#ref hi

I've played around with the LM3914 before - courtesy of @johns_az.  At the time, I didn't need to worry about much besides the fact that it worked!

http://tymkrs.tumblr.com/post/10444281077/generalities-learned-from-the-lm3914n-courtesy-of

In any case, I'm working on a calibration heavy board right now (well heavier than what I've dealt with previously) and realized I didn't know what Pins 6-8 really did.

Thanks to the ever patient folks at the Tymkrs IRC, longhornengineer, @johns_az, I've come to a new understanding of the role of pin 6, 7, and 8 and why they're connected to each other.

So pin 6 is Ref Hi.  If you think of the LED Bar driver as dividing up the amount of voltage you're allotting it into 10 equal parts, and then letting the LED bar represent those divisions via light, Ref Hi is the highest voltage.  

So let's say Ref Hi is set to 10 volts.  That's 1volt per comparator.  And therefore, if you see 5 LEDs light up on your LED bar, that means it's reading 5volts.

Well here, and in many other examples, you'll see Ref HI connected to Ref Out.  So let's look at Ref Out (7) and Ref Adj (8).  Essentially, Ref Out is 1.25v higher than whatever Ref Adj is, hence its name (I think).  If you set Ref Adj to 5 volts, then Ref Out is 6.25v.  Likewise if you set Ref out to ground, then Ref Out is 1.25v.

Now if you connect Ref Hi to Ref Out, which is what a normal use case would call for, Ref Hi would equal whatever Ref Out is, which is 1.25v over Ref Adj.

EDIT: @Rednaxander, my sensei in all things electronic, said that:

"Connecting RHi to things besides Ref Out is not particularly abnormal.  The 1.25V reference is really just there for 1) a convenient voltage reference and 2) setting LED brightness.  There is also one more case use seen here:

http://www.biltek.tubitak.gov.tr/gelisim/elektronik/dosyalar/21/LM3914.pdf on page 7 under "Internal Voltage Reference"  Basically you can feed Ref Out through a voltage divider to create the reference value for Ref Adj.  

Ignoring the Iadj current which is relatively small, if R1 and R2 are equal for example, when RefAdj is 1.25v, Ref Out will be 2.5V, which divided in half is 1.25V, so the voltage is stable.  It balances out according to the formula."

Let RefOut be A. Let RefAdj be B. Let current out of RefOut be C

A = B + 1.25V according to the voltage reference

C = A / (R1 + R2)

B = C * R2

(those last two from V=IR rules) and from there, it's simple algebra. (that's a simplification ignoring current from RefAdj, but close enough)

If anyone else has any other input about this, let me know!