Missing Freewheeling Diode on a Fuel Shut-Off Valve Relay
A 28 V DC fuel shut-off valve is energised by a relay whose coil measures 220 Ω and 120 mH. The coil is switched on the low side by an NPN transistor, and a suppression diode is fitted at the relay socket with its cathode to the coil positive terminal. The valve has been stuck energised; the transistor is found failed short circuit between collector and emitter. Tested in circuit on diode range, the suppression diode reads open in both directions, and the socket shows blackening around its pins. The coil itself measures 219 Ω and the supply is 27.8 V. Certification requires the valve to be de-energised within 300 ms of switch release. The technician proposes fitting a transistor with a higher Vceo rating and leaving the diode out, arguing that a clamp diode slows relay release and that similar aircraft circuits have run for years without one.
1. Calculate the steady-state coil current and the stored magnetic energy, and explain why the missing diode destroys the switching transistor.
Read the worked answer
Current: I = V/R = 27.8 / 219 = 0.127 A (about 127 mA). Energy: W = ½LI² = 0.5 × 0.120 × 0.127² = 0.97 mJ.
The 0.97 mJ is small, which is exactly why the 'it is only a small relay' argument is misleading: the damage is done by voltage, not by energy. Inductor current cannot change instantaneously, so v = −L di/dt. If the transistor attempts to interrupt 127 mA in 1 µs, the ideal induced emf is 0.120 × 127/10⁻⁶ ≈ 15 kV; stray winding capacitance and breakdown limit it in practice to a few hundred volts - still far above the 40–60 V Vceo of a small NPN. The junction avalanches, and a short-circuit failure between collector and emitter is the classic outcome, matching the finding.
With the diode fitted, the coil current freewheels through the diode and the collector is clamped to roughly one diode drop above the 28 V rail, about 28.7 V, within the transistor's rating. Fitting a higher-voltage transistor without the diode only postpones the failure: the energy is still dumped in the device every cycle, the socket keeps arcing, and EMI increases.
2. The diode reads open in both directions in circuit. Is that conclusive evidence that the diode has failed? State how you would prove it and what ratings and orientation you would check on replacement.
Read the worked answer
No. A multimeter on diode range applies only about 1–2 V through a limited current; with the 220 Ω coil in parallel the current is diverted and the forward drop across the junction may never be developed, so a healthy diode can read open in both directions. The blackened, corroded socket is a second confounder: poor probe contact alone produces an open reading.
Proof: isolate the circuit and lift one end of the diode (or unplug it), then re-test. A good silicon diode gives roughly 0.5–0.7 V forward and infinite reverse on the meter; a shorted diode reads near zero both ways, an open one reads open both ways. Compare against the second half of the relay coil if it is a dual-diode package.
Ratings to verify before fitting: reverse repetitive voltage well above the 28 V rail (100 V is typical practice), forward current at least the 127 mA coil current with margin, and correct orientation - cathode to the coil terminal fed from +28 V, so the diode is reverse-biased in normal operation and becomes forward-biased only at switch-off. Renew the socket and pins as well; blackening means arcing has already eroded the contact surfaces and a new diode in a damaged socket will fail again.
3. Evaluate numerically the technician's claim that a clamp diode delays valve release beyond the 300 ms requirement, and describe the release-time verification.
Read the worked answer
With the diode fitted, the coil current decays through the coil resistance and the forward-biased diode: τ = L/R = 0.120 / 220 = 0.55 ms. A relay typically drops out at 20–50% of rated coil current, so the electrical delay is t = τ ln(I₀/I_t) = 0.55 ms × ln(1/0.3) ≈ 0.66 ms, i.e. 1–3 ms including a couple of extra time constants.
300 ms is over two orders of magnitude larger. Even adding the valve's mechanical actuation time (tens of milliseconds at most), the diode's effect on release is negligible. The claim is quantitatively wrong.
Verification: with the relay and valve installed, use a current probe and oscilloscope, or the manufacturer's relay test set, to measure from switch release to coil current falling below the drop-out threshold, and confirm the valve actually closes (position indication, flow or pressure switch) within 300 ms. Repeat over several cycles and at minimum supply voltage (the worst case for drop-out), and record the result. Restore the suppression diode as part of the same work; the transistor failure is a consequence of its absence, not a coincidence.