operational amplifiers

I have been given the following problem for a college assignment. I was hoping that someone with more knowledge than I could help me out;

Operational amplifier (op-amps), which have negative feedback applied, can have their overall gain fixed using a simple resistor network in the feedback path. Explain why this is so.

Cheers

Reply to
Simon Ward
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What have you done to determine the answer to date?

Reply to
Sam Wormley

Theoretically, an op amp with no feedback would have infinite gain.

FrediFizzx

Reply to
FrediFizzx

Take a look at application notes for Op-Amps. Here are a couple links to get you started.

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Michael A. Terrell
Central Florida
Reply to
Michael A. Terrell

have you heard of the parable of the ass and the carrot and the stick?

--
Achilles: I wish my wish would not be granted.
           < an undescribable event occurs >
Achilles: What happened?  Where's my Genie?
Tortoise: Our context got restored incorrectly.
Achilles: What does that cryptic comment mean?
Tortoise: The system crashed.


To email me send to :

rb  AT ieee DOT org
Reply to
Roy McCammon

On Mon, 28 Jul 2003 18:21:22 -0700, "FrediFizzx" Gave us:

And no control. And "infinite" has starvation points elsewhere in the chip and circuit. It would be a single, always on device. Hell, a transistor works in that instance. Feedback loops are critical to control circuitry. They don't just guess. They have to be responsive, active control elements, and that requires data from what they are controlling. We call it feedback. Why does that word scare some people? One monitors a point, and utilizes the data to control the system that controls the point being monitored.

Feed forward should even be more scary, then. ;]

If one lives in the basement of a house or building in a wet region of the world, are they considered to be "critically damped"?

Reply to
DarkMatter

What's your college attendance record?

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Jim Backus  OS/2 user
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Reply to
Jim Backus

Douglas Eagleson snipped-for-privacy@yahoo.com

You are asking the question called why does the op-amp's internal circuit implement the defined external ampilfier. And it is designed that way using transistors.

And when the instructor claims infinite causality of gain by the added feedback short, it is an incorrect claim in physics. The gain is always limited by the transistor's hfe.

So the defined op-amp is an abstract theory of its specific circuit.

Reply to
Douglas Eagleson

Simon, you do understand that this group has little use for students using them in lieu of doing their "homework"? Having said that, picture an amplifier with two inputs. One of these inverts the input signal and the other does not. If the output signal, or some part of it, is fed back to the inverting input, this is called negative feedback. If all of it is fed back, the gain is 1. In the history of electronics, negative feedback looms as one of big, big ideas.

Anyway, if the amplifier has lots of gain and if the entire signal is fed back to the inverting input, the result is an amplifier that follows the input with no gain. Most times, only a part of the output is fed back to the input.

The classic model used to understand and analyze op-amps is that the difference between the two inputs is zero when the amplifier is dealing with normal signals. So, if you connect a voltage divider between the output and the inverting input, you get an easy to analyze circuit that predicts the gain to be equal to ratio of two resistors. If the feedback resistor is

10,000 ohms and the input resistor is 1,000 ohms the gain is Rf/Ri or 10 in this case.

Hope that helps.

Reply to
Charles Schuler

The amp on it's own has huge gain so in normal use the two inputs are at more or less the same voltage. In the classic circuit the + input is connected to 0V so the circuit behaves as if the -ve input was also at 0V.

Now think about current flows in the circuit....

The current going through the input resistor (Rin) is Vin/Rin

The current flowing in the feedback resistor (Rfb) is Vout/Rfb

Since the input impedance of the opamp is very high these two currents are equal (it has nowhere else to go..

Vin/Rin = Vout/Rfb

Rearrange this to give

Vout/Vin = Rfb/Rin

eg Gain = Rfb/Rin

Reply to
CWatters

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