Noninverting Amplifier Gain in Micro Supply - Page 1
Forum section: electronics / beginners

author: poindexterity
author_rank: Contributor
timestamp: July 19, 2016, 01:36:11 pm
content: Hi all,I had a basic understanding of electronics from high school, but have only recently started to learn more fully. To this end I have assembled a bench to perform experiments.Unfortunately, I don't yet have an adjustable power supply! I came across Dave's YouTube series and watched through to the PCB layout last week.I'm trying to fully understand the schematics for the MicroSupply, RevC, as seen here:http://www.eevblog.com/files/uSupplyBenchRevC.pdfI'm looking at the noninverting amplifier between VSet and the Set pin of the LT3080. The notes say that this amp's gain is 10, but I can't get that result when I run the numbers.I understand that the gain formula is Gain = R2/R1 + 1. When I count up the resistance for those values, I getR1 = 1000R2 = 1000 + 100 + 18000 / 2 = 10100So, Gain = 10100 / 1000 + 1 = 11.1What am I getting wrong here?p.s.: It's great that Dave did not release boards or a BOM. It's forcing me to understand the circuit before I go about building it. Kudos, Dave!

author: RandallMcRee
author_rank: Frequent Contributor
timestamp: July 19, 2016, 02:53:50 pm
content: The feedback network is the two 18K 0.1% resistors in parallel and the 1k resistor to ground.18K in parallel is 9k. 9K/1K = 9. Hence gain is 1+9 = 10.I think you are perhaps confused by the other components connected to Vout of the opamp.

author: poindexterity
author_rank: Contributor
timestamp: July 19, 2016, 02:59:25 pm
content: But those 2 18k resistors have 1.1k between them and the op amp output. Are they ignored somehow?The feedback path I see is:Output -> 1k -> 100R -> (9k) -> Inverting InputIs there another path I'm not understanding?

author: Kilrah
author_rank: Unknown
timestamp: July 19, 2016, 03:08:02 pm
content: Quote from: poindexterity on July 19, 2016, 02:59:25 pmYes. The point of interest we want to set the voltage of is that SET line at the right of R12, not the output of the opamp chip, so the feedback circuit starts there.The output of the opamp will be whatever it has to be so that SET is 10x the voltage at the positive input and will thus "automatically" compensate for any voltage drop across R11/R12.

author: poindexterity
author_rank: Contributor
timestamp: July 19, 2016, 03:31:27 pm
content: So really then, it's a question of semantics. Correct me if I'm wrong:The true gain of the op amp (measured from Noninverting Input to Output) is 11.1, but we're dividing the output voltage and measuring from a point where gain is 10. So the effective gain of the circuit at the point of measurement is 10.

author: newbrain
author_rank: Super Contributor
timestamp: July 19, 2016, 03:57:53 pm
content: Quote from: poindexterity on July 19, 2016, 03:31:27 pm

author: Kilrah
author_rank: Unknown
timestamp: July 19, 2016, 08:54:11 pm
content: Quote from: poindexterity on July 19, 2016, 03:31:27 pmNo, it would be variable because if Q2 starts conducting say 2mA the gain between non-inverting input and the SET node will still be 10, but between noninverting input and output pin 1 of the opamp will be more than 11.1 (that would be 11.1x the input, PLUS the 2V drop in R11).

author: StillTrying
author_rank: Super Contributor
timestamp: July 19, 2016, 09:39:35 pm
content: An op amp does what it has to, to keep both of it's inputs at the same voltage, the exact voltage at it's output doesn't matter, it's not interested in that.When UA3 it gets the correct voltage across R14, this voltage across (R14+R15) is 10X higher which is where the output is taken from.The voltage follower U12A LM358 looks like a mistake with it's inputs at V+, is there a RevDversion.

author: poindexterity
author_rank: Contributor
timestamp: July 20, 2016, 12:04:52 pm
content: Quote from: Kilrah on July 19, 2016, 08:54:11 pmAh, good point!