Question 4 of 6: Problem 3 — basic chopper, completing the load table

Question 4: Problem 3 — basic chopper, completing the load table (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.

Find. The one missing entry in each row, taking "maximum allowed current" to mean that the peak of the steady-state ripple reaches exactly 20 A.

Approach. In steady-state continuous conduction the current rises exponentially towards $V_i/R$ during the on-time and decays exponentially to zero during the free-wheel interval; imposing periodicity gives one closed-form expression for $I_{max}$ in terms of $V_i$, R, $T_{on}$, T and $\tau$, and each case inverts that one equation for a different unknown.

1. Derive the peak-current expression. During the on-time the switch connects the source, so $i(t)=(V_i/R)+\bigl[I_{min}-(V_i/R)\bigr]e^{-t/\tau}$ and after $T_{on}$ the current has reached $I_{max}$. During the free-wheel interval there is no source, so the current simply decays, $i(t)=I_{max}e^{-t/\tau}$, and after $T_{off}=T-T_{on}$ it must have returned to $I_{min}$ for the waveform to be periodic. Eliminating $I_{min}$ between the two gives $$\boxed{\;I_{max}=\frac{V_i}{R}\cdot \frac{1-e^{-T_{on}/\tau}}{1-e^{-T/\tau}}\;}$$ and the trough follows from the free-wheel decay alone, since that interval has no source: $I_{min}=I_{max}\,e^{-(T-T_{on})/\tau}$. Every row of the table is this one relation solved for a different symbol, and because $\tau=L/R$ is given directly, L never has to be found first.
2. Case 1 — solve for the load resistance. Here $T_{on}/\tau=2.00/1.25=1.6$ and $T/\tau=2.50/1.25=2.0$, so $$R=\frac{V_i}{I_{max}}\cdot\frac{1-e^{-1.6}}{1-e^{-2.0}} =\frac{24}{20}\times\frac{0.798103}{0.864665}=1.2\times 0.923019$$ $$\boxed{\;R_1=1.1076\ \Omega\;}$$
3. Check Case 1 against the mean. The duty ratio is $\delta=2.00/2.50=0.800$, so the free-wheel interval leaves $I_{min}=20\,e^{-0.5/1.25}=13.41$ A and the mean current should be $\delta V_i/R=0.800\times 24/1.1076=17.33$ A. That lies between $I_{min}$ and $I_{max}$, as it must. For completeness the inductance is $L=\tau R=1.25\times 1.1076=1.385$ mH.
4. Case 2 — solve for the chopper period. Rearranging the same expression for the only unknown, $$1-e^{-T/\tau}=\frac{V_i}{R\,I_{max}}\bigl(1-e^{-T_{on}/\tau}\bigr) =\frac{24}{1.15\times 20}\times\bigl(1-e^{-2.5}\bigr)=1.043478\times 0.917915=0.957824$$ so $e^{-T/\tau}=0.042176$ and $$T=-\tau\ln(0.042176)=1.00\times 3.16591$$ $$\boxed{\;T_2=3.1659\ \text{ms}\;}$$ The result must satisfy $T>T_{on}$, and 3.166 ms against 2.50 ms does; had the algebra returned a period shorter than the on-time, or a bracket outside the interval from 0 to 1, the stated data would have been inconsistent.
5. Check Case 2. The duty ratio is $2.50/3.1659=0.790$, the switching frequency $1/T=315.9$ Hz, and $I_{min}=20\,e^{-0.6659/1.00}=10.28$ A. The mean $\delta V_i/R=0.790\times 24/1.15=16.48$ A again sits between the extremes. $L=\tau R=1.15$ mH.
6. Case 3 — solve for the on-time. Now $$1-e^{-T_{on}/\tau}=\frac{R\,I_{max}}{V_i}\bigl(1-e^{-T/\tau}\bigr) =\frac{0.9\times 20}{24}\times\bigl(1-e^{-1.2/1.5}\bigr)=0.75\times 0.550671=0.413003$$ hence $e^{-T_{on}/\tau}=0.586997$ and $$T_{on}=-1.5\ln(0.586997)=1.5\times 0.532735$$ $$\boxed{\;T_{on,3}=0.7991\ \text{ms}\;}$$
7. Check Case 3 and compare the three rows. The duty ratio is $0.7991/1.2=0.666$, $I_{min}=20\,e^{-0.4009/1.5}=15.31$ A, and the mean $0.666\times 24/0.9=17.76$ A lies between them. Comparing the rows shows what the table is teaching: Case 3 has the shortest period relative to its time constant ($T/\tau=0.80$), so its ripple is the smallest — 4.7 A peak-to-peak against 6.6 A in Case 1 and 9.7 A in Case 2 — even though its steady-state ceiling $V_i/R=26.7$ A is the highest of the three. Ripple is governed by $T/\tau$, not by the duty ratio.
8. Confirm every row against the original relation. Substituting each completed row back into $I_{max}=(V_i/R)(1-e^{-T_{on}/\tau})/(1-e^{-T/\tau})$ returns 20.000 A in all three cases, which is the closing check the table asks for.