Relay Driver Transistor Operating in Active Region
An assembly line reports intermittent relay operation. A 5 V CMOS gate drives an MMBT3904 NPN through a 10 kΩ base resistor. The transistor switches a 12 V relay whose coil resistance is 160 Ω, drawing 75 mA at 12 V. The relay contacts chatter, and the SOT-23 transistor runs hot enough to discolor the board. The transistor datasheet gives hFE(min) = 100 at 10 mA, 60 at 100 mA, VCE(sat) = 0.3 V at IC = 75 mA and IB = 7.5 mA, VCEO = 40 V, and PD = 350 mW at 25 °C. The CMOS gate can source 8 mA while maintaining VOH = 4.6 V. There is no diode across the coil. The technician measures 4.9 V on the gate output and 8.5 V across the relay coil when energized. The relay must pull in reliably at 9 V minimum.
1. Using the measured voltages, show quantitatively why the relay chatters and the transistor overheats.
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Base current: IB = (VOH − VBE)/RB = (4.9 − 0.75)/10k ≈ 0.415 mA. The relay needs 75 mA, so the forced beta is 75/0.415 = 181. The transistor's guaranteed hFE at 75 mA is between 60 and 100, so 181 is far above the guaranteed gain and the device cannot saturate. It operates in the active region with VCE = 12 − 8.5 = 3.5 V. Dissipation = 3.5 × 0.075 = 0.26 W, plus base drive, in a SOT-23 rated 350 mW at 25 °C. Self-heating and ambient rise push the junction near or beyond its rating, explaining the discoloration. The coil receives only 8.5 V, below the 9 V pull-in minimum, so the relay chatters. The 10 kΩ resistor is sized for a logic signal, not a 75 mA load.
2. Select a new base resistor that guarantees saturation with a forced beta of 10, and verify it against the gate's 8 mA source capability.
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Target IB = IC/10 = 75/10 = 7.5 mA. With VOH = 4.6 V at 8 mA and VBE ≈ 0.8 V at 7.5 mA, RB = (4.6 − 0.8)/7.5 mA = 507 Ω. Use 510 Ω, giving IB = 3.8/510 = 7.45 mA and forced beta = 75/7.45 = 10.1. The gate must source 7.45 mA, which is within its 8 mA capability, but verify VOH at that current in the gate datasheet because the 4.6 V value is only guaranteed at 8 mA. Check VCE(sat): at 75 mA and 7.5 mA base drive it is 0.3 V, so the coil sees 11.7 V and the transistor dissipates only 22 mW. If the gate cannot supply 7.5 mA in practice, use a small logic-level MOSFET or an intermediate buffer.
3. Explain the consequence of the missing coil diode and specify its correct orientation.
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The relay coil is inductive. When the transistor turns off, coil current cannot change instantly, and L·di/dt produces a large positive voltage at the collector, often tens to hundreds of volts depending on stray capacitance. The MMBT3904 has VCEO = 40 V, so the spike can avalanche the transistor and destroy it; even if it survives, the spike can couple into nearby logic and cause erratic operation. Add a flyback diode such as a 1N4148 or 1N4007 across the coil, with the cathode to +12 V and the anode to the collector. During turn-off the diode clamps the collector to about 12.7 V and circulates the coil current. Do not reverse it: cathode to ground would short the 12 V supply through the diode when the transistor is on. The diode slightly slows relay release; if fast release is required, use a zener plus diode.