Discharge time for a 300 V reservoir
An avionics power supply uses a 470 µF electrolytic reservoir capacitor charged to 300 V DC. Before the unit may be opened, the capacitor must be discharged below 50 V within 60 s of power removal, and must stay below that level while the unit is on the bench. The workshop stocks 2 W and 5 W resistors in 22 kΩ, 47 kΩ, 68 kΩ, 100 kΩ, 220 kΩ and 470 kΩ. A colleague proposes the 470 kΩ part 'because it wastes the least power'. The capacitor is known good, the rail is exactly 300 V, and after switch-off the only path across the capacitor is the bleed resistor that is fitted.
1. Select a bleed resistor that meets the 60 s requirement and specify its power rating, showing the calculation.
Read the worked answer
Discharge follows V = V₀·e^(−t/RC), with τ = RC. For V ≤ 50 V at t = 60 s: e^(−60/τ) ≤ 50/300 = 0.1667 → 60/τ ≥ 1.7918 → τ ≤ 33.5 s → R ≤ 33.5/470 µF = 71.2 kΩ.
Check the candidates (τ = R × 470 µF):
| R | τ | V at 60 s | Dissipation at 300 V |
|---|---|---|---|
| 470 kΩ | 221 s | 300·e^(−0.272) = 228 V | 0.19 W |
| 100 kΩ | 47 s | 300·e^(−1.277) = 83.7 V | 0.90 W |
| 68 kΩ | 32.0 s | 300·e^(−1.877) = 45.9 V ✔ | 1.32 W |
| 47 kΩ | 22.1 s | 300·e^(−2.716) = 19.8 V ✔ | 1.92 W |
68 kΩ is the smallest value that just satisfies the limit, giving a sensible margin (45.9 V at 60 s, 17.0 V at 120 s). Power = 300²/68 000 = 1.32 W, so a 5 W part is required; a 2 W part would be running at 66 % of rating at full voltage and is not acceptable for continuous duty. Stored energy is ½CV² = 21 J, which is potentially lethal - the resistor is a backup, not a substitute for proving dead with a meter.
2. Show quantitatively why the colleague's 470 kΩ proposal is unacceptable.
Read the worked answer
τ = 470 kΩ × 470 µF = 221 s.
At 60 s: V = 300·e^(−60/221) = 300 × 0.762 = 229 V - nearly 80 % of full charge, and far above the 50 V limit.
To reach 50 V would take t = τ·ln(300/50) = 221 × 1.792 = 396 s, about 6.6 minutes. During that time the unit is open with a lethal voltage present on a component that looks de-energised.
The 0.19 W saving is trivial; the safety case dominates. Judgement point: the bleed resistor is a safety device sized by a time criterion, not by a power-loss criterion. Also note that a failed-open bleed resistor gives no indication - the discharge time should be verified by measurement on first build and after any resistor replacement.
3. The 68 kΩ resistor will be permanently across the 300 V rail. Comment on this and suggest an alternative.
Read the worked answer
Permanent dissipation = 1.32 W, current = 300/68 000 = 4.4 mA. On a low-power avionics rail this is a significant continuous waste and a local hot spot on the PCB; it also derates the 5 W resistor to about 26 % of rating, which is thermally acceptable but adds to enclosure temperature rise.
Alternatives:
- Switched bleeder: a relay or transistor connects the resistor only when the supply is off (or when an access panel is opened), giving fast discharge with near-zero standing loss.
- Indicator + bleeder combined: an LED or neon plus resistor that shows when the rail is live.
- Two-stage bleeder: a low-value resistor for the first few seconds followed by a high-value one.
Judgement point: the design must satisfy both the 60 s discharge limit and the standing-power budget; the cheapest solution (one fixed resistor) is often acceptable but should be a conscious decision, supported by a dissipation and temperature-rise check, not an accident.