Can anybody find out the Hysteresis of the attached circuits. Consider the vltage source will ingreese from 0 to 270V.

 

Hello,

Here is the circuit you posted in the PDF:

I assume the output of the comparator will go to 0 Volts when low and to + 5 Volts (=V2) when high.
The upper switch limit will be when the output is high:
The resistor R5 will be parallel to R2.
The voltage at the input will be 5 * R1 / (R2||R5+R1)
The lower switch limit will be when the output is low:
The resistor R5 will be parallel to R1.
The voltage at the input will be 5 * (R1||R5) / (R2 + (R1||R5))

The hysteresys will be the difference between these values.

Greetings,
Bertus

 

Hello,

The calculations I gave will not work.
The comparator has open drain output, so a resistor to the powersupply is needid.
See datasheet.
This resistor will add to the R5 when the high switch voltage is calculated.

Greetings,
Bertus

 

Thanks Bertus...

 

\(V_{\small{(-H)}}=\Large {{\frac{\frac{5}{20}}{{\frac{1}{20}}+{\frac{1}{1}}+{\frac{1}{68}}}}}\, \normal {=\,0.234807\,Volts}\)

\(V_{\small{(+H)}}=\Large \frac{{\frac{5}{20}}+{\frac{5}{(68+R_{pu})}}} {{\frac{1}{20}}+{\frac{1}{1}}+{\frac{1}{(68+R_{pu})}}}\)

Here are the two Millman's Theorem equations.

Once you choose a value for Rpu, you can plug it into the second equation and solve for the positive hysteresis setpoint.

hgmjr

 

To restate what has been said a different way, think in terms of voltage dividers. When the two inputs are at the same voltage, that is when it will switch.

Each leg with 2 resistors will divide the voltages to fixed values, which sets the hysteresis.