OpAmp Inputs

Hello,

I have a question regarding operational amplifier. How does exactly the inverting and non-inverting terminals make a difference in calculations, the output voltage for example? Lets say I ignored the fact there's inverting and non-inverting inputs, so no signs provided, how should my calculations be affected? I'm asking this because I have not noticed anything that takes into account the sign of input terminals in performing calculations.

Thanks.
analoger

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Joined: Sun Dec 23, 2012 11:11 pm

Re: OpAmp Inputs

Sorry I think my question is unclear. Let me reward my question this way:

Where do we take into consideration the inverting/non-inverting input in an OpAmp terminal behavior analysis/calculations?
analoger

Posts: 78
Joined: Sun Dec 23, 2012 11:11 pm

Re: OpAmp Inputs

analoger

Posts: 78
Joined: Sun Dec 23, 2012 11:11 pm

Re: OpAmp Inputs

Hi Analoger,

Because OpAmps are such basic and versatile devices, your question is somewhat ambiguous.
analoger wrote:How does exactly the inverting and non-inverting terminals make a difference in calculations, the output voltage for example?

Equations for OpAmps vary depending on the additional components connected to the inputs & outputs.
The most common circuit configuration is possibly the inverting amplifier.
However, there are also these configurations: non-inverting, unity follower, transconductance, transimpedance, summing, difference, integrator, differentiator, peak detector, clipper (both inverting & non).
So you see, it depends.
If you care to share how you might use the OpAmp, maybe we can be more helpful.

John john444

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Location: Claremore, Ok

Re: OpAmp Inputs

Well I'm talking here very general so no specific application or configuration. What I'm asking is when deriving these formulas for a certain configuration, how does switching the (-) and (+) of the input terminals change the way we do the derivation math (KVL/KCL, nodal analysis, ...etc.)? For me, I could not find any difference, so I assume so far it's just a convention that has no meaning in studying the terminal behavior of the OpAmp. However the real physical element distinguishes between the two input terminals by (+) and (-) which makes me suspicious about my assumption.
analoger

Posts: 78
Joined: Sun Dec 23, 2012 11:11 pm

Re: OpAmp Inputs

analoger wrote:Well I'm talking here very general so no specific application or configuration. What I'm asking is when deriving these formulas for a certain configuration, how does switching the (-) and (+) of the input terminals change the way we do the derivation math (KVL/KCL, nodal analysis, ...etc.)? For me, I could not find any difference, so I assume so far it's just a convention that has no meaning in studying the terminal behavior of the OpAmp. However the real physical element distinguishes between the two input terminals by (+) and (-) which makes me suspicious about my assumption.

Switching the inverting and non-inverting terminals will not change your derivations. Using an inverting op-amp as an example, feedback may be applied to the inverting pin (negative feedback) or the non-inverting pin (positive feedback) and you will derive the same transfer function of Vout/Vin = -Rf/Rg. ImaginaryAxis

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Re: OpAmp Inputs

Perhaps this will help.

Regardless of whatever components are connected, the opamp will attempt to drive the output so that the voltage at the inverting input is equal to the voltage at the non-inverting input. Feedback is taken at the inverting input.

That one fact is probably the most important thing to understand about an op-amp and it drives all the derived equations. lyndon

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Re: OpAmp Inputs

An op amp is an active device, so KVL and KCL won't help you much. In fact, the usual practice is to assume the input impedance is infinite so it's 'invisible' to KVL (doesn't cause any voltage drop) and KCL (no current flows in).

By itself, an op amp is just a very high-gain amplifier. The voltage at the output will be the difference between IN+ and IN- times, oh, 50,000 or so. If IN+ = 1v and IN- = .5v, the output will want to be (1v - .5v) * 50,000, or 25,000v. In practice, the output will just hit VCC and stop there. If IN+ = .5v and IN- = 1v, the output will want to be (.5v - 1v) * 50,000, or -25,000v. Again, in practice the output will just go to the negative rail.

The only input values that don't slam the output into one rail or the other are the ones that are very close together.. if IN+ is one microvolt higher than IN-, the output will be (1uV * 50,000) = 50mV higher than IN+. If IN+ is 1 microvolt below IN-, the output will be 50mV less than IN+.

It's more or less impossible to keep two voltages that closely in sync independently of each other, so we use passive components around the op amp so the output controls the input.

Let's say we short the output directly to IN-.. it can't do any harm since we assume both IN+ and IN- have infinite impedance. If the voltage at IN- is less than the voltage at IN+, the output will shoot towards VCC as before, but now it will drag IN- with it. You can't go from "below IN+" to "as high as the output can go" without crossing the IN+ voltage along the way, so eventually IN- will sit at exactly the same voltage as IN+. When that happens there's no reason for the output voltage to rise any more (IN+ - IN- = 0), so the upshot is that we have a circuit that pulls IN- and OUT as close to IN+ as it possibly can.

The output will resist moving away from the value of IN+ as strongly as it can.. if you try to move the voltage at IN- up 1uV, the output will pull 50mV the opposite way. That 50,000:1 gain is now fighting any attempt to move IN- away from the value of IN+.

Now let's make things a little more interesting: instead of shorting the output to IN-, let's put a 1k resistor between them. Left to itself, the circuit will still balance so IN- and OUT are exactly equal to IN+. If we pour 1mA of current into the node where the 1k resistor meets IN- though, things get interesting:

IN- has infinite impedance, so no current leaves the node through IN-. That means all the current has to leave the node through the resistor, and the far end of the resistor is tied to OUT. 1mA through 1k of resistance produces 1v, so if OUT doesn't move, the voltage at IN- will rise by 1v. We've already established that OUT is 50,000 times easier to move than IN- though, so OUT will drop until IN- is exactly equal to IN+ again.. in this case, 1v below IN+.

Now, I sort of hand-waved that 1mA into existence, so let's introduce an actual component: let's connect one end of a 100 ohm resistor to the node where IN- meets the 1k feedback resistor. To get 1mA through it, we need a voltage drop of 100mV, so the free end has to sit 100mV higher than IN- (which equals IN+). If we lower the free end to 100mV below IN+, we'll get 1mA flowing out of the IN- node, and the only thing that can supply that current is the 1k feedback resistor. OUT will rise until IN- equals IN+ and the currents flowing through both resistors are equal, and crunching those numbers puts OUT 1v above IN+. Whenever the free end of the 100R input resistor changes, the output of the op amp will move ten times as far the opposite way.. making the circuit an inverting amplifier with a voltage gain of 10.

So: by itself an op amp does little more than compare the voltage of IN+ and IN- and send the output to whichever rail leaves IN+ between IN- and OUT. The components around the op amp are what make it do interesting things. Those external components do obey KVL and KCL, but you need the assumption "if there's feedback to IN-, the voltage at IN- will always be the same as the voltage at IN+" to make the math work.
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