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I came across this question in my textbook:

The first stage of a 2-stage amplifier has a voltage gain of \$150\$, an input resistance of \$1500 \Omega\$, an equivalent noise resistance of \$700 \Omega\$ and an output resistance of \$20 k\Omega\$. For the second stage, these values are \$500\$, \$50 k\Omega\$, \$1200 \Omega\$, and \$1 M\Omega\$ respectively.

  1. Sketch the circuit diagram illustrating this scenario

  2. Calculate the equivalent noise resistance of this 2-stage amplifier

I'm having a bit of a confusion regarding this question. Shouldn't the output resistance of the first stage be same as the input resistance of the second stage of the amplifier i.e. \$R_2\$?

enter image description here

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    \$\begingroup\$ Your last question would be right if you were trying to transfer maximum power. That might be the case for a final stage feeding a transducer where power transfer is important. Nothing in your description shows this to be the case, though. \$\endgroup\$
    – jonk
    Commented Sep 13, 2018 at 20:52
  • \$\begingroup\$ @jonk I don't understand what you're trying to say. What would be the output resistance of the first stage in terms of \$R_1,R_2,R_3\$? And what would be the input resistance of the second stage ? \$\endgroup\$
    – user193992
    Commented Sep 14, 2018 at 5:09
  • \$\begingroup\$ I was just reading your last sentence in the context of the quoted question and answering that. I can't address your hand-crafted diagram, because it's not readily meaningful to the textbook question context and it would take me too long to straighten you out. \$\endgroup\$
    – jonk
    Commented Sep 14, 2018 at 6:34
  • \$\begingroup\$ Please tell us where this comes from. Noise impedance or resistance almost always means the ratio of a voltage noise source to a current noise source. But you already told us that the input noise resistance is 700 Ohms, so there is nothing to calculate. I’m puzzled. \$\endgroup\$
    – 10ppb
    Commented Mar 14, 2021 at 0:15

2 Answers 2

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The impedances need to match for maximum power transfer, but you probably don't care so much about that for the first stage output. But the input impedance of the second stage should probably be higher than the output impedance of the first stage, or the second stage will load the first stage and limit your operating range.

An ideal amplifier has infinite input impedance and zero output impedance, or at least a reasonable facsimile of those conditions. On the other hand, if your first stage tries to drive 10V through an output impedance is 1k and into an input impedance of 10 ohms, then the second stage will receive $$\frac{10V\cdot10}{(1k+10)ohms} = 0.1V.$$

In your case, you're driving through 20k into 50k. So your second stage will see $$\frac{50}{(50+20)}$$ or 5/7 of whatever the first stage puts out.

Sorry I don't know Mathjax so ugly equations. BTW, what is this textbook you're using?

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  • \$\begingroup\$ You can learn some basic Mathjax by using the edit button to view how I modified your answer to include Mathjax. \$\endgroup\$
    – user105652
    Commented Sep 14, 2018 at 3:49
  • \$\begingroup\$ I saw that, thanks very much. Looks a lot like $$\LaTeX$$, actually. (Cool, it works in comments too!). I suppose the real problem is that I am lazy! :P \$\endgroup\$
    – Jim
    Commented Sep 14, 2018 at 4:38
  • \$\begingroup\$ This site will accept Mathjax, LaTex, basic HTML such as <sup>1</sup>. \$\endgroup\$
    – user105652
    Commented Sep 14, 2018 at 5:06
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No, those are two different resistances if you draw the diagram correctly. The question intends that your diagram contain ideal voltage amplifiers, which you should represent as triangles pointing to the right. The input resistances should be resistances from the amplifier inputs to ground. The output resistances should be resistances in series with the outputs. Input current noise generators should be from the inputs of the ideal amplifiers to ground. Input voltage generators should be in series with the ideal amplifier inputs.

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