Teachable moment - Voltage Divdier?

[TechJeff]

Hello Everyone,

So while I am not a total beginner, I still don't really understand why things sometimes don't just blowup and meltdown.  So I thought I might ask a question from this diagram about theory and practice.

The first diagram shows a voltage divider that leads back to the negative- but it is connected to the positive voltage in! 

My immediate observation is that this is a short circuit.  Why doesn't it just dump all the current into a short? 

Two diagrams to look at.  One shows a close up of the voltage divider.  Ignore the arrows, I used this to ask a different question.

Second diagram shows the whole thing.

Thanks guys for helping me understand what's going on here.

TJ

[wasedadoc]

I don't see any "short".

[themadhippy]

i see a repeat of yesterdays posting

[SanctePieDecime]

A short circuit is (in effect) a 0 Ohm path from a voltage source to ground (from postive to negative to use your terminology). In practice we consider a short circuit to a very low reistance path between the voltage source an ground (because nothing has 0 Ohms of resistance - not even superconductors).

The reason for this is Ohm's law which states that current is voltage divided by resistane. Because of this, when there is 0 resistance the function becomes undefined (and approaches infinity). "Infinite" current is undesireable both in terms of power consumption and in terms of component ratings.

Why is this not a short circuit, well there are two resistors between the voltage source and ground (one of 1.5 kOhm and one of 1 kOhm). Ignoring the minimal current draw from the voltage divider part, the current would be the input voltage over the resistance. In this case, using a 12 V input voltage the current would come out to be about 4.8 mA which is a fairly safe value for most components.

How does this divide voltage? Well, again it is Ohm's Law. We know the current now and the resistance of each resistor which means that we can measure the voltage drop across each resistor using a little algebra. We can rearrange the equation to be: voltage equals current time resistance. The voltage across the 1.5 kOhm resistor would be 1.5k (1500) times 4.8m (0.0048) which is 7.2 V and the voltage drop across the 1 kOhm resistor is 1k (1000) time 4.8m (0.0048) which is 4.8 V. This second value is the value that you will be able to read out of your voltage divide because that is the voltage drop between that point and ground.

There are plenty of videos out there that explained it better than I did. Here is one I thought was fitting considering that this is the EEVBlog forum after all:

[xvr]

The first diagram shows a voltage divider that leads back to the negative- but it is connected to the positive voltage in!

Negative lead of divider is a lead that connected to the negative power source. Not right one or down one (on scheme)

[TechJeff]

True, I am the same guy.  But I got my answer on the practical level, but I don't understand why it worked.  That's my question here.  Why doesn't my quad copter burst into flames?  I mean it did not, but puzzles like this bug me.

I see the next post may answer.

TJ

[TechJeff]

Ah!  Ok.  I get it.  The resistors are a load and therefore are a valid circuit!  I suddenly realized that after I saw him drawing it out on his white board. 

The resistors in my circuit finally made sense.  For some reason, I was thinking that the voltage had to flow into the Arduino first- or it was a short circuit.  But the Arduino is not the only thing doing work in that first drawing. 

Being a Systems Admin, I use to work up at the top level down to the board level- but no components.  So I forgot my basic training in electronics class in High school and early work career- that was a long time ago.   

Thank you for pointing me to the Ozzy, he did the trick.   

TJ

[TechJeff]

Wow, I really like this guy- thank you SanctePieDecime!

[BillyO]

Does this help?

[TechJeff]

Now THAT is clear and makes sense.  The way it was on his drawing, It said to come off the negative with the resistors and go to the positive. 

That really bothered me, because it just seemed backwards from diagram theory, though I guess it would still work. 

Had his diagram looked like yours, there would have been no reason for this post.

However, it's good that we have his diagram, because it teaches me and anyone else that reads this, to be wary and check over diagrams carefully. 

In this case I think the gentleman that posted this online was long on practical experience, but short on making a diagram to explain his wiring.  He did say that he put the system together for those experienced in electronics, not those slightly above beginners.  My experience is limited to High School electronics, ie, what is an electron, Mr. Ohm and what are the components of a circuit.  And then some more training in my early work life, working with computer controlled assembly stations, PLCs and finally the early Arduino.  That left me with the ability to read basic diagrams like this, but not enough to call a ball from a strike- as I needed to here.

I knew this looked wrong, but could not tell why it was wrong.   

Thank you everyone and thank you BillyO! 

TJ