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I know that, if I have a voltage amplifier, the output resistance Rout should be as small as possible: indeed, if the circuit is replaced by its equivalent Thevenin's circuit (1st picture), the voltage on a (generic) load is maximized if Rout goes to zero.

1st question: if I replace the voltage amplifier with Norton's circuit (2nd picture), the output resistance Rout doesn't change, then if Rout goes to zero, I should be able to see the same behaviour of the voltage amplifier: I would expect, indeed, that the more Rout goes to zero, the more the current flows in the load resistance RL, thus increasing the output voltage (that is, a voltage amplifier). But I'm not able to show this: if Rout goes to zero, I get a current source which is shorted by itself, and all the current flows in Rout=0 (and the output voltage is of course 0).

2nd question Is it correct to say that, in general, if the output resistance Rout of a generic circuit is very great, a circuit behaves like a current amplifier (if I put it in Norton's circuit this is evident, if I put it in Thevenin's circuit I don't know how to show this) and if the output resistance is very small, a circuit behaves like a voltage amplifier (if I put it in the Thevenin's circuit this is evident, if I put it in the Norton's circuit I don't know how to to show this)



2 Answers 2


Andy's answer is on point. I thought I give you a different insight.

The problem with the extreme cases is that it can make no sense in hurry if you just plug in the values (0 or infinity). For this reason, it makes more sense to look at it as a limit: Either a value that approaches zero or a value that approaches infinity.

For your first question, think of it that way. You can find the current through the load by using the current divider equation. That will leave you with:

$$ I_{R_L}=\dfrac{R_oAV_i}{(R_o+R_L)R_o} \tag1$$

Just like in a math class, you'd simplify the equation as much as possible first and then you apply the limit. In (1), you can get rid of the \$R_o\$ factors and get:

$$ I_{R_L}=\dfrac{AV_i}{(R_o+R_L)} \tag2 $$

Apply the limit:

$$ \lim_{R_o \to 0}I_{R_L}=\dfrac{AV_i}{(R_o+R_L)} =\dfrac{AV_i}{R_L}\tag3 $$

You can take it from there (find \$V_o=I_{R_L}R_L\$ and divide by \$V_i\$). This should prove out that you get the same \$\dfrac{V_o}{V_i}\$ like in your first image.

For your second question, Andy's answers covers it. Forget about amplifiers for a moment—if you have a source that has a high output impedance, then it may make more sense to represent as a current source (Norton). If the impedance is low, it may make more sense to model it as a voltage source (Thevenin). That's why your problem gets confusing if you try to transform your non ideal voltage source (with \$R_o\to 0\$) to a Norton equivalent (the opposite would be true too—try changing a non ideal current source to a Thevenin equivalent with \$R_o\to \infty\$).

But in general, when you have extreme cases that will produce nonsensical results if you directly plug in values, you're better off looking at it as a limit and see what you get.


1st question

You are just getting confused. A voltage source with zero output resistance is an infinite current source across zero ohms. You can't just do a conversion from a voltage source with finite output resistance to a finite current source with zero parallel impedance because that makes no sense.

2nd question

If the output resistance is very great you CANNOT say it behaves like any amplifier because you are only talking about one component. Ditto for the 2nd part of your 2nd question.

Extra bit (mine)

I see you have asked 10 questions now and got reasonably good answers to several of them but you haven't formally accepted any answers yet. I would also conclude that you probably may not not have upvoted any answers yet. Now here's the deal with stack exchange. It runs on a form of currency, call it good-will if you want, and that good-will is upvotes and answer acceptance.

So, please take the tour and do your kind duty.


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