# [SOLVED] Difference (?) of voltage not more than 2 volt ?

#### vanderghast

Joined Jun 14, 2018
63
I have just observed how hard it is to come with a design of "absolute" value.

The pristine problem is to get a "simple" (less than 4 opamps ! ) solution which will light on a green LED if the difference between two rails is less than, say, 2 V. Evenbetter if the "2' is "adjustable" by design. I first thought that is was a piece of cake, but even with four standard op-amp (which unfortunately most behave as they do with negative voltage even if their low supplied voltage is the "ground", while some Analog Devices OpAmps claim to do otherwise, these failed to be ... "simple" and commonly available in simulators) I have a hard time to come with a solution. Humble me! Here are the specs: with two rails which can be at +12V or -12V or absent (battery polarity inversed, or one is at an open circuit), light on a standard green LED (can use a reliable available source at an unspecified voltage, independently of the rails source... else, how can you light on a LED with no power at all? ) if the difference of voltage ( what is a voltage if not a difference ? ) between the two rails is less then 2.5 to 2 volt (absolute value, that is). I have spent too much time trying to solve that apparent simple riddle that I resolved ... to post the question here. Once again, humble me with a "simple" solution. TINA (Texas Instruments) or LTspice (Analog Devices) circuit simulation really appreciated, with triangular rails variations for -12V to +12V (different non congruent frequency) appreciated, but any insight thought of "old timer" not deprecated either (and doubly appreciated, in truth). With respect dues, after all, that sounds so simple that I have a strong feeling that I miss something so elementary... Thanks in advance.

#### MrChips

Joined Oct 2, 2009
26,790
Here are hopefully some helpful tips in posting a question.

1) It is difficult to read such a long paragraph with no line or paragraph breaks.
2) Sometimes less is more. Try to state the problem in concise sentences.
3) A picture is worth a thousand words. It would be better to describe your circuit with a circuit diagram.
4) Establish the specifications of the circuit using a table showing voltages.

Now for some tips on the technical details.

1) If all your voltages are positive, you don't need a negative supply. There are analog ICs that will work on single supply voltage.
2) It would be easier to use an analog comparator instead of an op-amp.
3) There is no such thing as exactly 2.0V. There is always noise in voltages. Hence you have to take into account some error margin, for example. 2.00 ± 0.05V.
4) Always allow for some hysteresis where the threshold voltage is different for increasing and decreasing voltages. If you don't have hysteresis then your output will oscillate.

• DickCappels and xox

#### LowQCab

Joined Nov 6, 2012
2,329
What You missed is stating the problem that You are trying to solve,
rather than presenting your attempts at solving that problem, as the problem.
What's broke ?
.
.
.

• DickCappels

#### MrSalts

Joined Apr 2, 2020
2,060
Here are hopefully some helpful tips in posting a question.

1) It is difficult to read such a long paragraph with no line or paragraph breaks.
2) Sometimes less is more. Try to state the problem in concise sentences.
3) A picture is worth a thousand words. It would be better to describe your circuit with a circuit diagram.
4) Establish the specifications of the circuit using a table showing voltages.

Now for some tips on the technical details.

1) If all your voltages are positive, you don't need a negative supply. There are analog ICs that will work on single supply voltage.
2) It would be easier to use an analog comparator instead of an op-amp.
3) There is no such thing as exactly 2.0V. There is always noise in voltages. Hence you have to take into account some error margin, for example. 2.00 ± 0.05V.
4) Always allow for some hysteresis where the threshold voltage is different for increasing and decreasing voltages. If you don't have hysteresis then your output will oscillate.
The problem is easier if bipolar supply rails are used so you don't need to bias with a virtual ground to know when rail A is a higher voltage than rail B and when rail B is higher than rail A.

#### vanderghast

Joined Jun 14, 2018
63
The problem is easier if bipolar supply rails are used so you don't need to bias with a virtual ground to know when rail A is a higher voltage than rail B and when rail B is higher than rail A.
That makes perfect sense and acceptable in the specific context. And indeed, that setting allows a much simpler circuit!

#### vanderghast

Joined Jun 14, 2018
63
Here are hopefully some helpful tips in posting a question.

1) It is difficult to read such a long paragraph with no line or paragraph breaks.
2) Sometimes less is more. Try to state the problem in concise sentences.
3) A picture is worth a thousand words. It would be better to describe your circuit with a circuit diagram.
4) Establish the specifications of the circuit using a table showing voltages.

Now for some tips on the technical details.

1) If all your voltages are positive, you don't need a negative supply. There are analog ICs that will work on single supply voltage.
2) It would be easier to use an analog comparator instead of an op-amp.
3) There is no such thing as exactly 2.0V. There is always noise in voltages. Hence you have to take into account some error margin, for example. 2.00 ± 0.05V.
4) Always allow for some hysteresis where the threshold voltage is different for increasing and decreasing voltages. If you don't have hysteresis then your output will oscillate.
Definitively, the missing context seems important. The intended use is to signal that a voltage is becoming too high to plug a given component between the two rails. So the exact crossover at 2V is not of importance.

Post #4 just gave me a new idea about how to simplify greatly the circuit.

Thanks for your time, I consider the problem solved, practically.

#### vanderghast

Joined Jun 14, 2018
63
What You missed is stating the problem that You are trying to solve,
rather than presenting your attempts at solving that problem, as the problem.
What's broke ?
.
.
.
The main problem was due to the possibility of negative voltage at any of the two rails.

So following along with the suggestion of post #4, that should allow me to produce a much simpler solution.

I consider the problem solved, for all practical ends. Thanks for your help.

• MrSalts