When the difference input voltage of Op Amp is equal to zero

Discussion in 'General Electronics Chat' started by Hassan mahmoud, Jan 23, 2016.

1. Hassan mahmoud Thread Starter New Member

Jan 23, 2016
19
0
I'm confused, when I assume that the difference input voltage of the op amp is equal to zero? I asked a friend and he told me that it's always zero. But Assume this: if we have an input on inverting pin equals to 5 V and an input on non inverting input equals to 10 V how we could say that the difference equals to zero? it should be 5 !

2. Papabravo Expert

Feb 24, 2006
10,340
1,850
You must distinguish between an opamp in the open loop configuration with a gain in excess of 100,000 where a millivolt of difference will send the output to the supply rail, and an opamp in the closed loop configuration where the output always moves in such a way as to make the input difference equal to zero. Which one are you talking about. As always a schematic diagram is worth 1000 words.

3. Alec_t AAC Fanatic!

Sep 17, 2013
5,979
1,138
He's wrong.
Opamps are often (not always) used with negative feedback, the result of which is that the output self-adjusts until the inverting and non-inverting inputs are approximately equal. The greater the opamp inherent (open-loop) gain the closer the two inputs will be. However, there's no rule saying you have to use negative feedback .

4. Hassan mahmoud Thread Starter New Member

Jan 23, 2016
19
0
I'm just making sure that the input voltage difference is approximately zero if and only if the op amp has a negative feedback. In case the saturation mode and positive feedback the input voltage difference will not be equal to zero, unless the V+ and V- have the same value. Is this true?

5. Hassan mahmoud Thread Starter New Member

Jan 23, 2016
19
0
What if there are a negative and positive feedback? shall we consider the difference equal to zero? Or we should determine which feedback "dominate" the other?

File size:
36.7 KB
Views:
4
6. Lestraveled Well-Known Member

May 19, 2014
1,957
1,218
When speaking of a op-amps input differential voltage being zero due to negative feedback, you must accept that there are errors due to gain and offsets that make the differential voltage not exactly zero, but very very close.

In the example circuit in post #5, the differential voltage will still resolve to zero if it can, based on the actual component values. The positive feedback may be high enough to destabilize the op-amp.

absf likes this.
7. dannyf Well-Known Member

Sep 13, 2015
2,196
417
He's mostly correct. They are cases where the two are difrerent: in an open loop configuration (including phase errors in a negative feedback), and being driven into clipping for example.

8. ErnieM AAC Fanatic!

Apr 24, 2011
7,442
1,628
Feedback be dammed, here is why the inputs of an op amp are assumed to be "equal or very close" in many cases.

Take a traditional op amp with a +/- 15 volt power supply. We will look at it when the output is NOT in saturation, or the output is in the range of
-15V <= Vout <= +15V

The output of an op amp is given by the equation:

Vout = Av * (V+ - V-)
where Vout is the output voltage, V+ the + input, V- the - input, and Av is the open loop gain.

Now even with an old fashioned 741 type the open loop gain is 50,000. So let us see what is going on at the input for the largest signal the amp can make:

Vout = +15V = 50,000 * (V+ - V-)

(V+ - V-) = 15V / 50,000 = 0.3 mV

So the largest signal the amp can make is produced by a voltage difference of only 0.0003 volts.

Hey, that's damn close to zero so let's just assume it really is zero and get some work done.

absf and anhnha like this.
9. Papabravo Expert

Feb 24, 2006
10,340
1,850
Don't see how you can do useful work with an opamp in the open loop configuration. Maybe you're just more creative than I am.

10. Lestraveled Well-Known Member

May 19, 2014
1,957
1,218
It's called a comparator.

absf and ErnieM like this.

Apr 5, 2008
15,806
2,389
Hello,

Also keep in mind that opamp can have an offset.
In the datasheet it is often given for some temperatures, as it can be temperature related.

I have here the specs of the old 741 for the offset:

Bertus

absf likes this.
12. WBahn Moderator

Mar 31, 2012
18,089
4,917
You make this assumption whenever you are assuming you are working with an ideal opamp that operating in its active region (i.e., is not saturated).

If you apply 5 V on the inverting pin and 10 V on the non-inverting pin, then you have an opamp that is saturated.

13. Papabravo Expert

Feb 24, 2006
10,340
1,850
If you need a comparator you use a comparator -- not an opamp. The output is inherently digital and the open collector output is arguably more useful.

14. #12 Expert

Nov 30, 2010
16,705
7,355
See me stay out of this one? I've been in this argument before. Op-amp input voltages seem to bring out the OCD in people as if nobody phrases their definition exactly the same as anybody else, therefore they are wrong and need to be corrected. Been there, done that, choose to merely design circuits using op-amps instead of arguing about it.

Salute to ErnieM in post #8

Edit: Misspelling

Last edited: Jan 23, 2016
ErnieM likes this.
15. WBahn Moderator

Mar 31, 2012
18,089
4,917
From a practical design standpoint, you assume it is close enough to zero to be called zero as long as the opamp is in the active region. We made this assumption even when designing circuits on IC's using really, really crappy opamps that had an open-loop gain of less than ten at times. But of course we back-up the design computations with simulation results and sometimes (not all the time) had to tweak component values to reflect the very finite gain.

16. Lestraveled Well-Known Member

May 19, 2014
1,957
1,218
Generally I agree, but I have used an op-amp as a comparator because I had spare op-amps in a circuit. There is no law that says you can't use an op-amp as a comparator. You just have to understand what you are doing.

17. Lestraveled Well-Known Member

May 19, 2014
1,957
1,218
Yup, there is a lot of OCD in this thread.

ErnieM and #12 like this.
18. hp1729 Well-Known Member

Nov 23, 2015
2,105
235
Consider an inverting amplifier with the non-inverting input at ground. As you apply a voltage at the input resistor it upsets the balance of the differential amplifier. This error signal is what is amplified. The feedback resistor is selected to drive the output to the necessary voltage to bring the inverting input equal to the non-inverting input.
Differential amplifier. We typically have a constant current source feeding two lines. In each line is a transistor that is our inputs (inverting and non-inverting). At the other end is a current mirror. Current through one side mirrors current through the other side. Non-inverting side is the mirror of the inverting side. With no input the current through both legs are equal (ideally). When we apply a voltage across the inputs this balance is upset and an "error signal" is generated. This is our signal.

We can put the constant current source and the current mirror at either end of the differential amplifier.

• Design 610 Transistor Op Amp Circuits.pdf
File size:
8.1 KB
Views:
4
absf likes this.
19. ErnieM AAC Fanatic!

Apr 24, 2011
7,442
1,628
Hey, it's no where as bad as when we start trying to figure out what the term "current" means.

20. dannyf Well-Known Member

Sep 13, 2015
2,196
417
For dc applications, the reason that the input differential is / approaches zero is simple: high gain.

The opamp doesn't know if it is in an open loop or a closed loop circuit. All it does is to take the difference between the non-inverting input and the inverting input and amplified it, at a very high gain. The opamp's gain is always the same, regardless of it is open loop or closed loop.

In a closed loop application, the output is fed back to the inverting input so as the output goes up / down, the potential on the inverting input goes down / up. That reduces the potential difference between the non-inverting input and the inverting input.

That process would not stop until the difference is sufficiently small, or practically zero.

Thus, your friend is "mostly correct".