# Schmitt trigger offset voltage

I found this circuit which generates an offset voltage to the hystersis (so the threshold voltages are for example not -3V and 3V but -2V and 4V). It would be the same as if you remove R1 and put a voltage source between R2 and GND, right? The circuit above just realizes that?

How does this exactly work and how can I calculate the resistors? I don't understand how it works with R1. The voltage where R1 is connected to is the same as the positive battery voltage of the op-amp. The output will change between 5V and -5V.

For my logic, 2.5V should drop on R1 and 2.5V on R2. However, there are also 5V (or -5V) that need to drop on R_hyst and R2. How does that work? What will be the final voltages and how to calculate it? An explanation would be awesome.

Thank you

I found this circuit which generates an offset voltage to the hystersis (so the threshold voltages are for example not -3V and 3V but -2V and 4V).

No, they won't be +/- 3 volts because to get symmetrical hysteresis voltages (when you have a split +/- 5 volt supply) you will need R2 taking down to -5 volts. At the moment it is connected to 0 volts and this won't give symmetrical-about-ground hysteresis voltages. Connect R2 to -5 volts.

With a feedback resistor of 576 kohm as shown in the diagram AND R2 taken to -5 volts, the positive hysteresis voltage will be: -

$$\text{10 volts}\cdot \dfrac{\text{R2}}{\text{R1 || 576 k + R2}} - \text{5 volts}$$

And, using R1 = R2 = 100 kohm the positive hysteresis is +400 mV. By the same token, the negative hysteresis voltage will be -400 mV (due to symmetry now that R2 is connected to the -5 volt supply).

The above calculation assumes that the comparator output can swing rail-to-rail.

It would be the same as if you remove R1 and put a voltage source between R2 and GND, right?

Here's where you words are confusing - if you put a voltage source between R2 and ground, that could mean two things: -

• disconnect R2 from ground and insert a voltage source or
• add a voltage source across R2

I think you mean the first one and, if you do, then that is an acceptable way to set levels of hysteresis.

No it would not be the same if you remove R1 and use a voltage source to drive V+ pin. The V+ threshold voltage would be fixed and there is no way the output feedback resistor can change the threshold voltage.

To calculate the voltages, you just have two cases. When output is high, the feedback resistor is pulling high to positive supply voltage so it is in parallel with R1. The other case is when output is low, it will pull the node down to negative supply voltage. You can use Thevenin to find out that the R1 and R2 boil down to 2.5V source with 50k impedance.