Question 4 of 6: Problem 3 — basic dc chopper with an R–L load

Question 4: Problem 3 — basic dc chopper with an R–L load (15 points)

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 chopper switches a 24 V source into a series R–L load at a fixed 2 ms period, and the duty ratio is specified indirectly through the ripple, as the ratio of minimum to maximum load current.

Find. (a) the load time constant and the on-time; (b) the peak and valley load currents; (c) the time-domain current expressions for both sub-intervals and the current at $t=1$ ms and $t=1.5$ ms.

Approach. During the free-wheel interval no source is present, so the current is a pure decaying exponential and the stated current ratio fixes the off-time on its own; only then do the steady-state boundary conditions fix the current levels.

1. Compute the load time constant. $$\tau=\frac{L}{R}=\frac{0.45\times10^{-3}}{1.8}=\boxed{0.250\ \text{ms}}$$
2. Extract the off-time from the ratio alone. With the switch open the load free-wheels through the diode with no driving source, so $i(t)=I_{max}e^{-t/\tau}$ and $$\frac{I_{min}}{I_{max}}=e^{-t_{off}/\tau}\ \Rightarrow\ t_{off}=-\tau\ln(0.75)=0.250\times0.28768=0.0719\ \text{ms}$$ The current magnitudes never enter this step.
3. Hence the on-time and the duty ratio. $$t_{on}=T-t_{off}=2.000-0.0719=\boxed{1.9281\ \text{ms}}\qquad \delta=\frac{t_{on}}{T}=0.9640$$

With the levels still unknown, the periodic steady state supplies them. Matching the end of the rising interval to the start of the decaying one and requiring the waveform to repeat gives the standard pair of boundary conditions.

4. Solve the steady-state peak. $$I_{max}=\frac{V_{i}}{R}\cdot\frac{1-e^{-t_{on}/\tau}}{1-e^{-T/\tau}}=13.333\times\frac{1-e^{-7.7123}}{1-e^{-8}}=\boxed{13.332\ \text{A}}$$
5. And the valley from the given ratio. $$I_{min}=0.75\,I_{max}=0.75\times13.3318=\boxed{9.999\ \text{A}}$$ The peak-to-peak ripple is therefore 3.333 A.

Two checks close part (b). The mean current predicted by the duty ratio, $\delta V_{i}/R=0.9640\times24/1.8=12.85$ A, lies between the valley and the peak as it must. And the widely used linear-ripple estimate $\Delta I\approx V_{i}\delta(1-\delta)T/L$ returns 3.698 A against the true 3.333 A, an 11 per cent error, because that shortcut assumes $\tau\gg T$ whereas here $\tau$ is an eighth of $T$. The exponential forms must be used.

6. Write the on-interval expression. Measuring $t$ from the switch turning on, with initial value $I_{min}$: $$i_{on}(t)=\frac{V_{i}}{R}\left(1-e^{-t/\tau}\right)+I_{min}e^{-t/\tau}=13.332-3.333\,e^{-t/0.25\ \text{ms}}$$ valid for $0\le t\le 1.9281$ ms.
7. Write the off-interval expression. Measuring $t^{\prime}$ from turn-off, $$i_{off}(t^{\prime})=I_{max}e^{-t^{\prime}/\tau}=13.332\,e^{-t^{\prime}/0.25\ \text{ms}}$$ valid for $0\le t^{\prime}\le0.0719$ ms.
8. Screen each requested instant against the on-time before choosing a branch. Both 1 ms and 1.5 ms are less than $t_{on}=1.9281$ ms, so both lie in the on-interval and $i_{on}$ applies to each.
9. Evaluate. $e^{-4}=0.018316$ and $e^{-6}=0.0024788$, so $$i(1\ \text{ms})=13.332-3.333\times0.018316=\boxed{13.272\ \text{A}}$$ $$i(1.5\ \text{ms})=13.332-3.333\times0.0024788=\boxed{13.325\ \text{A}}$$

Both values sit within half a per cent of $V_{i}/R=13.333$ A. That flatness is the correct answer, not a slip: four time constants have already elapsed by 1 ms, so the current has settled to its resistive limit long before the switch opens.