Learning AoE: output impedance calculation

[Aeternam]

This is from Learning AoE. Context: the R1,R2 "circuit" drives the R3,R4 "circuit". I'm looking for values of R1 and R2 so that R3,R4 is a light enough load to not attenuate the signal.

Problem: Given R1 = R2, what's the value for R1 and R2 for a maximum "droop" of 10%?

This is on frickin' page 22 of 1100 and I can't figure it out. I feel like a complete idiot 

Help? Please?

[Andy Watson]

Ignore R3 and R4 and asses the equivalent resistance presented by R1 and R2 at their junction. Hint: You are not looking for the value and you should not need a calculator.
R3 and R4 present a load of 20k, this forms a potential divider with the equivalent resistance of R1 and R2. You also know that this load will represent 90% of the potential divider.

[BobsURuncle]

The point of this exercise is to apply simplifying assumptions and use the rule of thumb to specify the output impedance and input impedance of each part of a circuit so that the driving circuit is not significantly loaded down, i.e. voltage attenuation is less than 10%.   In this case you break up the circuit into 2 parts where R1R2 drive R3R4. So you are concerned with Rin and Rout of each.  Input of R1R2 is at the top of R1, the output is the node between R1 and R2; the output of R2R3 is the node between R2 and R3.

Now having established what is driving what....  The rule of thumb applied to this problem is  Rout for the R1R2 network must be <= 1/10 of Rin for the R3R4 network. The simplifying assumption is that the input impedance that R3R4 sees at its output will  be infinite (open circuit); and that the output impedance of the source that drives R1R2 (V) is zero.

What is the input impedance of R3R4 that R1R2 sees? 10K +10K = 20K.  So the Rout of R1R2 network must be <= 1/10(20K) =2K.   Therefore R needs to be 4K or less because the thevenin equivalent of these 4K || 4K is 2K, which is R1R2's output impedance.

[Aeternam]

It all makes sense now.

Rin of R3R4 is 20k, so Rout of R1R2 must be equal to or less than 2k.

The thevenin equivalent of a 2k Rout is 4k || 4k hence R = 4k or less. Easy

Thanks

[rstofer]

Side issue:  Does the Learning AoE book give the answers?
Second side issue:  Does Learning AoE follow along with AoE?

I have neither book but I am curious.  I haven't done any of this stuff in quite a long time and it might be fun to play with circuits again.  But it won't be as much fun if I can't check my answers.

[MrSlack]

Side issue:  Does the Learning AoE book give the answers?
Second side issue:  Does Learning AoE follow along with AoE?

I have neither book but I am curious.  I haven't done any of this stuff in quite a long time and it might be fun to play with circuits again.  But it won't be as much fun if I can't check my answers.

I have both. Most of the answers are in the footnotes or on the instruments in the labs. I've scanned through it so will help anyone I see mentioning it. AoE is much more detailed than Learning AoE. AoE is a cross of reference and qualitative knowledge. Learning AoE is a staged tutorial.

I learned how to do all this stuff from the predecessor, the student manual back in the 1990s. Was a valuable journey.

[Aeternam]

The different chapters start out with a textual explanation of what you're looking at. This is sprinkled with tiny exercises like the one I mentioned in post #1, for these no solution is provided but the text nudges you in the right direction. Then comes a lab section (where you build and measure "stuff") and supplementary notes on the subject at hand. Finally there is a section with worked examples, retouching all the problems and exercises mentioned in the first part of the chapter. For these, solutions and further explanations are provided.

References to AoE are aplenty and each chapter gives a recommended reading list in AoE.

Both books are EXCELLENT and well worth their money.