opamp model


by jd12345
Tags: model, opamp
jd12345
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#1
Jan28-13, 10:16 AM
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An ideal opamp is said to have infinite gain and high input resistance.
But why do the designers want to have infinite gain? This makes it impossible to use it as differential amplifier as even small voltages will have very large values. We have to add a negative feedback to use the opamp properly

why cant they just design the opamp with lesser gain so thats its easier to use without any feedback
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f95toli
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#2
Jan28-13, 10:21 AM
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An op-amp designed to be used for differential signals without feedback would be a instrumentation amplifier...They've been around for a very long time.

Note that you can build a "discrete" instrumentation amplifier using a couple of op-amps.
jd12345
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#3
Jan28-13, 12:17 PM
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I dont understand how making the gain incredibly high and then adding feedback to compensate it help.
Can you explain

skeptic2
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#4
Jan28-13, 12:37 PM
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opamp model


Here is a basic reason. Suppose you want to use the opamp as an impedance buffer. You have a high impedance source that you want to drive a low impedance load. The opamp has a gain of 10. You connect the source directly to the non-inverting input and connect the output directly to the inverting input. The output will be equal to the voltage at the non-inverting input plus the gain times the difference between the two inputs. So with a voltage of 5V on the non-inverting input and a gain of 10, what will the output voltage be? It will be 4.5V - not a very good buffer.

But if the gain is 100K, then the output voltage will be 4.99995 volts.
skeptic2
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#5
Jan28-13, 12:49 PM
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I note that when I set the cruise control on my car exactly at 50mph, the car's speed stabilizes at 49mph. Does that mean that the gain in my cruise control is 50?
jim hardy
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#6
Jan28-13, 01:22 PM
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I dont understand how making the gain incredibly high and then adding feedback to compensate it help.
There's plenty of tutorials around
http://www.electronics-tutorials.ws/opamp/opamp_2.html

the mental exercise of working feedback in your head needs to become second nature.

Let's use skeptic's cruise control as an example. (Edit - sorry skeptic i thought that was OP- my face is red , old jim)
One toe power on accelerator pedal controls maybe five hundred horsepower, if you have an old Plymouth hemi from early 60's.
That's a lot of gain.
Your cruise control has two inputs, measured speed and desired speed, and one output, engine power.
Engine power becomes speed after overcoming friction and inertia. That's the system being controlled.

Measuring the system's output(speed) and feeding it back to accelerator pedal keeps engine at just right power to match car's need for power, probably thirty hp or so.
Result is that measured and desired speed are held equal.

It is the duty of the circuit designer to surround the op-amp with a circuit that enables it to hold its inputs equal.

Now you have at the math and see why steady state error is 1/( open loop gain).
High open loop gain gives small error, and small error is good.. That's why it is used.
the_emi_guy
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#7
Jan28-13, 02:01 PM
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Quote Quote by jd12345 View Post
An ideal opamp is said to have infinite gain and high input resistance.
But why do the designers want to have infinite gain?
The idea of infinite gain op amp is to allow gain with feedback to be set arbitrarily and precisely with external resistors. In other words the gain depends largely only on resistor values which can be very precise.

If I want an amplifier with a gain of, say, 20, I could design a custom transistor circuit that has an open loop gain of 20 and has no feedback. Problem is:
1 - I would only be able to use it at a gain of 20.
2 - Its gain would vary over temperature, process, and applied voltage.
jd12345
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#8
Jan28-13, 09:01 PM
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Got it. Thank you guys
rbj
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#9
Jan28-13, 09:57 PM
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Quote Quote by jd12345 View Post
I dont understand how making the gain incredibly high and then adding feedback to compensate it help.
Can you explain
the reason they do that is so that the parameters of the op-amp circuit depends virtually entirely on the parts around the op-amp and not the op-amp itself. that way it doesn't matter if the op-amp has a gain of 100,000 or 150,000 or 90,000, the behavior of the circuit will not be much different if the surrounding components are the same.
jd12345
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#10
Jan29-13, 05:26 AM
P: 260
Ok so this way the opam functioning only depends on the parts surrounding it.

But I dont understand one thing - in inverting amplifier the voltage gain is -r2/r1. We dont really need an amplifier to achieve this result right? We can directly connect input voltage to output voltage with some resistance combination to achieve the same result.
I think i am missing some concept
NascentOxygen
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#11
Jan29-13, 05:30 AM
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Quote Quote by jd12345 View Post
But I dont understand one thing - in inverting amplifier the voltage gain is -r2/r1. We dont really need an amplifier to achieve this result right? We can directly connect input voltage to output voltage with some resistance combination to achieve the same result.
I think i am missing some concept
I think you can answer this yourself. Step 1: sketch the resistor network that you propose will provide a gain of x10 (or is it -10; you choose). Step 2: write some equations based on Ohm's or Kirchoff's laws. Step 3: determine Vo/Vin
NascentOxygen
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#12
Jan29-13, 05:40 AM
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Quote Quote by jd12345 View Post
I dont understand how making the gain incredibly high and then adding feedback to compensate it help.
Can you explain
Don't forget that the feedback network can include reactance (e.g., capacitors) as well as resistance, so the feedback can be frequency dependent, thus allowing you to tailor the gain to have peaks and troughs (i.e., the gain varies as the frequency varies). The feedback can also include non-linear elements, allowing a designer to invent non-linear circuits.
jd12345
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#13
Jan29-13, 07:18 AM
P: 260
I couldnt really come up with a circuit that has voltage gain as the opamp but i am pretty sure you can get the same result using some resistors and voltage dource. You dont need an opamp with feedback to achieve that.
Can you please tell me an application of inverting opamp
jim hardy
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#14
Jan29-13, 12:05 PM
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Quote Quote by jd12345 View Post
I couldnt really come up with a circuit that has voltage gain as the opamp but i am pretty sure you can get the same result using some resistors and voltage dource.
That's a voltage divider.

You dont need an opamp with feedback to achieve that.
Can you please tell me an application of inverting opamp
try and build a voltage divider with gain > 1.
jd12345
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#15
Jan29-13, 01:32 PM
P: 260
Oh yeah ok silly me!
So let me see if I got it right - the characteristics of an opamp help us to gets its output depend only on the circuit surrounding and not its internal working, temperature etc.
NascentOxygen
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#16
Jan29-13, 06:25 PM
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Quote Quote by jd12345 View Post
So let me see if I got it right - the characteristics of an opamp help us to gets its output depend only on the circuit surrounding and not its internal working, temperature etc.
That's the general idea. But in the real world everything is a compromise, so you still need to keep in mind the limitations of the op-amp itself on your overall design.

An op-amp also allows you to realize functions that cannot be constructed using a network of passive components alone (including, for example, negative resistance).


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