Freewheeling Diode and Brake Release Time in a 220 V DC Circuit
During commissioning of a 220 V DC station-battery panel, a holding brake is switched by an IGBT with only a freewheeling diode connected across the brake coil. Coil data: R = 44 Ω, L = 0.80 H, steady current 5 A. The mechanical release specification requires the coil current to fall below 10% of its initial value within 120 ms; otherwise the brake drags on the shaft during start-up and the drive trips on overcurrent. The IGBT is rated 600 V (Vces) and is fed from the 220 V battery through a loop with about 5 µH of stray inductance; treat the coil as an ideal inductance in series with the resistance. Stores hold one 100 Ω, 50 W wirewound resistor and one 47 Ω, 25 W wirewound resistor, and nothing else can be procured before the panel is energised in two days. The brake operates once every 20 s in service.
1. With only the freewheeling diode in place, compute the decay time constant and decide whether the 120 ms release specification is met. State clearly where the stored energy goes.
Read the worked answer
With the diode fitted, the coil current freewheels through L, R and the diode, so the loop resistance is essentially R = 44 Ω (the ~1 V diode drop is negligible):
τ = L/R = 0.80/44 = 18.2 ms
i(t) = 5·e^(−t/18.2 ms), so i = 0.5 A (10%) at t = τ·ln 10 = 2.303 × 18.2 = 41.9 ms.
The 120 ms requirement is met with roughly 78 ms of margin, so the plain diode is sufficient and no resistor is required. Stored energy is ½LI² = ½ × 0.8 × 25 = 10 J, dissipated almost entirely in the coil resistance (~9.9 J) with about 0.1 J in the diode, spread over roughly 100 ms. At one operation per 20 s the average dissipation is only 0.5 W, which is thermally trivial.
The diode must block 220 V with margin (a 600 V part) and carry a 5 A forward surge; its average current is negligible because it conducts for only ~42 ms per 20 s.
2. The mechanical engineer asks for faster release and proposes putting one of the spare resistors in series with the diode. Evaluate both resistors for peak voltage across the IGBT, new decay time and acceptability. What do you recommend?
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A series resistor R_a raises the turn-off peak to approximately I₀(R + R_a):
- 47 Ω: peak = 5 × (44 + 47) = 455 V, a 24% margin to the 600 V rating. New τ = 0.8/91 = 8.8 ms, reaching 10% in 2.303 × 8.8 = 20 ms.
- 100 Ω: peak = 5 × (44 + 100) = 720 V, which exceeds the IGBT's 600 V. The device would go into avalanche, and the coil insulation and resistor would also see 720 V. Reject.
Recommendation: since the 120 ms specification is already met at 41.9 ms, fitting the 47 Ω resistor buys response time the machine does not need while eroding voltage margin by 24% and adding a 455 V stress on the coil. Keep the diode alone. If faster release is genuinely wanted later, the 47 Ω is the only feasible spare, and its pulse-voltage rating and the coil's insulation level must be verified before fitting.
3. The specification is later tightened to 15 ms. Show whether any resistor combination can meet it within the 600 V IGBT rating, and propose a workable alternative.
Read the worked answer
15 ms requires τ = 15/2.303 = 6.5 ms, i.e. R_total = 0.8/0.0065 = 123 Ω, so R_a ≈ 79 Ω. The peak would then be 5 × 123 = 615 V, just above the 600 V IGBT rating and outside any sensible margin, and the coil would see the same stress. A simple resistor therefore cannot meet 15 ms within this switch rating.
A voltage clamp is the correct tool: with a TVS/zener clamp at, say, 400 V, the decay is almost linear with t = LI/V = 0.8 × 5/400 = 10 ms, comfortably inside 15 ms. The clamp absorbs essentially the whole 10 J each operation and must be selected for that pulse energy (a 400–440 V, 1500 W-class transient suppressor), with clamping voltage kept below about 500 V and leakage at 220 V checked.
The 5 µH stray inductance contributes only L·di/dt ≈ 5 µH × (5 A/10 ms) ≈ 2.5 mV here, so it is not a constraint. Add coil-current monitoring so a failed-open diode is annunciated rather than discovered when the IGBT fails.