Three-sensor alarm that sounds with all sensors open
A Plus Two team builds a three-sensor alarm using 7400 NAND gates. The buzzer must sound when at least two of A, B, C are HIGH. Each sensor is a switch to +5 V; a 10 kΩ resistor pulls the gate input down to ground when the switch is open. On the bench, with all three switches open, the buzzer sounds. When B is disconnected and left floating, the buzzer also sounds, and a voltmeter at that input reads 1.4 V. The final gate output is measured at 2.4 V when it should be LOW. The buzzer coil draws 40 mA and is connected directly between the final gate output and +5 V. The team has only 7400 and 74HC00 ICs, 10 kΩ resistors, and a 5 V supply. The 7400 datasheet gives IIL(max) = 1.6 mA, IOL(max) = 16 mA, IOH(max) = 0.4 mA, VIL(max) = 0.8 V, and VIH(min) = 2.0 V.
1. Explain why the 10 kΩ pull-down is unsuitable for the 7400. Calculate a suitable pull-down resistor.
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A 7400 TTL input sources current when it is pulled LOW; IIL(max) is 1.6 mA. The pull-down resistor must carry that current while keeping the input below VIL(max) = 0.8 V. The maximum value is R = VIL / IIL = 0.8 V / 1.6 mA = 500 Ω. A 10 kΩ resistor would need 16 V across it to sink 1.6 mA, which is impossible from a 5 V supply, so the input does not reach a valid LOW. It sits at an indeterminate voltage, measured here as 1.4 V, and the gate can interpret it as HIGH or oscillate. Use 470 Ω pull-downs with the 7400; then the LOW voltage is at most 1.6 mA × 470 Ω ≈ 0.75 V. The resistor dissipates about 5² / 470 ≈ 53 mW when the switch is closed.
2. Why is the final gate output 2.4 V instead of a valid LOW? Design a proper buzzer interface.
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The buzzer draws 40 mA, but the 7400 can sink only IOL(max) = 16 mA. Connecting the buzzer directly from the output to +5 V asks the gate to sink 40 mA. The output transistor cannot do this, so the output voltage rises to 2.4 V and the gate dissipates excessive power. Use a low-side NPN transistor such as a BC547. Connect the buzzer between +12 V or +5 V and the collector, emitter to ground, and drive the base through a resistor. For a forced beta of 10, IB = 40 mA / 10 = 4 mA; RB ≈ (5 − 0.7) / 4 mA ≈ 1.1 kΩ, so use 1 kΩ. Add a diode across the buzzer coil if it is inductive. The gate then only drives the base current, well within its ratings.
3. Would replacing the 7400 with a 74HC00 allow the 10 kΩ pull-downs to be kept? State the precautions.
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Yes. A 74HC00 is CMOS and has input leakage currents in the nanoampere range, so a 10 kΩ pull-down easily holds an unused or open input LOW. However, CMOS inputs must never be left floating, because they can drift to an intermediate voltage, cause excessive supply current and oscillate. Keep the 10 kΩ pull-downs on every used input and tie every unused input firmly to ground or +5 V. The 74HC00 output current is also limited, so the 40 mA buzzer should still be driven through a transistor rather than directly from the gate output.