Thinking of V Divider as Amplifier Stages (Art of Electronics) - Page 1
Forum section: electronics / beginners

author: baeckel
author_rank: Newbie
timestamp: July 11, 2017, 03:00:28 pm
content: I just started working through the Art of Electronics 3rd Ed. Page 7 saysQuoteI'm struggling to make sense of this.Asumptions I am making:The"upper resistance"refers to R1in a generic voltage divider and not the larger of two resistors.The voltage divider is being readright to leftin this instance.If those assumptions are correct it makes sense to me that the upper resistance and Vincould be thought of as an amplifier output, but I don't understand how the lower resistance would represent the"input of the following stage". It seems like it should be the "output of the preceding stage".Can anyone help clarify for this newb?Thanks!

author: rstofer
author_rank: Super Contributor
timestamp: July 11, 2017, 04:07:35 pm
content: The upper resistor could represent the output impedance of the first stage and the lower resistor could represent the input impedance of the second stage.The input impedance of the second stage loads the first stage and, intuitively, this is obvious. But if we look at it as a voltage divider, we can put numbers to the effect.Suppose you had a first stage with 1M of output impedance and a second stage with 1k of input impedance. You wouldn't have a very large signal at the input.This will come up again when you use microcontroller ADCs where the input impedance is fairly low and to get a proper conversion, the source impedance must be less than 10k (or less).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 11, 2017, 09:22:35 pm
content: They're speaking of the Thevenin equivalent source and load:The source voltage and resistance, and the load voltage and resistance, are in series. You can rearrange this so there is one voltage (the source and load voltages subtract), and two resistors in series vertically, as a resistor divider.Note that it's symmetrical too, which isn't always important (most loads are understood to be unpowered, so that VL = 0). But this becomes relevant for bidirectional circuits (many communications channels), and at RF, where it's an important analytical tool, used to abstract away a load's complicated behavior (i.e., the load reflects power back towards the source), by leaving a placeholder to make that connection in a separate step.Tim

author: baeckel
author_rank: Newbie
timestamp: July 12, 2017, 03:04:03 am
content: Thanks Rstofer, and thanks in advance for dealing with my struggle to comprehend.My assumption that the schematic is being read right to left is not correct. I was forcing myself to think of the "second stage" as if it were a stage within an amplifier circuit when it was actually the load on the amplifier. I did not think of the "next stage" as the next stage in the circuit, I thought of it as the next stage in the amplifier (e.g. audio amplifier'sphono>preamp>ampstages.Thinking of the lower resistor as a load on the amplifier circuit makes sense to me right now, thank you.

author: baeckel
author_rank: Newbie
timestamp: July 12, 2017, 03:13:57 am
content: T3sl4c01l, cool name.They don't get to VTh/RThuntil the next section, but you're right in that they are describing the voltage divider with one eye on the Thevenin equivelents. I never realized that loads could be powered, I'm looking forward to the part where I start to learn about bidirectional and RF circuits!Thanks

author: rstofer
author_rank: Super Contributor
timestamp: July 12, 2017, 04:16:43 pm
content: Once you get Thevinin (and Norton) out of the way, I suspect you will get to mesh and nodal analysis. This is where you will see multiple sources and odd looking circuits. Mesh analysis is pretty easy to understand but it will almost always have the disadvantage of more equations than nodal analysis. As I recall, nodal analysis was not given much emphasis when I took the class but there are times when it results in smaller matrices and smaller is good!I am not enough of a math major to prove that there are always less nodes than meshes but I think it is true. Of course, somebody will probably come up with a circuit to disprove my guess.I can work a 3x3 matrix by hand but 4x4 gets difficult. The right answer, of course, is to plug the matrix into a solver like Matlab or wxMaxima. I'm a huge fan of both but wxMaxima is free.Here's a thread related to Kirchoff's Laws. There are two batteries and a T configuration of resistors. Reply 6 is where I post the wxMaxima solution.https://www.eevblog.com/forum/beginners/kirchoffs-law/Here is an even better thread related to AC circuit analysis. It goes on for 7 pages but it is still worth reading. When we start getting into complex numbers, a program like wxMaxima is invaluable.https://www.eevblog.com/forum/beginners/mesh-analysis/

author: baeckel
author_rank: Newbie
timestamp: July 13, 2017, 07:08:14 pm
content: Thanks for the links! I imagine I'm going to try and do everything in Python, fail, try/learn wxMaxima.