Throughout this thread you mention voltages of 13, 12.8, 12.9, 14, etc.

1. What are the max and min voltages when the system is operating?

2. Can the control circuit be powered by the voltage being monitored, or does it have a separate power source?

3. How accurate do the turn on and turn off trip points have to be:
Output transistor turns on when the voltage is greater than xx.xx V, +/- x.xx V
Output transistor turns off when the voltage is less than yy.yy V, +/- y.yy V

In its most simple form, your task can be done by adding one zener diode to the schematic in post #6. However, this will not be very accurate, and not adjustable. For a better answer, give better information.

ak

 

1. voltages is from around 12 to 14.7 ( Car )
2. Same power source. to both controll and operate of relay.
3. voltages is not that important, just that when voltages comes above aprox 12.9v the circuit activate the relay.
it then should hold the relay activated until voltages drops below aprox 12.9.V

some people in denmark wants taillight on car allways when driving,

This curcuit helps doing that.

When car starts to charge, the voltage gets over 12.9 V and the taillights are on,( 1-10 seconds after start )
When car is stopped the charge stoppes and voltages goes under 12,9 and taillights are off. ( can takes around 10-20 sec to drop all the way,)

The problem is when car battery slowly drops in voltages, and you cut off the "load"( taillights) the voltages can raise a little bit again, causing the relay to on/off many times.

 

The problem is when car battery slowly drops in voltages, and you cut off the "load"( taillights) the voltages can raise a little bit again, causing the relay to on/off many times.

That is solved with a circuit addition called hysteresis. It sets two different trip points, one for when the voltage is increasing and one for decreasing. For example, if the upper trip point is 13.1 V and the lower is 12.9 V, than the relay won't chatter when the system voltage is 13.0 V. If the relay is off, the voltage has to exceed 13.1 V to turn on; if the relay is on, the voltage has to decrease below 12.9 v for it to turn off. If the system is in the off state and the voltage slowly rises above 13.1 V, the relay will come on. So you have to pick the two trip points to be far enough apart to prevent any false trips yet close enough together to catch all true trips. Once you have those numbers, the circuit is pretty straightforward.

https://images.search.yahoo.com/sea...p=hysteresis+comparator+schematic&fr2=piv-web

http://www.ti.com/lit/ug/tidu020a/tidu020a.pdf

ak

 

Are you using the TL431 circuit in post #15?
If so, just add a resistor from the relay taillight output to the sense (R) input of the TL431 to add some hysteresis.
Start out with about 50kΩ.

 

Are you using the TL431 circuit in post #15?
If so, just add a resistor from the relay taillight output to the sense (R) input of the TL431 to add some hysteresis.
Start out with about 50kΩ.

Yes . Ok Will try with 50k

 

just to be clear Crutschow
Drawing 1 or 2 or 3 ?

 

just to be clear Crutschow
Drawing 1 or 2 or 3 ?

3

 

testet IRL. with 53K, and that did not work,
cut in voltages = ((10000/2400)+1)*2,5 = 12,91 V
Cut off voltages = ((10000*53000)/(10000+53000)) = 8412 Ohm = ((8412/2400)+1)*2,5 = 11,26 V ( BAD )

Need to change R1 to 10k2 and keep R2 at 2k4 and add the "new" value as 500k.
Gives me a cut in voltages 13,125 V
Cut off again at 12,91 V

 

Gives me a cut in voltages 13,125 V
Cut off again at 12,91 V

So are those values acceptable?

 

Should be, have 300 pcs i have to change now.

 

Should be, have 300 pcs i have to change now.

You built 300 before you got a breadboard circuit working?

 

Yes.
Have build 2000 before just with a transistor instead and a extra wire to turn on

Another question. Will the relay output not get powered thrue both resistor?
I know it's 510200 ohm and the output volt Will be low but still ?

 

.........Another question. Will the relay output not get powered thrue both resistor?
I know it's 510200 ohm and the output volt Will be low but still ?

Yes, it will give an open circuit voltage of 2.5V through the resistor
But if it's a 10A load, the 5μA or so through the resistor will have no adverse effect and the voltage will be very close to zero.