# Simple question about op amps

Discussion in 'General Electronics Chat' started by blah2222, Aug 8, 2011.

1. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
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36
Just a curious question: Is there a real difference between the inputs of an op-amp functionally (noninverting or inverting) and why are they given those names?

Thanks,
JP

2. ### #12 Expert

Nov 30, 2010
17,842
9,179
When the non-inverting input of an op-amp is at a higher (more positive) voltage than the inverting input, the output goes positive. The amp does not invert the stimulus. Positive in, positive out.

When the inverting input is at a more positive voltage, the amp outputs a more negative voltage. Thus, the stimulus is inverted.

3. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
573
36
Uh, isn't the voltage ideally supposed to be roughly the same between both inputs with zero current going into them?

I don't know how that would affect the polarities of the outputs with such small differences in voltages at the inputs.

4. ### #12 Expert

Nov 30, 2010
17,842
9,179
When an opamp is operating correctly in closed loop mode, there is very little voltage difference between the input terminals, usually millivots to microvolts. Opamps don't always work in closed loop mode. It depends on how you design the circuit around them. Still, the logic of what I said applies. You can work with those ideas to determine how to design the feedback loop for closed loop operation. If the input change logically dictates that the opamp will respond by making its output more positive, you design the feedback loop in a way that the positive going output will be able to satisfy the opamps need to get its inputs to resemble each other, and the other way around.

As for how tiny input differences cause change? Opamps have voltage amplification factors in the millions. A millivolt of difference can cause a mathematical result of more than a thousand volts of output. Of course the opamp can not output a thousand volts unless you provide a thousand volts for it to work with, and there is no opamp chip that can survive a thousand volts. You only calculate the results and deal with the limitations of survivable voltage.

5. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
573
36
Hm, now it seems like I'm more confused. What is a open loop or closed loop system for an op amp?

I really don't see how a slight voltage difference could change the polarity of the output voltage.

6. ### #12 Expert

Nov 30, 2010
17,842
9,179
A slight difference in the input voltage doesn't change the polarity of the output. The polarity of the slight difference changes the polarity of the output.

Open loop means the opamp is not able to satisfy the condition that the two inputs be very near the same voltage. When the opamp is in open loop condition, its output will be either as far positive or as far negative as it can go, considering the supply voltages it has available.

When the opamp is in closed loop mode, its output has a circuit connected in a way that allows the amp to achieve making its inputs very close to the same voltage as each other.

7. ### MrChips Moderator

Oct 2, 2009
14,298
4,198
You are referring to an ideal opamp.

An ideal opamp will have infinite gain, infinite input impedance, zero output impedance and infinite bandwidth.
Since the input impedance is infinite, the input current would be zero.

A real opamp may come pretty darn close with very high gain and very high input impedance and very low input current. As such it would be very unstable and not of much use. This is what is referred to as the open-loop condition.

In a real application we limit the gain by adding negative feedback using a resistor network from the output to the inverting input. Thus we close the loop. This way we can choose the gain we want. Hence the voltage difference between the non-inverting input and the inverting input becomes amplified by the gain while respecting the polarity of the voltage difference. Hence it is called a differential or difference amplifier.

Jan 28, 2005
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9. ### ErnieM AAC Fanatic!

Apr 24, 2011
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The names are anything but arbitrary. An op amp has an output of:

Vo = Av * (V+ - V-)

Where V+ is the voltage at the + input and V- is the voltage at the - input, Vo is the output voltage and Av is the amplifier's gain.

It is this difference of input voltages that is amplified. When you are computing (A-B) then which is A and which is B is important.

Dec 26, 2010
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In most if not all op-amp circuits, making the connections to the inverting and non-inverting inputs the wrong way round will prevent proper operation. People have been known to get the two inputs confused while building a circuit, so that it refuses to work.

If you have not yet got to grips with the concept of a closed-loop system, you may have to take this largely on trust, but it really is a mistake on a par with turning a steering wheel the wrong way.

11. ### JMac3108 Active Member

Aug 16, 2010
349
66
Here is another way to see it.

All of the ideal op-amp properties that you are reading about, such as the inverting and non-inverting inputs being at the same voltage, and zero input current, these only apply when the op-amp is used in a closed-loop configuration.

Closed-loop means that there is feedback. You can recognize this by seeing some component (most often a resistor) connecting between the output of the op-amp and the inverting input. Look up the standard op-amp inverting amplifier and non-inverting amplifier circuits and you will see that they each have a resistor from the output of the op-amp back to the inverting input.

Now, here is an easy way to mentally picture what is going on with the feedback and how it makes the op-amp inputs equal.

The op-amp will drive its output to whatever voltage is required to make the inputs equal.

Now, in an open-loop configuration, where the op-amp has no feedback, nothing is holding the inputs equal. Since an op-amp has very high open-loop gain, a very small voltage difference at the input will drive the output one way or the other, depending on which input is higher. If the inverting input is higher, the op-amp output drives low. If the non-inverting input is higher, the op-amp output drives high.

So, to get to the heart of your question, yes the inverting an non-inverting inputs are different and it does matter.

Hope this helps.

blah2222 likes this.
12. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
573
36
I was just getting confused because the oscillators too have feedback components, except on their non-inverting inputs. They look identical though:

Inverting Amplifier

Schmitt Trigger - 1

---------------------------------------------------------------

Non-Inverting Amplifier

Schmitt Trigger - 2

13. ### Audioguru AAC Fanatic!

Dec 20, 2007
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Your Schmitt-trigger opamp circuits are not oscillators.
To oscillate they need a capacitor at the inverting input to ground and a resistor from the output to charge and discharge the capacitor.

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14. ### MrChips Moderator

Oct 2, 2009
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Here is how I look at opamps:

This is the basic differential amplifier configuration.

Opamps use feedback, positive and negative feedback.
Observe that in this circuit, R2 provides negative feedback from the output pin to the inverting input.

Negative feedback provides many benefits, mainly, it reduces the open-loop gain to something manageable and it stabilizes the circuit.

Using this basic configuration, you may choose to apply your signal input to either V1 or V2 while connecting the remaining input to a constant voltage such as GND or a voltage reference. Note whether you choose V1 or V2 will affect the sign of the output.

Applying positive feedback causes the opposite effect. A small amount of positive feedback is used to create a comparator with hysteresis or a Schmitt-trigger. With sufficient positive feedback the circuit will oscillate.

15. ### Audioguru AAC Fanatic!

Dec 20, 2007
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Absolutely NOT!
It will not oscillate, instead the output will latch at DC as high as it can go or as low as it can go. See my previous post about the circuit for a Schmitt trigger.

16. ### MrChips Moderator

Oct 2, 2009
14,298
4,198
Sorry, I didn't mean to contradict you. You are correct. With a bit of delay in the feedback the circuit will oscillate.