so now help me
maybe I don't understand something
but let say the Vin=10V
then on PIN2 I will get 5V - right?
and on PIN1 I will get 10V ,no?

what am I missing?

 

so now help me
maybe I don't understand something
but let say the Vin=10V
then on PIN2 I will get 5V - right?
and on PIN1 I will get 10V ,no?

what am I missing?

Well, if you use an comparator (the LM393) -.- the output will swing low when Vin is > 10.5VDC

So you may need to invert this to suit your requirements. This can be done using an transistor.

Basically, with an comparator, if the inverting input is greater than the non inverting input, then the output will swing low.

 

this I know and understand,
this is what I don't understand
phase1:
battery is 12V -
PIN1 I have 0.9V?
PIN2 I have 6V ?
correct?

phase2:
battery is 10V
PIN1 -0.00005 V
PIN2 -5V

I can see I'm doing something wrong ... but what?

 

this I know and understand,
this is what I don't understand
phase1:
battery is 12V -
PIN1 I have 0.9V?
PIN2 I have 6V ?
correct?

phase2:
battery is 10V
PIN1 -0.00005 V
PIN2 -5V

I can see I'm doing something wrong ... but what?

You basically just need Vref @10.5VDC. This Vref can be obtained using resistors, and connected to the non inverting input. The inverting input is where you connect Vmeasure, and if the voltage being measured is above Vref, then the output of the comparator swings low (0 volts)

I think that you are over complicating an building blocl that has been around since 1966 (or something like that)

 

o.k.
but again the Vref which is connected to PIN1(non invert input) and
Vmeasure connected to PIN2(Invert input) came from the same source
so when the source goes down-->V1 goes down-->V2 goes down
so there will be no compering , or the compering will be wrong
this is what I understand - correct me

 

o.k.
but again the Vref which is connected to PIN1(non invert input) and
Vmeasure connected to PIN2(Invert input) came from the same source
so when the source goes down-->V1 goes down-->V2 goes down
so there will be no compering , or the compering will be wrong
this is what I understand - correct me

Yes well then you need to regulate Vref.

 

I know this and this was my question
how can I do this? using just one power source?
how the compare works I know

 

Hello,

Did you not look at the image Dodgydave showed you?

Adjust the potentiometer so that it reacts on the wanted voltage.

Bertus

 

I know this and this was my question
how can I do this? using just one power source?
how the compare works I know

Precision voltage ref source IC would suffice.
I don't have part numbers off the top of my head though.

<SNIP>

Google for precision voltage reference IC.

 

Hello,

Did you not look at the image Dodgydave showed you?

Adjust the potentiometer so that it reacts on the wanted voltage.

Bertus

Or the zener diode would do the trick too.

Notice the zener as Vref?

 

to see I understand the circuit
and let say I put the potentiometer on 60K - so I get V3=Vin*60/100 =Vin*0.6
phase 1 :
battery=12, V2=6.2V V3=7.2V
battery=11, V2=6.2V V3=6.6
battery=10.5, V2=6.2V V3=6.3
battery=10.3, V2=6.2V V3=6.18
battery=10, V2=6.2V V3=6

from the point of Battery=10.3 I will get 0V at the output , right?

also I understand that R1 is for lower the current , right?
but I can use 50K if have in stock , and it will still work the same

Thanks ,

 

I liked alfacliff's suggestion- zener and resistor- simple, elegant (and much easier for me to understand! )

 

is my logic an calculation is correct?

 

You can use any value of resistance for the potentiometer, from 10k to 100k.

 

is my logic an calculation is correct?

You have the right idea. But I haven't pulled an calculator out to verify the answers.

 

I can't use any value
I need it to make the correct voltage divider , no?

 

I can't use any value
I need it to make the correct voltage divider , no?

Yes. Use OHMs law logic though.

a. Decide on IT (total current)
b. Calaculate RT (total resistance)
c. Calculate resistor values for required drops

*For IT just pick an number in the range of mA*

 

so the potentiometer value I ahve calculate is good ,no?
also the value of R1- if I take it as 50K
I will get Imax=12/50=240uA
minimum lose of battery power

 

so the potentiometer value I ahve calculate is good ,no?
also the value of R1- if I take it as 50K
I will get Imax=12/50=240uA
minimum lose of battery power

Yes. Not mA but uA. Min lose battery power.

Just check the datasheet to make sure things O.K

 

data sheet of what?
the zener diode? or the 741?
both of them use less then 240uA -s oit should work o.k.