ t
 iL  I2 |   /\        /\
        |  /  \      /  \
     I1 | /    \____/    \___ t
        0 Ton  T

Average output voltage

Rise of current in $T_{on}$ equals the fall in $T_{off}$ (steady state):

$$\begin{aligned} \Delta I &= \frac{V_s}{L}T_{on} = \frac{V_0 - V_s}{L}T_{off} \\ V_0\,T_{off} &= V_s (T_{on} + T_{off}) = V_s T \\ V_0 &= \frac{V_s\,T}{T - T_{on}} = \frac{V_s}{1-k} \end{aligned}$$

Equivalently, the switch-node voltage $v_x$ is 0 for $kT$ and $V_0$ for $(1-k)T$; its average must equal $V_s$ (no average voltage across L): $(1-k)V_0 = V_s$.

For $k = 0 \to 1$, $V_0$ varies from $V_s$ to a very large value; e.g. $k = 0.5$ gives $V_0 = 2V_s$.

RMS value of output voltage

• With a large filter capacitor the load voltage is nearly ripple-free, so $V_{0,rms} \approx V_0 = \dfrac{V_s}{1-k}$.
• The chopped voltage at the chopper terminals ($v_x$, the voltage the load sees before filtering) is a pulse of height $V_0$ for $(1-k)T$:

$$\begin{aligned} V_{x,rms} &= \left[\frac{1}{T}\int_{kT}^{T}V_0^2\,dt\right]^{1/2} = V_0\sqrt{1-k} \\ &= \frac{V_s}{1-k}\sqrt{1-k} = \frac{V_s}{\sqrt{1-k}} \end{aligned}$$

and its average is $V_{x,avg} = (1-k)V_0 = V_s$.

Example: $V_s = 100$ V, $k = 0.6$: $V_0 = 100/0.4 = 250$ V; $V_{x,rms} = 100/\sqrt{0.4} = 158.1$ V.

Uses: regenerative braking of dc motors, battery charging from a lower voltage, boost SMPS.

A step-down (Type-A) chopper controls the speed of a separately excited dc motor by varying the average armature voltage $V_0 = kV_s$. The motor armature is an RLE load: resistance $R_a$, inductance $L_a$ and back EMF $E_b$.

Circuit

  +----CH----+-------------+
  |          |             |
  |          |            Ra
 Vs        FD ^            La
  |          |            (Eb)  motor
  |          |             |
  +----------+-------------+

Modes of operation

Mode 1: CH ON ($0 < t < T_{on}$). Source connects to the armature, $v_0 = V_s$:

$$V_s = R_a i_a + L_a\frac{di_a}{dt} + E_b$$

The current rises exponentially from $I_{min}$ to $I_{max}$; energy is stored in $L_a$.

Mode 2: CH OFF ($T_{on} < t < T$). The armature current freewheels through FD, $v_0 = 0$:

$$0 = R_a i_a + L_a\frac{di_a}{dt} + E_b$$

The current decays from $I_{max}$ to $I_{min}$. If $L_a$ is small or $k$ is low, it can reach zero before the next ON period (discontinuous conduction); then $v_0 = E_b$ for the remaining time.

 v0  Vs |____      ____
        |    |    |    |
      0 |____|____|____|____ t
 ia Imax|   /\        /\
        |  /  \      /  \
    Imin| /    \____/    \__ t
         Ton Toff

Steady-state relations (continuous conduction)

$$\begin{aligned} V_0 &= kV_s = E_b + I_a R_a \\ E_b &= K\phi\,\omega \\ \omega &= \frac{kV_s - I_a R_a}{K\phi} \end{aligned}$$

So speed is controlled by the duty ratio $k$: increasing $k$ increases $V_0$ and the speed. Torque $T = K\phi I_a$ depends on the load current.

Current limits:

$$I_{max} = \frac{V_s}{R_a}\cdot\frac{1-e^{-kT/T_a}}{1-e^{-T/T_a}} - \frac{E_b}{R_a}, \quad I_{min} = \frac{V_s}{R_a}\cdot\frac{e^{kT/T_a}-1}{e^{T/T_a}-1} - \frac{E_b}{R_a}$$

with $T_a = L_a/R_a$. Conduction is continuous if $I_{min} > 0$.

Features

• Operates in the first quadrant only (motoring, forward); braking needs a Class B or C chopper.
• High chopping frequency or an added series inductor reduces current ripple and torque pulsation.
• Smooth, efficient (no rheostat loss) control; widely used in battery vehicles, forklifts and traction.

In variable-frequency (frequency modulation) control, $T_{on}$ is kept constant and the duty cycle is changed by varying the chopping period $T$ (hence $T_{off}$).

Given: $V_{dc} = 200$ V, $T_{on} = 3$ ms (constant), $k = 0.3$, resistive load.

OFF time

$$\begin{aligned} k &= \frac{T_{on}}{T} \;\Rightarrow\; T = \frac{T_{on}}{k} = \frac{3}{0.3} = 10\ \text{ms} \\ T_{off} &= T - T_{on} = 10 - 3 = 7\ \text{ms} \end{aligned}$$

Chopping frequency $f = 1/T = 100$ Hz.

Average output voltage

$$V_o = kV_{dc} = \frac{T_{on}}{T_{on}+T_{off}}V_{dc} = 0.3\times200 = 60\ \text{V}$$

 vo  200|___         ___
        |   |       |   |
      0 |___|_______|___|_______ t
         3ms  7ms
        |<--- T = 10 ms --->|

Answer: $T_{off} = 7$ ms; average output voltage $V_o = 60$ V.

Note: variable-frequency control is simple, but a wide frequency range makes filter design hard and may cause interference; for low $k$ the OFF time becomes very long, so the load current may become discontinuous.

In a step-down chopper the average output is $V_o = kV_s$ with $k = T_{on}/T$. Variable-frequency (frequency modulation) control keeps either $T_{on}$ or $T_{off}$ constant and changes the period $T$ to change $k$.

Operation in variable-frequency mode

  +---CH---+--------+
  |        |        |
 Vs      FD^      Load  vo
  |        |        |
  +--------+--------+

 constant Ton, larger T -> lower k
 vo |__      __      __
    |  |____|  |____|  |____
 constant Ton, smaller T -> higher k
 vo |__  __  __  __  __
    |  ||  ||  ||  ||  |

• The switch is turned ON for a fixed time $T_{on}$; the next turn-on is delayed (or advanced) to set $T$.
• Increasing $T$ (lower frequency) lowers $k$ and $V_o$; decreasing $T$ raises them.
• Drawbacks: the chopping frequency varies over a wide range, so input/output filter design is difficult; low frequency at small $k$ increases current ripple and may make load current discontinuous; possible interference with signalling and telephone lines. Hence constant-frequency (PWM) control is usually preferred.

Numerical

Given: $T_{on} = 5$ ms (constant), $k = 0.4$.

$$\begin{aligned} T &= \frac{T_{on}}{k} = \frac{5}{0.4} = 12.5\ \text{ms} \quad (f = 80\ \text{Hz}) \\ T_{off} &= T - T_{on} = 12.5 - 5 = 7.5\ \text{ms} \end{aligned}$$

Average output voltage:

$$V_o = kV_s = 0.4\,V_s$$

The supply voltage is not stated; taking $V_s = 200$ V (the value used in the same textbook circuit):

$$V_o = 0.4\times200 = 80\ \text{V}$$

Answer: $T_{off} = 7.5$ ms; $V_o = 0.4V_s$ ($= 80$ V for $V_s = 200$ V).

Choppers are classified by the quadrants of the $v_0$–$i_0$ plane they work in. Type B works in the second quadrant only (regeneration); Type C works in the first and second quadrants (motoring and regeneration).

Type-B chopper (second quadrant)

     D
 +---|<|----+--------+
 |          |        |
 Vs        CH        L
 |          |       (E)  load
 |          |        |
 +----------+--------+

The load must contain an EMF $E$ (e.g. a dc motor during braking or a battery).

• CH ON: the load is short-circuited through CH; $v_0 = 0$. $E$ drives current through L and CH (current direction is out of the load's positive terminal, i.e. $i_0$ negative); energy is stored in L: $E = L\,di/dt + Ri$.
• CH OFF: the inductor voltage $L\,di/dt$ adds to $E$, so $v_0 = E + L\,di/dt$ exceeds $V_s$; diode D conducts and the current flows back into the source. Energy is returned to $V_s$.
• Load voltage is positive, load current is negative, so power flows from load to source.
• Average output voltage: $V_0 = (1-k)V_s$, where $k$ is the duty cycle of CH. It works as a step-up chopper from $E$ to $V_s$.

 v0  Vs |    ____      ____
        |   |    |    |    |
      0 |___|    |____|    |__ t
 |i0|   |   /\        /\
        |  /  \      /  \
        | /    \____/    \__ t
         Ton Toff

Use: regenerative braking of dc motors.

Type-C chopper (first and second quadrants)

 +-----+--------------+
 |     |              |
 |    CH1   ^D2       |
 Vs    |    |         |
 |     +----+-----+  load
 |     |    |     |  R,L,E
 |    D1^  CH2    |   |
 |     |    |     +---+
 +-----+----+

It combines a Type-A chopper (CH1, D1) and a Type-B chopper (CH2, D2) in parallel.

• First quadrant (motoring): CH1 is switched. CH1 ON: $v_0 = V_s$, $i_0$ positive, power to load. CH1 OFF: $i_0$ freewheels through D1, $v_0 = 0$.
• Second quadrant (braking): CH2 is switched. CH2 ON: load shorted, $E$ drives negative current through CH2, energy stored in L. CH2 OFF: negative current flows through D2 into the source ($v_0 = V_s$), returning energy.
• Load voltage is always positive; load current reverses. CH1 and CH2 must never be ON together (supply short).
• Use: dc motor drives with motoring and regenerative braking in one direction (traction, electric vehicles).

With an ON-state drop of $V_{ch}$, the load receives $(V_s - V_{ch})$ for $kT$ and zero for the rest of the period.

Given: $V_s = 230$ V, $R = 10\ \Omega$, $V_{ch} = 2$ V, $k = 0.4$.

 v0   228|_____       _____
         |     |     |     |
       0 |_____|_____|_____|____ t
          kT    T

Average output voltage

$$\begin{aligned} V_a &= \frac{1}{T}\int_0^{kT}(V_s - V_{ch})\,dt = k\,(V_s - V_{ch}) \\ &= 0.4\times(230 - 2) = 0.4\times228 = 91.2\ \text{V} \end{aligned}$$

RMS output voltage

$$\begin{aligned} V_{rms} &= \left[\frac{1}{T}\int_0^{kT}(V_s - V_{ch})^2\,dt\right]^{1/2} = \sqrt{k}\,(V_s - V_{ch}) \\ &= \sqrt{0.4}\times228 = 0.63246\times228 = 144.20\ \text{V} \end{aligned}$$

Corresponding load currents: $I_a = 91.2/10 = 9.12$ A, $I_{rms} = 14.42$ A. Chopper efficiency would be $(V_s - V_{ch})/V_s = 228/230 = 99.13\%$.

Answer: $V_a = 91.2$ V, $V_{rms} = 144.20$ V.