A Frost Alarm That Cannot Work: Loading a Potential Divider with a Relay Coil
A student is building a frost alarm. The sensor is an NTC thermistor whose resistance is 1000 Ω at 25 °C, 700 Ω at 40 °C and 470 Ω at 46 °C. It is wired as the lower arm of a potential divider: a fixed 220 Ω resistor from the +12.0 V rail (internal resistance negligible) to the thermistor, with the lower end of the thermistor at 0 V. A 12 V relay is connected directly across the thermistor. Its coil has resistance 400 Ω, needs at least 6.0 V to stay energised and releases below that, and needs about 8.5 V to pull in from the released state. The student calculates 12 × 1000/(1000 + 220) = 9.8 V at 25 °C and concludes there is a generous safety margin. On testing, the relay releases at about 45 °C; when the room cools again the relay never re-energises, even in a cold room. The student wants release at 40 °C and asks which resistor value to use.
1. Explain the difference between the student's 9.8 V prediction and the observed behaviour, including why the relay never re-energises.
Read the worked answer
The 9.8 V figure is the unloaded divider output, but the coil is in parallel with the thermistor and loads the divider.
At 25 °C: R_p = (1000 × 400)/(1000 + 400) = 286 Ω V_coil = 12.0 × 286/(220 + 286) = 6.8 V
That is only 0.8 V above the release level, not a generous margin - the loading has removed 3 V.
Release point: V_coil = 6.0 V requires R_p = R₁ = 220 Ω (because V = 12R_p/(R₁ + R_p) = 6 ⇒ R_p = R₁). 400R_T/(400 + R_T) = 220 ⇒ 180R_T = 88 000 ⇒ R_T = 489 Ω, which from the table lies between 44 °C and 46 °C. The observed release at about 45 °C is exactly as predicted.
Why it never returns: the relay needs about 8.5 V to pull in. The largest voltage the divider can ever place on the coil occurs when the thermistor is cold and R_T → ∞: V_max = 12.0 × 400/(220 + 400) = 7.7 V
7.7 V < 8.5 V, so the pull-in condition can never be met, however cold the thermistor becomes. The relay has dropped out permanently. This is a design fault, not a faulty relay.
2. Determine whether any single value of the 220 Ω resistor could give release at 40 °C with this relay.
Read the worked answer
Two conditions must hold simultaneously.
Release at 40 °C (R_T = 700 Ω), where V_coil = 6.0 V requires R_p = R₁: R₁ = 400 × 700/1100 = 255 Ω
Pull-in at 25 °C requires V_coil ≥ 8.5 V: 12R_p/(R₁ + R_p) ≥ 8.5 ⇒ 12R_p ≥ 8.5R₁ + 8.5R_p ⇒ 3.5R_p ≥ 8.5R₁ ⇒ R_p ≥ 2.43R₁
But R_p can never exceed the coil resistance of 400 Ω, so pull-in demands R₁ ≤ 400/2.43 = 165 Ω.
The release condition demands R₁ = 255 Ω and the pull-in condition demands R₁ ≤ 165 Ω. These are incompatible: no value of R₁ can satisfy both.
The underlying reason is worth stating: the divider output must change by the relay's pull-in/release ratio, 8.5/6.0 = 1.42, whereas the sensor resistance changes only by 1000/700 = 1.43, and resistive loading compresses that further (with negligible loading the output ratio here is 7.1/6.0 = 1.18). A passive divider driving this coil is over-constrained before any resistor is chosen.
3. Propose and justify a working design, giving numbers.
Read the worked answer
The coil current must be decoupled from the sensing divider. Two workable solutions:
(a) Op-amp comparator plus transistor (recommended). The divider now drives only a high-impedance input, so the unloaded values apply: 9.84 V at 25 °C and 7.08 V at 40 °C. Set a reference of 8.0 V (a second divider or a Zener) on the other comparator input. At 25 °C, 9.84 > 8.0, the output is high and the relay energises through an npn transistor; at 40 °C, 7.08 < 8.0, the output swings low and the relay releases. A positive-feedback resistor of a few hundred kilohms gives about 0.5 V of hysteresis, preventing chatter when the thermistor sits near the threshold. A flyback diode across the coil is essential.
(b) Change the actuator. Use a solid-state relay or a low-current relay whose pull-in current is a few milliamps, so that the divider is not loaded.
Why not simply fit an op-amp without a transistor: the comparator output typically sources 20 mA or less, and a 400 Ω coil at 8.5 V needs 21 mA - marginal, and a buffer is safer.
A further refinement: the coil's copper resistance rises about 0.4% per kelvin, so a warm coil draws less current than a cold one; the threshold should be set from a warm measurement, not a cold one.