NAND Gate Bistable with Thermistor Input

A NAND gate bistable latch with thermistor input is a simple circuit found in many home appliances where there is a need to control logic function depending upon temperature. This type of memory could be useful to use in a kettle. For example, when the user presses the SET switch, the output Q=1, which could control the heating element to heat water. When the thermistor heats up, its resistance drops, hence if it were to be connected to a potential divider circuit, it could be used for providing the RESET signal, which would make Q=0, thereby cutting the power to the heating element.

Reading Logarithmic Scales

When you look at the documentation of a thermistor component, there is usually a graph showing resistance of the component in the vertical axis, and the temperature in the horizontal axis. They usually give resistance in logarithmic scale, which is why the intervals are not evenly spaced.

Each section is 10 times greater than the previous, hence you see, 10 to the power 3, 10 to the power 4, 10 to the power 5... However, you need to know how the uneven spaced lines between divide out! To make it simple to understand, here are the numbers for the first three sections. It is actually quite simple when you think about it.

1. 1000, 2000, 3000, 4000, 5000, 6000, 7000, 8000, 9000, 10,000
2. 10,000 20,000 30,000 40,000 50,000 60,000 70,000, 80,000 90,000 100,000
3. 100,000 200,000, 300,000 400,000 500,000 600,000 700,000 800,000 900,000 1000,000

Let us say that we looked at such a graph and determined that a typical thermistor has a resistance of 20 kΩ when the temperature is 50 °C, which is exactly how hot I like my Earl Grey tea!

Thermistor Operation

When the temperature is greater than 50 °C the RESET input becomes low and therefore this initiates the RESET function making Q=0. When Q=0, the power to the heating element is cut.

For NAND gates, the threshold for logic 1 is 2.5 V or more, and the threshold for logic 0 is 2.5 V or less. Nevertheless, we need to set the potential divider network to this threshold. Since the resistance of the thermistor will be 20 kΩ at the desired temperature of 50 °C, the resistor R also has to be 20 kΩ, that way the midpoint will be at 2.5 V when the thermistor is 50 °C and the flip-flop will change state.

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