# Output Resistance Inverting Op-Amp Config.

Discussion in 'Homework Help' started by jegues, Jan 26, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
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43
Hello all,

Today in class we were talking about the non-inverting op-amp configuration and how we would like our op-amps to have a high input resistance, without sacrificing any gain.

The circuit attached is the provided solution to this dilemma and the professor went on to talk about input resistances and output resistances of the non-inverting voltage amplifer (circuit attached) and the non-inverting current amplifier (similiar circuit).

We defined the input resistance for the voltage amplifer as simply,

$R_{in} = \frac{v_{1}}{i_{1}}$

Where the two are the voltage divided by the current across the first resistor, $R_{1}$.

He then attempted to explain the output resistance of the voltage amplifier as well as the current amplifier. (non-inverting configuration)

He scribbled a few things on the board, erased them and was constantly rearranging his explanation. As a result, I'm extremly confused as to how we find the output resistance of these two circuits.

Someone care to explain?

Also, we've looked very briefly at the "inside" of the amplifier. From what we've since it's simply an resistor $R_{in}$ connected to ground, where the voltage across said resistor ($V_{in}$) is the controlling voltage of a voltage controlled voltage source, $A \cdot V_{in}$ in series with one last resistor, $R_{out}$.

Could someone please clarify how we obtain the output resistance of each amplifier (voltage and current) and why they are important?

Thanks again!

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2. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
Reason output impedance is important: High output impedance cannot source much current, in a DC perspective (think TTL Logic). With AC, maximum power transfer and linearity occurs when impedances are matched. The latter is important in areas such as audio preamps and RF amplifiers.

The same can be said for input impedance as well, it needs to be known as well as the output impedance, otherwise it is hard to work around what will happen at different operating points. The Miller Effect is an important factor in the case of amplifier impedance.

You should ask your prof for clarification also, as the answers we may give may not match the answers he thought he was telling you when the test shows up. That's a part of the new "college ways" I've seen that I don't like (Do it the prof's way or else).

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
The Rin is equal to Rin = Vin/Iin and the input resistance can reduce the voltage gain.
Rout = Vout/Iout for Vin=0V

And in electronic circuits, we always hooking the output of something (output of a amplifier, output of a voltage input source...) to the input of something else (load resistor, input stage of a next amplifier...).

So in our case, the signal source (5V) is the output of a first amplifier with series internal impedance Rout, driving the load resistance R_Load=1K ( or next stage input impedance Rin).
After we draw this in a simplified schematics we see our old friend the voltage divider.

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