What does the Hysteresis do in a UVLO (under Low Voltage Lockout)

[Daniel Floyd]

As fare as I have researched UVLO Under Low Voltage Lockout is a circuit which is used to turn off and on the output power of a electronic device according to voltage selected. For instance in a charging circuit the power will only flow to the load once the capacitor has charged to a certain point, upon discharge the UVLO will stop the flow of power after this value of voltage.

Am i correct in assumption of this operation.

A circuit diagram of this has bee attached.

Can anybody Explain what the Hysteresis does as i can not find any material which explains this?

although In my understanding of hysteresis is when a ferromagnetic material wen put with a electric field will hit the remenent saturation point at this point if the field is reversed the ferromagnetic material will never return to zero in fact it will return to the opposite saturation point thus forming a loop effect. I don't understand how this would relate to the UVLO.

 

[Daniel Floyd]

With Some more research it has become apparent that the hysteresis in fact sets a sort of boundary line when using a schmitt trigger for example:Under Voltage Lockout is 3.0V and the Hysteresis is 500mV
so because of this the device will turn off at 3.0V allowing the charge to reacumulate and the device will not turn back on until the input goes above 3.5 V ( Vuvlo + Vhyst ) because of this hysteresis window, this is so that it dosent turn back on at 3.0V and then discharge below 3.0.

am I correct.

 

[TurtleMeister]

Yes, you are correct. It prevents the circuit from being in an undecided state where the load is switched on and off at a rapid rate.

 

[Jony130]

As for your Under Low Voltage Lockout circuit. If R6 = 30K and R8 = 150K. The circuit will saturate Q2 for Vin_on > 2.5V *(1 + R6/R7) = 10V
And Hysteresis is provide by R8. So Q2 is in saturation (ON) until Vin drop to
IR8 = ((Vz2 - Vd2) - 2.5V)/R8 = (5V - 2.5V)/150kΩ = 16.7μA
IR7 = 2.5V/10kΩ = 250μA
IR6 = IR7 - IR8 = 233.3μA

Vin_off = 2.5V + IR6 * R6 = 2.5V + 7V = 9.5V

So as you can see we have an extra 16.7uA hysterisys current into TL431 reference node due to the 150kohm resistor feedback which ensures that Q2 does not switch (OFF) while entering into conduction due to small fluctuations in Vin voltage.

And this 2.5V is a TL431 reference voltage.

 

[Daniel Floyd]

Jony.

infact i have posted the wrong picture by mistake i do appologise, that is a picture of an over voltage system although i assume that it works the same anyway.

I have reattached the Under Voltage system
Thank you for the mathematics you have provided do you know the equation to work out the hysteresis which I am guessing is using the R5, the place where gathered this circuit from has told me that this is actually has an under voltage Lockout of 3.0V and the Hysteresis window of 500mV

but I am now confused in how they caculated both of these values do you have the equation for this.

 

[Daniel Floyd]

UVLO Circuit, Appologies for the error.

 

[Jony130]

But how can this circuit even work at such a low voltage input voltage if we have a 4.7V Zener diode in series with the Q1 base?

 

[Daniel Floyd]

i got this slightly wrong, i now believe the circuit before the UVLO is a charging circuit and when the voltage rises above 4.7V the UVLO passes this on to the output.

I then believe the R5 and C1 create the time constant thus creating hysteresis which means the voltage will be locked on until the capacitor has discharged which holds around 500mV thus the voltage flowing at 4.7V but turning off at 4.7V - 500mv = 4.2V allowing to recharge back to 4.7V. Dan.