Relay driver that chatters and overheats
A Plus Two project uses a BC547 NPN transistor to switch a 12 V relay from a 5 V logic output. The relay coil measures 400 Ω. The base is driven through a 22 kΩ resistor; no base-emitter resistor is fitted. The relay should energise when the output is 5.0 V. On the bench the relay sometimes chatters, and the transistor becomes too hot to touch after a minute. With the output high, measurements are VBE = 0.72 V, VCE = 4.3 V and coil voltage = 7.7 V. The 12 V supply holds steady at 12.0 V. The BC547 datasheet gives hFE(min) = 110 at IC = 30 mA and VCE(sat) = 0.25 V at IB = 3 mA. No freewheeling diode is fitted across the coil. The student proposes replacing the BC547 with a higher-current transistor and keeping the same 22 kΩ base resistor.
1. Calculate the base current and the coil current. Is the transistor in saturation? Explain the measured VCE and the heating.
Read the worked answer
Coil current with the transistor partly on is IC = 7.7 V / 400 Ω = 19.25 mA. Base current is IB = (5.0 − 0.72) V / 22 kΩ = 0.195 mA. The forced beta is therefore IC / IB = 19.25 / 0.195 ≈ 99. At saturation the coil current would be 12 V / 400 Ω = 30 mA. The datasheet hFE(min) is 110, so the base current is below the value needed to guarantee saturation; the transistor is operating in the active region, not as a switch. The load line explains the readings: 12 V − 7.7 V = 4.3 V across the transistor, and 4.3 V × 19.25 mA ≈ 83 mW is dissipated in a small plastic transistor, enough to make it hot. The relay chatters because the coil voltage is only 7.7 V, not the full 12 V.
2. Select a base resistor that will guarantee saturation with an overdrive factor of about 10. Show the calculation.
Read the worked answer
For saturation, design for IC = 30 mA and a forced beta of about 10, so IB(required) ≈ 30 mA / 10 = 3 mA. Using VBE ≈ 0.8 V, RB = (5.0 − 0.8) V / 3 mA = 1.4 kΩ. A standard 1.2 kΩ resistor gives IB = 4.2 V / 1.2 kΩ = 3.5 mA, a forced beta of 30 / 3.5 ≈ 8.6, which is safely below hFE(min) = 110. Check base resistor power: (4.2 V)² / 1.2 kΩ ≈ 15 mW, so a 0.25 W resistor is fine. The higher-current transistor alone would not fix the problem because the existing 22 kΩ resistor still supplies too little base current.
3. Explain what happens when the relay turns off without a freewheeling diode. Does the proposed transistor replacement solve this?
Read the worked answer
The relay coil is inductive. When the transistor turns off, the coil current cannot change instantly, so the coil generates a large back-EMF, V = L di/dt. This adds to the 12 V supply and appears as a high-voltage spike at the collector. The BC547 has a limited VCEO, typically 45 V, so the spike can drive it into avalanche breakdown and damage it. A 1N4007 diode connected across the coil, reverse-biased during normal operation, gives the coil current a safe path and clamps the spike to about 0.7 V above the supply. A transistor with a higher current rating does not remove the spike and does not correct the insufficient base drive; it may survive longer, but the circuit would still be unreliable.