Low voltage switch high voltage cut out

[Fraser32]

Hi guys New to this forum, i am after a device that will operate a switch at 11.8v and then turn off at around 13.8v any ideas or suggestions would be greatly appreciated

Edit: volatges are rough and dont need to be exactly on at 11.8 or off at exactly 13.8

 

[ronsimpson]

Question:
The is a voltage you want to monitor.
There is a switch.
When the voltage is from 0 to 11.8 volts the switch open or closed???
When the voltage is from 11.8 to 12.8 the switch is closed?
When the voltage is above 13.8 volts the switch is off?

Please explain again. I don't understand.

 

[Fraser32]

0-11.8v the switch will be closed and remain closed until the voltage reaches 13.8v at 13.8v the switch will then open and remain open until the voltage falls to approx 11.8v ath which point the switch will close again and the cycle will continue

 

[ronsimpson]

In this circuit VRef is a voltage in between your Low Voltage and High Voltage. Example 12.8V
R3,4 adds hysteresis. (window)
How does it work:
Vref=12.8
If Vin is low then Vout=high. R4 pulls up VRef1 by 1 volt. (13.8)
If Vin is high then Vout= 0 volts. R4 pulls down VRef1 by 1 volt. (11.8)

Here, this might help.
**broken link removed**

 

[Fraser32]

Thanks mate will look into it way take a little bit for me to wrap my head around it

 

[MikeMl]

Better to use a Window Comparator (which requires two separate IC comparators)

One way is to use a 555 timer chip. It has two comparators which set and reset a flip-flop. I just built one of these for a golf cart battery charger, so the trip points are 37.5V and 44V. With a bit of work, you could train it to work at 11.8V and 13.8V. It is on a different computer, so if you would like to see it, I'll post it tomorrow.

 

[Fraser32]

Would love to see it Mike!

 

[MikeMl]

Here is the basis of how to use a 555 as a CutIn, CutOut voltage compare circuit, especially useful for battery monitoring. First, look at what a 555 intrinsically does when its supply voltage Vcc is connected to a fixed reference voltage, and a slowly varying input is applied directly to the Thrs and Trig pins. The 555's in-built trip points are at 1/3Vcc (at Trig), and 2/3Vcc (at Thrs).

The 555's Out pin is driven from an internal latch, which is set when the voltage at the Trig pin goes below 1/3Vcc (CutIn), and is reset when when the voltage at the Thrs pin goes above 2/3Vcc (CutOut). The Trig condition overrides the Thrs condition, so for Out to reset, the voltage at Trig must be above 1/3Vcc. This is just how the 555 works...

Here is a simple example or the basic idea: Vcc = V(ref) = 9.0V, so CutIn = 1/3*V(ref) = 3.0V and CutOut = 2/3*V(ref) = 6.0V.

More later:

 

[MikeMl]

Here is how to move the CutIn and CutOut voltages to something more useful for a battery monitor. The addition of two voltage dividers: R1/R2 and R3/R4 lets us move the CutIn and CutOut to 11.8V and 13.8V, respectively.

First, calculate R2 such that the Trig pin drops below 3.0V as the battery discharges below 11.8V. Using the familiar voltage divider equation (ratio and proportion) Vtrig/Vin = R2/(R1+R2).
Solving for R2: (Vtrig/Vin)*(R1+R2) = R2
Simplify: R2 = R1*(Vtrig/Vin) + R2*(Vtrig/Vin)
Gather terms: R2 - R2*(Vtrig/Vin) = R1*(Vtrig/Vin)
Simplify: R2 = R1*(Vtrig/Vin)/(1-Vtrig/Vin))

R2=499000*(3/11.8)/(1-3/11.8)) = 499000*(3/11.8)/(1-3/11.8)) = 170K

Second, calculate R4 such that the Thrs pin rises above 6.0V as the battery charges above 13.8V.
R2=499000*(6/13.8)/(1-6/13.8)) = 499000*(6/13.8)/(1-6/13.8)) = 384K

Remaining to do:
Where do you get a 9.00V reference?
Trip points of a 555 are close to 1/3Vcc and 2/3Vcc, but not exact.
Dealing with resistor tolerances.
Making the circuit immune to noise and glitches...