You need to implement a comparator circuit with hysteresis, and you pull up some app notes. They look something like this, credit to Analog Devices:

If you have a circuit in hand, there are a dozen or so discussions online to determine the comparator’s characteristics: high threshold `$V_{TH}$`

, low threshold `$V_{TL}$`

, and hysteresis `$V_{HYST}$`

. We asume that `$R_{PULL-UP}$`

is reasonably small compared to the feedback network so that `$V_{OH}$`

is fixed. These equations are pretty easy to derive yourself from basic circuit principles, but it’s a time saver to have them available. Now, if you want to design your comparator circuit, and already know it’s characteristics, you have to either guess and check values, implement a solver, or do some basic algebra to solve the equations to get the resistor values in terms of the comparator’s characteristics. Every time I do this, I lose my notes, so this time, they’re going online.

Assumptions

Going into these solutions, we’re going to assume that the circuit designer knows the following:

- \(V_{TH}\)
- \(V_{TL}\)
- \(V_{HYST}\) – the difference between
`$V_{TH}$`

and`$V_{TL}$`

`$V_{OH}$`

– the voltage at pin 2 when the output is high. For comparators with an open-comparator output and a small`$R_{PULL-UP}$`

with respect to the feedback network, this will be`$V_{PULL-UP}$`

`$V_{OL}$`

– the voltage at pin 2 when the output is low. For comparators with an open-ground output, this will be 0 or near 0. I’m going to assume that`$V_{PULL-UP}$`

is tied to`$V_{CC}$`

.

Also, you will be picking some resistor values to make the equations solvable. They are somewhat arbitrary, but should be chosen as a trade-off between wasting current making the feed-back network to small with respect to `$V_{PULL-UP}$`

. If you are designing on a PCB, choosing resistors that are too large may lead to PCB leakage having a significant effect on accuracy.

Single-supply Noninverting Hysteresis

This is the easiest of the two comparator types because the hysteresis equation is dependent only on `$R_3$`

and `$R_4$`

. Choosing an arbitrary `$R_3$`

, we have:

`$$R_4 = \frac{R_3(V_{OH}-V_{OL})}{R_4}$$`

Then, we can do some math to determine `$V_{REF}$`

as a function of the trip-point, which is halfway between `$V_{TH}$`

and `$V_{TL}$`

. Note that `$V_{REF}$`

is not the same as `$V_{TRIP}$`

.

`$$V_{REF} = \frac{ 2 R_4 V_{TRIP} + R_3(V_{OH}+V_{OL}) }{ 2R_3 + 2 R_4 }$$`

Once you have \(V\_{REF}\), finding \(R1\) and \(R2\) is just solving a voltage divider. Choosing an arbitrary \(R1\):

`$$R_2 = R_1\left(\frac{V_{CC}}{V_{REF}}-1\right)$$`

Single-supply Inverting Hysteresis

This one is more aggravating because the hysteresis feedback and trip voltage networks interact. The trick (there might be a better one) is to rearrange `$V_{TL}$`

and `$V_{TH}$`

like so:

```
$$ V_{TH} = \frac{R_1 R_3 V_{CC} + R_1 R_2 V_{OH}}{R_1 R_2 + R_1 R_3 + R_2 R_3} $$
$$ V_{TL} = \frac{R_1 R_3 V_{CC} + R_1 R_2 V_{OL}}{R_1 R_2 + R_1 R_3 + R_2 R_3} $$
```

Then you can divide these equations to end up with `$R_2$`

and `$R_3$`

being your only unknowns:

`$$ \frac{V_{TH}}{V_{TL}} = \frac{ R_3 V_{CC} + R_2 V_{OH} }{ R_3 V_{CC} + R_2 V_{OL} } $$`

You end up with:

`$$ R_3 = \frac{R_2(V_{TL}V_{OH}-V_{TH}V_{OL})}{ V_{CC}(V_{TH} – V_{TL}) } $$`

We can use this relationship to find `$R_2$`

in terms of only `$R_1$`

:

`$$ R_2 = R_1 \left[\frac{V_{CC}V_{HYST}(V_{OL}-V_{TL})}{V_{TL}(V_{TL}V_{OH}-V_{TH}V_{OL})}+\frac{V_{CC} – V_{TL}}{V_{TL}}\right] $$`

So, choose `$R_1$`

, then solve for `$R_2$`

and `$R_3$`

consecutively.

Here’s pictures with the work if anyone’s curious or finds mistakes.