I want to control a small 12V case fan. I will set values of R_{1}, R_{2} and R_{3} so that the fan will work above temperatures 40^{o}C.

I understand that in these kind of systems, there will be an indecisive region in which the comparator output will be fast changing between high and low. In this practical case, when the temperature is in the vicinity of 40^{o}C, there will be an unstable behavior.

Is there any way of make this circuit work in schmitt trigger mode (e.g.; stop under 38^{o}C, start above 42^{o}C, and keep previous state between 38^{o}C and 42^{o}C) by changing it as little as possible, and without using any schmitt trigger logic gate.

## Best Answer

To create a Schmitt-trigger you have to supply positive feedback, from the opamp's output to the non-inverting input. Usually this input will be the threshold voltage, and it will take one of two values (that's the hysteresis) depending on the opamp's output.

In your case you have the signal on the non-inverting input. You can also make it work this way, but I would suggest you switch both inputs, and also swap R1 and PTC still have the same behaviour: a higher PTC resistance will decrease the inverting input, and when it reaches the threshold the fan will be switched on. So let's do that, and add an R5 from output to the R2/R3 node.

You mention the hysteresis in °C, but we need the voltages. Let's do a theoretical calculation with a \$V_H\$ and \$V_L\$ as thresholds, and assume a rail-to-rail output opamp. Then we have two situations: the high and the low threshold, and three variables: R2, R3 and the added R5. So we can choose one of the resistors, let's fix R2.

Now, applying KCL (Kirchhoff's Current Law) for the R2/R3/R5 node:

\$ \dfrac{12 V - V_L}{R3} + \dfrac{0 V - V_L}{R5} = \dfrac{V_L}{R2} \$

and

\$ \dfrac{12 V - V_H}{R3} + \dfrac{12 V - V_H}{R5} = \dfrac{V_H}{R2} \$

This is a set of linear equations in two variables: R3 and R5, which is easy to solve if you can fill in actual voltages for \$V_H\$ and \$V_L\$ and a freely chosen R2.

Let's for the sake of argument suppose that at 38 °C you have 6 V on the inverting input, and at 42 °C you'll have 5 V. Let's pick a 10 k\$\Omega\$ value for R2. Then the above equations become

\$ \begin{cases} \dfrac{12 V - 5 V}{R3} + \dfrac{0 V - 5 V}{R5} = \dfrac{5 V}{10 k\Omega} \\ \\ \\ \dfrac{12 V - 6 V}{R3} + \dfrac{12 V - 6 V}{R5} = \dfrac{6 V}{10 k\Omega} \end{cases} \$

or

\$ \begin{cases} \dfrac{7 V}{R3} - \dfrac{5 V}{R5} = \dfrac{5 V}{10 k\Omega} \\ \\ \\ \dfrac{6 V}{R3} + \dfrac{6 V}{R5} = \dfrac{6 V}{10 k\Omega} \end{cases} \$

then after some replacing and shuffling we find

\$ \begin{cases} R3 = 12 k\Omega \\ R5 = 60 k\Omega \end{cases} \$

I already said it's less common, but you can also use the current schematic, and the calculations are similar. Again, add an R5 feedback resistor between output and non-inverting input. Now the reference input is fixed by the ratio R2/R3, and the hysteresis will shift your measured voltage up and down, which — at least for me — needs some getting used to.

Let's suppose we fix the reference voltage at 6 V by making R2 and R3 equal. Again we calculate the currents at the node PTC/R1/R5, where PTC\$_L\$ and PTC\$_H\$ are the PTC values at 38 °C and 42 °C resp., and R1 and R5 are our unknowns. Then

\$ \begin{cases} \dfrac{6 V}{PTC_H} = \dfrac{12 V - 6 V}{R1} + \dfrac{0 V - 6 V}{R5} \\ \\ \\ \dfrac{6 V}{PTC_L} = \dfrac{12 V - 6 V}{R1} + \dfrac{12 V - 6 V}{R5} \end{cases} \$

Again, solve for R1 and R5.