22-Elec-B8 Power Electronics and Drives · December 2014

Question 4 of 6: Basic DC Chopper — Ripple Ratio and Time-Domain Currents

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4: Basic DC Chopper — Ripple Ratio and Time-Domain Currents (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.

Part (a) — Why series smoothing reactors are used

A series reactor placed in the dc link of an inverter, or in series with the switching devices, is a deliberately slow element inserted into an otherwise very fast circuit, and it earns its place for five distinct reasons.

The first is device protection against $di/dt$. A thyristor turns on over a small initial area of its cathode and the conducting region then spreads laterally at a few tenths of a millimetre per microsecond; if the external circuit forces current to rise faster than the plasma can spread, the local current density destroys the device. Every thyristor therefore carries a $di/dt$ rating, and a series inductance of a few microhenries is the standard way of guaranteeing it, since $di/dt = v/L$ at turn-on.

The second is to stiffen the dc source. In a current-source inverter the link reactor is not a refinement but the defining component: it makes the link current sensibly constant over a switching period so that the inverter delivers rectangular current blocks to the machine, independent of the machine's counter-EMF. Even in a voltage-source inverter, a modest link reactor limits the ripple current returned to the rectifier and to the supply, reducing capacitor ripple heating and extending life.

The third is fault-current limiting. A commutation failure or a shoot-through across a bridge leg presents an almost unlimited rate of rise of current; the series reactor slows that rise enough for a fuse or an electronic trip to act before the devices reach their $I^2t$ limit. The fourth is commutation itself: in forced-commutated circuits, series and commutating inductance supply the energy that reverse-biases the outgoing device and holds it off for its full turn-off interval, so the reactor is part of the commutation loop, not merely a filter. The fifth is harmonic and interference reduction — the reactor presents an impedance rising with frequency, attenuating the switching-frequency currents that would otherwise be conducted into the supply and radiated from the wiring, and decoupling the inverter's fast transients from neighbouring equipment.

The cost of all this is stored energy that must go somewhere at turn-off, which is why series reactors and snubber networks are designed together, and a slower response to genuine demands for a change in current, which is why the reactor is sized to the smallest value that meets the $di/dt$ and ripple requirements.

Parts (b), (c) and (d) — Chopper analysis

Given. A basic chopper works from a 30 V source into a series $R$–$L$ load at a 3.0 ms period, and the ratio of minimum to maximum output current is held at 0.8.

Find. The load time constant and the on-time, the maximum and minimum load currents, and the time-domain expression for the current in each interval together with its value at $t = 1$ ms and $t = 2.8$ ms.

Approach. Exploit the fact that the free-wheeling decay contains no source, so the ratio of the two current limits fixes the off-time by itself; the on-time then follows from the period, the steady-state limits from the periodic boundary condition, and the sampled values from whichever interval each instant falls in.

1. Part (b) — time constant. The load is a plain series $R$–$L$ combination, so $$\tau = \frac{L}{R} = \frac{0.5 \times 10^{-3}}{0.25} = \boxed{2.0\ \text{ms}}$$ This is comparable with the 3.0 ms period, which warns immediately that the current will not be smooth and that no linear-ripple approximation will be safe.
2. Part (b) — the off-time follows from the ratio alone. When the switch opens the current free-wheels through the diode with no source in the loop, so it decays purely exponentially from $I_{max}$: $$i(t') = I_{max}e^{-t'/\tau}$$ Evaluating at $t' = t_{off}$ and dividing by $I_{max}$ removes the current magnitudes entirely: $$\frac{I_{min}}{I_{max}} = e^{-t_{off}/\tau} = 0.8$$
3. Part (b) — solve for the two intervals. Taking logarithms, $$t_{off} = -\tau\ln 0.8 = 2.0 \times 0.22314 = 0.4463\ \text{ms}$$ and therefore $$t_{on} = T - t_{off} = 3.0 - 0.4463 = \boxed{2.5537\ \text{ms}}$$ corresponding to a duty ratio $\delta = t_{on}/T = 0.851$. The ripple specification alone has fixed the timing, before any current level is known.
4. Part (c) — impose the periodic boundary condition. During the on-time the current climbs from $I_{min}$ towards $V_i/R$: $$i(t) = \frac{V_i}{R}\left(1 - e^{-t/\tau}\right) + I_{min}e^{-t/\tau}$$ In the steady state $i(t_{on}) = I_{max}$ and $I_{min} = I_{max}e^{-t_{off}/\tau}$; eliminating $I_{min}$ and using $e^{-t_{on}/\tau}e^{-t_{off}/\tau} = e^{-T/\tau}$ gives $$I_{max} = \frac{V_i}{R}\cdot\frac{1 - e^{-t_{on}/\tau}}{1 - e^{-T/\tau}}$$
5. Part (c) — substitute. With $e^{-t_{on}/\tau} = e^{-1.27686} = 0.27892$ and $e^{-T/\tau} = e^{-1.5} = 0.22313$, and $V_i/R = 120$ A, $$I_{max} = 120 \times \frac{0.72108}{0.77687} = \boxed{111.38\ \text{A}}$$ $$I_{min} = 0.8\,I_{max} = \boxed{89.11\ \text{A}}$$ The peak-to-peak ripple is 22.28 A about a mean of roughly 100 A.
6. Part (d) — write the two branches of the waveform. Substituting the numbers into the two interval solutions, with $t$ measured in milliseconds from the start of each interval, $$i_{on}(t) = 120 - 30.89\,e^{-t/2.0}, \qquad 0 \le t \le 2.5537\ \text{ms}$$ $$i_{off}(t') = 111.38\,e^{-t'/2.0}, \qquad 0 \le t' \le 0.4463\ \text{ms}$$ where $t' = t - t_{on}$ is measured from the instant of turn-off, not from the start of the period.
7. Part (d) — sample at $t = 1$ ms. Since $1 < t_{on}$, this instant lies in the rising interval: $$i(1\ \text{ms}) = 120 - 30.89\,e^{-0.5} = 120 - 18.74 = \boxed{101.26\ \text{A}}$$
8. Part (d) — sample at $t = 2.8$ ms. Here $2.8 > t_{on} = 2.5537$ ms, so the switch has already opened and the current is free-wheeling. Measuring from turn-off, $t' = 2.8 - 2.5537 = 0.2463$ ms, and $$i(2.8\ \text{ms}) = 111.38\,e^{-0.2463/2.0} = 111.38 \times 0.88414 = \boxed{98.48\ \text{A}}$$ Both sampled values lie between $I_{min}$ and $I_{max}$, as they must.
9. Check that the exponential treatment was necessary. The linearised ripple estimate $\Delta I \approx (V_i - I_{avg}R)t_{on}/L$ gives $(30 - 25.06)(2.5537 \times 10^{-3})/(0.5 \times 10^{-3}) = 25.3$ A against the true 22.28 A — 13 per cent high, because $\tau$ is only two thirds of the period and the current visibly curves within each interval. At a switching frequency ten times higher the two would agree closely.