Question 2 of 6: Basic Chopper — Critical On-Time and Peak Load Current

Problem 2: Basic Chopper — Critical On-Time and Peak Load Current (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.

Find. (a) how a chopper controls the mean load voltage and what changing t_on does to its operating mode; (b) the on-time at which the current just reaches zero at the end of each cycle; (c) the peak current in that boundary condition; (d) the peak current when t_on = 0.5T.

Approach. Answer (a) in prose, then apply the standard steady-state chopper current expressions — exponential rise toward (V − E_c)/R while the switch is on and exponential decay toward −E_c/R through the free-wheeling diode while it is off — solving the boundary condition I_min = 0 for the critical on-time and then evaluating I_max at both duty ratios.

Part (a) — principle of operation. A basic chopper is a d.c.-to-d.c. switch-mode converter: a fully controllable switch (a GTO, IGBT or force-commutated thyristor) is placed in series between a fixed d.c. source and the load, and a free-wheeling diode is connected across the load. The switch is closed for a time t_on and opened for t_off = T − t_on. While it is closed the full supply is impressed on the load and the current rises exponentially toward (V − E_c)/R with time constant τ = L/R; while it is open the inductance forces the current to continue through the diode, the load terminals are clamped near zero volts, and the current decays toward −E_c/R. The load therefore sees a rectangular voltage whose average value is V_o = δV with duty ratio δ = t_on/T, and the output is controlled smoothly from zero to V by varying δ alone — usually at constant frequency (time-ratio control), occasionally by varying T at fixed t_on (frequency modulation) or by current-limit control.

Varying the on-time changes not only the mean voltage but the mode of operation. When t_on is large enough that the current never falls to zero within a period, the chopper runs in continuous conduction: the current ripples between a non-zero I_min and I_max, the transfer characteristic is exactly linear (V_o = δV), and the load sees the smallest ripple. As t_on is reduced the ripple envelope drops until I_min just touches zero — the critical or boundary condition of part (b). Below that on-time the current extinguishes before the switch closes again and the chopper enters discontinuous conduction: the load terminals float up to E_c during the dead interval, so the mean voltage is no longer δV but is higher and depends on the load as well as the duty ratio, the transfer characteristic becomes non-linear, and closed-loop current control becomes markedly harder. Existence of the back e.m.f. is what creates the discontinuous region at all: with E_c = 0 the current would decay toward zero only asymptotically and continuous conduction would persist to very small duty ratios.

1. Part (b) — write the steady-state ripple limits. Matching the exponential rise and decay over one period gives the classical pair $$I_{max} = \frac{V}{R}\,\frac{1-e^{-t_{on}/\tau}}{1-e^{-T/\tau}} - \frac{E_c}{R}, \qquad I_{min} = \frac{V}{R}\,\frac{e^{t_{on}/\tau}-1}{e^{T/\tau}-1} - \frac{E_c}{R},$$ with $\tau = L/R = 12\times10^{-3}/9.6 = 1.25\ \text{ms}$ and $T/\tau = 0.25/1.25 = 0.2$.
2. Part (b) — impose I_min = 0 and solve for the critical on-time. Setting the second expression to zero, $$e^{t_{on}/\tau} - 1 = \frac{E_c}{V}\left(e^{T/\tau}-1\right) = \frac{20}{220}\,(1.22140 - 1) = 0.020128 .$$ Taking logarithms, $t_{on}/\tau = \ln(1.020128) = 0.019928$, so $$\boxed{t_{on,crit} = 0.019928 \times 1.25\ \text{ms} = 24.91\ \mu\text{s}}$$ corresponding to a duty ratio $\delta = 24.91/250 = 0.0996$, i.e. just under 10 per cent.
3. Part (c) — peak current at the boundary. Substituting the same on-time into the $I_{max}$ expression, with $e^{-0.019928} = 0.980270$ and $e^{-0.2} = 0.818731$: $$I_{max} = \frac{220}{9.6}\times\frac{0.019730}{0.181269} - \frac{20}{9.6} = 22.9167 \times 0.108842 - 2.0833 ,$$ $$\boxed{I_{max} = 0.411\ \text{A}}$$ The mean load current at this boundary is $(\delta V - E_c)/R = (0.0996\times220 - 20)/9.6 = 0.200\ \text{A}$, which sits sensibly between the limits 0 and 0.411 A and confirms the arithmetic.
4. Part (d) — peak current at 50 per cent duty ratio. Now $t_{on} = 0.5T = 0.125\ \text{ms}$, so $t_{on}/\tau = 0.1$ and $e^{-0.1} = 0.904837$: $$I_{max} = 22.9167\times\frac{0.095163}{0.181269} - 2.0833 = 22.9167\times0.524939 - 2.0833,$$ $$\boxed{I_{max} = 9.947\ \text{A}}$$ The companion minimum is $I_{min} = 22.9167\,(0.105171/0.221403) - 2.0833 = 8.803\ \text{A}$, so the operation is firmly continuous and the peak-to-peak ripple is only 1.145 A. The mean of the two limits, 9.375 A, agrees exactly with $(0.5\times220 - 20)/9.6 = 9.375\ \text{A}$.