Question 2 of 6: Basic Chopper with an R–L–Back-EMF Load

Question 2: Basic Chopper with an R–L–Back-EMF Load (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — principle of operation and operating modes

A basic (step-down, or buck) chopper is a controlled switch in series between a fixed d.c. supply and the load, with a free-wheel diode across the load. During the on-time $t_{on}$ the switch connects the load to the supply and the inductance stores energy; during the off-time $t_{off}=T-t_{on}$ the switch opens and the inductance drives the current round the free-wheel diode, so the load terminals sit at approximately zero volts. The load therefore sees a rectangular voltage of amplitude $V$ and duty ratio $\delta=t_{on}/T$, whose mean is $V_{o}=\delta V$. Because the switching is loss-free in principle, this is an efficient way to obtain a variable d.c. voltage from a fixed one — unlike a series dropping resistor, which wastes the difference as heat.

Control is exercised in one of two ways. In time-ratio or constant-frequency PWM control the period is fixed and $t_{on}$ is varied, which keeps the ripple frequency constant and makes filtering easy; this is by far the more common. In current-limit or variable-frequency control the on- and off-times are set by upper and lower current thresholds, which bounds the ripple directly but spreads the harmonic spectrum and complicates filtering.

Varying the on-time changes the mode as well as the mean. For a load with a back e.m.f., a long on-time keeps the current above zero throughout the period — continuous conduction, in which the mean current is $(\delta V-E_{c})/R$ and the transfer characteristic is linear in $\delta$. Shorten the on-time and the free-wheel decay reaches zero before the next turn-on: the current becomes discontinuous, the load terminals float up to $E_{c}$ during the dead band, the mean output voltage rises above $\delta V$, and the control characteristic becomes non-linear and load-dependent. The boundary between the two is exactly what part (b) asks for. In a drive this matters because discontinuous conduction raises the peak-to-mean current ratio and softens the speed regulation.

Parts (b), (c) and (d) — the critical on-time and the ripple band

Given.

Find. The on-time at which $I_{\min}=0$, the peak current at that on-time, and the two current extremes when the duty ratio is 0.5.

Approach. Write the steady-state periodic solutions of the first-order load over the on- and off-intervals, impose periodicity to get closed forms for $I_{\max}$ and $I_{\min}$, then set $I_{\min}=0$ for the boundary and evaluate both at $\delta=0.5$.

1. Note the time-constant ratio first. $\tau=L/R=15\times 10^{-3}/10=1.5$ ms against a period of 0.20 ms, so $T/\tau=0.1333$. The exponentials are shallow, but they are not negligible, and the linear-ripple approximation $\Delta I\approx V t/L$ is not accurate enough for the boundary condition — work with the exponentials throughout.
2. Solve the two intervals in the steady state. During the on-time the load sees $V-E_{c}$ across $R+sL$; during the off-time it free-wheels against $E_{c}$ alone. Imposing $i(0)=i(T)=I_{\min}$ gives the standard pair$$I_{\max}=\frac{V}{R}\,\frac{1-e^{-t_{on}/\tau}}{1-e^{-T/\tau}}-\frac{E_{c}}{R},\qquad I_{\min}=\frac{V}{R}\,\frac{e^{t_{on}/\tau}-1}{e^{T/\tau}-1}-\frac{E_{c}}{R}$$The back e.m.f. enters only as a constant offset, because it opposes the current in both intervals.
3. Impose the boundary condition for part (b). Setting $I_{\min}=0$ and solving for the on-time removes the current magnitudes entirely — only the voltage ratio and the period-to-time-constant ratio survive:$$t_{on,\text{crit}}=\tau\ln\!\left[1+\frac{E_{c}}{V}\left(e^{T/\tau}-1\right)\right]$$Substituting $E_{c}/V=18/220=0.08182$ and $e^{0.13333}-1=0.14263$:$$\boxed{t_{on,\text{crit}}=1.5\ \text{ms}\times\ln(1.011670)=17.40\ \mu\text{s}}$$a duty ratio of only 0.087. Below this the current is discontinuous.
4. Evaluate the peak current at that on-time (part c). With $I_{\min}=0$ the on-interval is a single exponential rise from zero towards $(V-E_{c})/R$, which is the quickest route:$$I_{\max}=\frac{V-E_{c}}{R}\left(1-e^{-t_{on}/\tau}\right)=\frac{220-18}{10}\left(1-e^{-0.011602}\right)$$$$\boxed{I_{\max}=20.2\times 0.011535=0.2330\ \text{A}}$$The general expression of Step 2 returns the same value, which confirms both the algebra and the boundary condition.
5. Move to $t_{on}=0.5\,T=0.10$ ms (part d). Now $t_{on}/\tau=0.06667$, and the two closed forms give$$I_{\max}=22\times\frac{0.064493}{0.124827}-1.8,\qquad I_{\min}=22\times\frac{0.068939}{0.142631}-1.8$$so that$$\boxed{I_{\max}=9.566\ \text{A},\qquad I_{\min}=8.834\ \text{A}}$$a peak-to-peak ripple of 0.733 A.
6. Check with the mean. In continuous conduction the mean load current must be $(\delta V-E_{c})/R=(110-18)/10=9.20$ A, and it must lie inside the ripple band: $8.834<9.20<9.566$. It does, and it sits almost exactly mid-band, as it should when $T\ll\tau$ makes the exponential segments nearly straight.