Question 3 of 6: Basic Chopper Feeding an R–L Load with Back EMF

Question 3: Basic Chopper Feeding an R–L Load with Back EMF (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.

Given. A step-down (buck) chopper switches a 220 V d.c. supply into a series R–L load that also contains a 20 V back emf, at a fixed period of 0.2 ms.

Find. The principles of chopper operation and the effect of on-time on the operating mode; then the critical on-time at which the minimum load current just reaches zero, the corresponding maximum current, and the current extremes at $t_{on}=0.5T$.

Approach. Solve the first-order load equation separately over the conduction and free-wheel intervals, match the two at the boundaries for periodic steady state, then impose $I_{min}=0$ for the critical case and evaluate the standard steady-state extremes for $t_{on}=0.5T$.

Part (a) — Principles of operation and the effect of on-time

A chopper is a controllable switch in series with the load, together with a free-wheeling diode connected across the load. When the switch is closed for a time $t_{on}$ the load is connected directly to the supply, the terminal voltage is $V_i$, and the current rises exponentially towards the value $(V_i-E_c)/R$ with time constant $\tau = L/R$. When the switch opens, the inductive current cannot stop, so it transfers to the free-wheeling diode; the load terminals are then clamped to approximately zero volts and the current decays exponentially towards $-E_c/R$. Repeating this at a fixed period $T$ produces a mean load voltage $V_o=\delta V_i$, where $\delta = t_{on}/T$ is the duty ratio, and a mean current $(\delta V_i-E_c)/R$. The switch itself is a forced-commutated thyristor in classical practice and a GTO, IGBT or MOSFET in modern equipment.

Three control strategies follow from the same circuit. Holding $T$ constant and varying $t_{on}$ is time-ratio (pulse-width) control, and it is by far the most common because the switching frequency, and hence the filter and the audible noise, are fixed. Holding $t_{on}$ constant and varying $T$ is frequency modulation, which appears in current-limit control; it complicates filtering because the harmonic spectrum moves. A hybrid of the two is used where a wide output range is needed.

Varying the on-time changes the operating mode as well as the mean output. As $t_{on}$ is increased the mean current rises in proportion, and once $t_{on}$ exceeds a critical value the current never falls to zero within a period: this is continuous conduction, in which the transfer characteristic is the clean linear relation $V_o = \delta V_i$ and the load behaves predictably. Below that critical on-time the current reaches zero before the switch next closes and the chopper enters discontinuous conduction. In this mode the load terminals float up to the back emf $E_c$ during the dead band, the mean output voltage is no longer $\delta V_i$ but depends on the load itself, and the small-signal gain of the converter changes sharply — a closed loop tuned in the continuous mode can become unstable in the discontinuous one. The peak-to-peak ripple is largest near $\delta = 0.5$ for a given period, and the practical lower limit on $t_{on}$ is set by the switch turn-on and turn-off times, while at very high $\delta$ the free-wheel interval may become too short for a forced commutation circuit to recover.

Parts (b) to (d) — Quantitative solution

1. Part (b) — write the two interval solutions. During conduction the load equation is $L\,di/dt+Ri = V_i-E_c$, and during free-wheeling it is $L\,di/dt+Ri=-E_c$. With $\tau = L/R = 15\times10^{-3}/10 = 1.5$ ms the steady-state extremes satisfy $$\begin{aligned} I_{max}&=\frac{V_i-E_c}{R}\left(1-e^{-t_{on}/\tau}\right)+I_{min}e^{-t_{on}/\tau},\ I_{min}&=-\frac{E_c}{R}\left(1-e^{-t_{off}/\tau}\right)+I_{max}e^{-t_{off}/\tau}. \end{aligned}$$
2. Impose the critical condition and eliminate the currents. Setting $I_{min}=0$ with $t_{off}=T-t_{on}$ and eliminating $I_{max}$ between the two relations gives a single logarithm in which the current magnitudes have disappeared entirely: $$t_{on,crit}=\tau\ln\left[1+\frac{E_c}{V_i}\left(e^{T/\tau}-1\right)\right].$$ Only the ratios $E_c/V_i$ and $T/\tau$ matter — the load current level is irrelevant to where the boundary lies.
3. Evaluate the critical on-time. With $E_c/V_i = 20/220 = 0.090909$ and $T/\tau = 0.2/1.5 = 0.133333$, so that $e^{T/\tau}-1 = 0.142631$: $$t_{on,crit}=1.5\ln\left(1+0.090909\times0.142631\right)=1.5\times 0.0128832 =\boxed{0.01932\ \text{ms}=19.32\ \mu\text{s}},$$ a critical duty ratio of $\delta_{crit}=0.01932/0.2 = 0.0966$.
4. Part (c) — evaluate the peak current at that boundary. With $I_{min}=0$ the conduction-interval relation reduces to $$I_{max}=\frac{V_i-E_c}{R}\left(1-e^{-t_{on,crit}/\tau}\right) =\frac{220-20}{10}\left(1-e^{-0.0128832}\right)=20\times 0.0128005 =\boxed{0.2560\ \text{A}}.$$ Cross-checking from the free-wheel interval instead, with $t_{off}=0.2-0.01932=0.18068$ ms, $$I_{max}=\frac{E_c}{R}\left(e^{t_{off}/\tau}-1\right)=2\left(e^{0.120450}-1\right)=0.2560\ \text{A},$$ which agrees exactly and confirms both interval solutions.
5. Part (d) — evaluate the extremes at $t_{on}=0.5T$. Solving the same pair of relations for general $t_{on}$ gives the standard steady-state results $$\begin{aligned} I_{max}&=\frac{V_i}{R}\,\frac{1-e^{-t_{on}/\tau}}{1-e^{-T/\tau}}-\frac{E_c}{R},\\ I_{min}&=\frac{V_i}{R}\,\frac{e^{t_{on}/\tau}-1}{e^{T/\tau}-1}-\frac{E_c}{R}. \end{aligned}$$ With $t_{on}=0.1$ ms, $t_{on}/\tau = 0.066667$ and $T/\tau = 0.133333$: $$\begin{aligned} I_{max}&=22\times\frac{0.0644929}{0.1248270}-2=\boxed{9.367\ \text{A}},\ I_{min}&=22\times\frac{0.0689394}{0.1426313}-2=\boxed{8.633\ \text{A}}. \end{aligned}$$
6. Check the mean and the ripple. The mean current must be $(\delta V_i-E_c)/R=(0.5\times220-20)/10 = 9.00$ A, and it does lie between the two extremes, almost exactly midway. The peak-to-peak ripple is 0.733 A, i.e. 8.1 per cent of the mean. Here the linear-ripple approximation $\Delta I \approx V_i\delta(1-\delta)T/L = 0.7333$ A is accurate to better than 0.1 per cent, because $\tau = 1.5$ ms is 7.5 times the period and the exponential segments are effectively straight. That approximation is not generally safe: when $\tau$ is comparable to or smaller than $T$ it can be in error by tens of per cent.