Q3Electric Drives

Question

Q.3. A switch-mode dc-dc converter is operating at a switching frequency of 20 kHz, and the input voltage (Vd) = 150 V. The average current being drawn by the dc motor is 8.0 A. In the equivalent circuit of the dc motor, back EMF (Ea) = 100 V, armature resistance (Ra) = 0.25 Ω, and armature inductance (La) = 4 mH. (a) Plot the output current and calculate the peak-to-peak ripple, and (b) plot the current on the DC side of the converter.

Answer 1

Given: switching frequency f=20kHz (switching period T=1/f=50μs), input (DC-link) voltage Vd=150V, average motor current Iavg=8.0A, motor back-EMF Ea=100V, armature resistance Ra=0.25Ω, armature inductance La=4mH.

(a) Required Average Output Voltage and Duty Cycle

In steady-state operation, the average voltage the chopper must deliver to the motor armature must exactly balance the sum of the motor's back-EMF and the average resistive voltage drop across the armature resistance (since, in steady state, the average voltage across the armature inductance is zero, as the average rate of current change over a full cycle is zero):

V_{o, a v g} = E_{a} + I_{a v g} R_{a} = 100 + 8.0 \times 0.25 = 100 + 2 = 102 V

The required duty cycle to achieve this average output voltage from the 150V DC-link supply is:

D = \frac{V _{o, a v g}}{V _{d}} = \frac{102}{150} = 0.68

(b) Peak-to-Peak Ripple Current Calculation

For a chopper-fed R-L-E load with switching period T much shorter than the load's electrical time constant (a valid approximation here, since L/R=4mH/0.25Ω=16ms, vastly longer than the 50μs switching period), the standard peak-to-peak ripple current formula (as introduced in an earlier answer in this paper) is:

Δ I = \frac{V _{d} \times D ( 1 - D )}{L _{a} \times f}

Substituting the given values:

Δ I = \frac{150 \times 0.68 \times ( 1 - 0.68 )}{4 \times 1 0 ^{- 3} \times 20 , 000} = \frac{150 \times 0.68 \times 0.32}{80} = \frac{32.64}{80} \approx 0.408 A

Result: the required duty cycle is D=0.68 (68%), and the resulting peak-to-peak ripple current is approximately 0.408 A.

Output Current Waveform (Plot Description)

The output (armature) current waveform is a periodic, roughly triangular (piecewise-exponential, closely approximated as triangular/linear for this problem's parameters, since the switching period is very short relative to the load's electrical time constant) ripple superimposed on the average DC current of 8.0A — during the on-interval (Ton=D×T=0.68×50μs=34μs), current rises from its minimum value (approximately 8.0-0.408/2=7.796A) to its maximum value (approximately 8.0+0.408/2=8.204A); during the off-interval (Toff=(1-D)×T=0.32×50μs=16μs), current falls back down from its maximum to its minimum value, repeating this cycle at the 20kHz switching frequency.

DC-Side Converter Current

The current on the DC (input) side of the converter — the current actually drawn from the 150V DC-link supply — is not continuous like the armature current, but instead flows only during the switch's on-interval (Ton), since during the off-interval the armature current instead circulates through the freewheeling diode without drawing any current from the DC supply at all. The DC-side current therefore appears as a series of pulses, each pulse having approximately the same magnitude as the (ripple-varying) armature current during that specific on-interval, with each pulse lasting Ton=34μs and separated by Toff=16μs gaps of zero current — the average value of this pulsed DC-side current equals D×Iavg=0.68×8.0=5.44A (confirming, by power balance, that input power Vd×Iavg,DCside=150×5.44≈816W approximately matches output power delivered to the motor circuit, Vo,avg×Iavg=102×8.0=816W, as expected for an ideal, lossless converter).

Cross-Check of the Ripple Result via the Direct On/Off Interval Method

The ripple result can be independently verified without using the compact ΔI=VdD(1-D)/(Lf) formula, by directly analyzing the rate of change of current during each sub-interval from first principles. During the on-interval, the armature circuit equation is La(di/dt)=Vd-Ea-iRa; since the resistive drop iRa (at most about 8.2×0.25≈2.05V) is small compared to Vd-Ea=150-100=50V, the rate of rise of current during Ton is approximately constant:

\frac{d i}{d t}_{o n} \approx \frac{V _{d} - E _{a} - I _{a v g} R _{a}}{L _{a}} = \frac{150 - 100 - 2}{4 \times 1 0 ^{- 3}} = \frac{48}{0.004} = 12, 000 A/s

Over the on-interval duration Ton=34μs, this gives a current rise of ΔI(on)=12,000×34×10⁻⁶≈0.408A, matching the value obtained from the standard ripple formula. During the off-interval, the armature circuit equation becomes La(di/dt)=-Ea-iRa (since the freewheeling diode clamps the chopper-side terminal to approximately zero), giving a fall rate of approximately (100+2)/0.004=25,500 A/s; over Toff=16μs this gives a current fall of 25,500×16×10⁻⁶≈0.408A — the rise during Ton and fall during Toff match exactly (as they must, for the current to return to its starting value at the end of each complete switching cycle in periodic steady state), independently confirming the ΔI≈0.408A result obtained from the standard formula.

Significance of the L/R Time Constant Relative to the Switching Period

The validity of both the compact ripple formula and the simplified constant-slope (triangular-ripple) approximation used above rests on the switching period T=50μs being much shorter than the armature's electrical time constant τ=La/Ra=4×10⁻³/0.25=16ms — in this problem, T/τ=50μs/16ms≈0.3%, an extremely small ratio, so the exponential charge/discharge curves that the current actually follows during each sub-interval are so close to their initial (linear) tangent lines that the piecewise-linear (triangular ripple) approximation introduces negligible error. If, instead, the switching frequency were reduced enough that T became comparable to τ (for example, if f were only a few hundred Hz rather than 20kHz), the current would traverse a much larger fraction of its full exponential rise/decay curve within each on/off interval, the triangular approximation would become inaccurate, and the true ripple would need to be computed using the full exponential solution of the first-order RL circuit equation rather than the simplified linear-slope formula used here.

Effect of Armature Inductance Value on Ripple

Since ΔI is inversely proportional to La in the ripple formula, halving the armature inductance (to La=2mH, all else unchanged) would double the ripple to approximately 0.816A, while doubling it (to La=8mH) would halve the ripple to approximately 0.204A — this inverse relationship is why chopper-fed DC motor drives with armatures of naturally low inductance (as is common in motors designed for low electrical time constant and fast torque response) often require an additional external series filter inductor to bring the total circuit inductance up to a level that keeps ripple within an acceptable percentage of the average current, typically specified as a design target of 5-15% of full-load current depending on the application's torque-smoothness and motor heating requirements. In this problem, the computed ripple of 0.408A relative to the 8.0A average is only about 5.1% — a comparatively small, well-controlled ripple level — confirming that the given 4mH armature inductance is adequate for this specific combination of voltage, frequency, and current without necessarily requiring a supplementary filter inductor, though the actual acceptability would also depend on the specific motor's rated current ripple tolerance and commutation requirements in a real design.

Practical Filter-Inductor Sizing Considerations

When an external filter inductor must be added (because the motor's own armature inductance is insufficient to keep ripple within the desired limit), several practical factors beyond the simple ripple formula must be considered in sizing it. First, the added inductor must be rated to carry the full peak armature current (here, approximately 8.204A) continuously without saturating its magnetic core — core saturation would cause the effective inductance to collapse at high current, defeating the purpose of the filter exactly when ripple control matters most (at high load). Second, the inductor's winding resistance adds directly to the total circuit resistance, increasing I²R losses and slightly reducing overall drive efficiency, so the added inductance must be sized as the minimum value that achieves the ripple target rather than being oversized. Third, the physical size, weight, and cost of the inductor all increase with required inductance and current rating, creating a direct design trade-off against simply increasing the chopper's switching frequency instead (since ripple is inversely proportional to both L and f) — modern IGBT/MOSFET-based choppers can often operate at high enough switching frequency (tens of kHz, as in this problem's 20kHz) that the required filter inductance, and hence its size/cost/loss penalty, is kept comparatively small, which is one of the key practical advantages that motivated the shift from lower-frequency thyristor choppers to higher-frequency transistor-based choppers in modern DC drive practice.