Question 5 of 5: Basic chopper — completing the design table

Question 5: Basic chopper — completing the design table (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.

Throughout, $V_i = 24\ \text{V}$ and the peak load current is held at its allowed maximum, $I_{\max} = 20\ \text{A}$. The load has no back e.m.f.

Find. The one missing entry in each row, with each row screened for feasibility before it is solved.

Approach. Every row is the same steady-state relation between $I_{\max}$, $V_i/R$, $T$, $T_{on}$ and $\tau$, inverted for a different unknown; derive it once, then apply it three times.

1. Derive the master relation once. During the on-interval the supply drives the R–L load, so the current climbs from $I_{\min}$ towards $V_i/R$; during the off-interval the free-wheel diode carries it and it decays towards zero: $$I_{\max} = \frac{V_i}{R}\left(1 - e^{-T_{on}/\tau}\right) + I_{\min}e^{-T_{on}/\tau}, \qquad I_{\min} = I_{\max}e^{-T_{off}/\tau}$$ Substituting the second into the first and using $T_{on} + T_{off} = T$ eliminates $I_{\min}$ and gives the relation that governs all three rows: $$\boxed{\;I_{\max} = \frac{V_i}{R}\cdot \frac{1 - e^{-T_{on}/\tau}}{1 - e^{-T/\tau}}\;}$$ Two facts follow immediately and are used as screens below. Because $T_{on} < T$, the fraction is always less than one, so $I_{\max}$ can never exceed $V_i/R$; and any row solved for the period must return a $T$ greater than its own $T_{on}$.
2. Case 1 — solve for the load resistance. Rearranging the master relation for $R$ with $T = 2.4$, $T_{on} = 1.8$ and $\tau = 1.5\ \text{ms}$: $$R = \frac{V_i}{I_{\max}}\cdot\frac{1 - e^{-T_{on}/\tau}}{1 - e^{-T/\tau}} = \frac{24}{20}\cdot\frac{1 - e^{-1.2}}{1 - e^{-1.6}} = 1.2 \times \frac{0.6988}{0.7981}$$ $$\boxed{\,R_1 = 1.051\ \Omega\,}$$ Back-substitution returns exactly 20.00 A, and the ceiling $V_i/R = 22.84\ \text{A}$ sits comfortably above the 20 A limit, so the row is consistent. The corresponding inductance is $L = R\tau = 1.58\ \text{mH}$.
3. Describe the resulting waveform. The duty ratio is $\delta = 1.8/2.4 = 0.75$, the minimum current is $I_{\min} = I_{\max}e^{-T_{off}/\tau} = 20e^{-0.4} = 13.41\ \text{A}$, and the mean current is $\delta V_i/R = 0.75(22.84) = 17.13\ \text{A}$ — which lies between $I_{\min}$ and $I_{\max}$ as it must. The ripple is 6.59 A, a third of the mean, because $\tau$ and $T$ are of the same order and the exponentials are visibly curved rather than straight.
Figure 5.1 — Two periods of the case-1 load current: exponential rise towards V/R = 22.84 A during Tₔₙ, exponential free-wheel decay towards zero during Tₒ₌₌. The vertical axis is broken to show the ripple band at scale.
4. Case 2 — screen the row before solving it. Here $R = 1.25\ \Omega$ is given, so the steady-state ceiling is $$\frac{V_i}{R} = \frac{24}{1.25} = 19.2\ \text{A}$$ which is below the required peak of 20 A. By the screen established in step 1, no chopper period can produce a 20 A peak into this load: even at $\delta \to 1$, which is continuous d.c. and not chopping at all, the current only approaches 19.2 A.
5. Confirm the infeasibility algebraically. Inverting the master relation for $T$ gives $$T = -\tau\ln\!\left[1 - \frac{V_i}{R\,I_{\max}}\left(1 - e^{-T_{on}/\tau}\right)\right] = -1.45\ln\left[1 - 0.9600(0.8089)\right] = -1.45\ln(0.2234) = 2.173\ \text{ms}$$ The bracket does lie in $(0,1)$, so the logarithm exists and a careless solution would stop here — but the answer, 2.173 ms, is less than the row's own on-time of 2.4 ms. A period shorter than its on-time is impossible, so this root must be rejected. The row is solvable only if $R < V_i/I_{\max} = 24/20 = 1.20\ \Omega$; the printed 1.25 Ω exceeds that bound.
Figure 5.2 — Peak current against chopper period for the case-2 data. The curve rises towards the asymptote V/R = 19.2 A as the period shortens towards Tₔₙ, and never reaches the required 20 A (shaded band).
6. Case 3 — solve for the on-time. With $T = 1.8$, $\tau = 1.5\ \text{ms}$ and $R = 0.9\ \Omega$ the ceiling is $V_i/R = 26.67\ \text{A} > 20\ \text{A}$, so the row passes the screen. Inverting the master relation for $T_{on}$: $$T_{on} = -\tau\ln\!\left[1 - \frac{I_{\max}R}{V_i}\left(1 - e^{-T/\tau}\right)\right] = -1.5\ln\left[1 - 0.7500(0.6988)\right] = -1.5\ln(0.4759)$$ $$\boxed{\,T_{on,3} = 1.114\ \text{ms}\,}$$ This is comfortably less than the 1.8 ms period, giving a duty ratio of $\delta = 0.619$.
7. Check case 3 the same way as case 1. Back-substitution returns 20.00 A exactly. The extremes are $I_{\min} = 20e^{-(1.8-1.114)/1.5} = 12.66\ \text{A}$ and $I_{\max} = 20\ \text{A}$, and the mean $\delta V_i/R = 0.619(26.67) = 16.50\ \text{A}$ lies between them. The ripple, 7.34 A, is larger than case 1's despite the shorter period, because the lower resistance raises the ceiling the current is chasing.

Final results — the completed table.

Case 2 requires $R < V_i/I_{\max} = 1.20\ \Omega$; at the printed $1.25\ \Omega$ the peak current cannot reach 20 A. Taking $R = 1.051\ \Omega$ instead would give $T = 3.74\ \text{ms}$.