A question regarding electronic negative feedback

[Faiq]

I have a confusion understanding one of the basic assumption used for operational amplifiers which is Vin = 0 or difference of inputs = 0.
Take for example an ideal amplifier with inputs V+ and V-
V+ = 1
V- = 0.5 * Vout
Supply Voltage = +-9
Vin = V+ - V-
At start V+ = 1 and V- = 0, providing us a Vin of 1V. Since gain is infinite, Vout = +9.
Now V+=1 and V- = 4.5, providing us a Vin of -3.5. Since gain is infinite, Vout = -9
Now V+=1 and V- = -4.5, providing us a Vin of 5.5. Since gain is infinite, Vout = +9
Now V+=1 and V- = 4.5, providing us a Vin of -3.5. Since gain is infinite, Vout = -9

My question is in this example Vin doesn't approaches 0. So can you explain how does an operational amplifier will work in this situation to make Vin = 0?

 

[Dale]

Don't set it as a discrete time series. Set it as a single equation at one moment in time and solve. Also leave the gain as a variable and then once you have the solution in terms of the gain take the limit as it goes to infinity.

 

[LvW]

My question is in this example Vin doesn't approaches 0. So can you explain how does an operational amplifier will work in this situation to make Vin = 0?

Why do you want Vin=0? This applies to an idealized (unrealistic) case only (open-loop gain Aol infinite).
If you really want to understand why - and at which voltage - the opamp settles, you must consider a real device with a finite Aol (large, but not infinite).
1.) You can start as in your example.
2.) On its way from -9V to +9V the voltage at the inv. input node tends to cross a positive value of V1=Vout/2 which fulfills the condition Vout=(1V-V1)*Aol.
This value of V1 will be slightly below the pos. input of 1V. When the difference Vd=(1V-V1) fulfills the condition Vd*Aol=Vout the system has reached its equilibrium and the system has settled.

 

[jim hardy]

My question is in this example Vin doesn't approaches 0. So can you explain how does an operational amplifier will work in this situation to make Vin = 0?

It takes a little backward thinking to get past this all too common mental stumbling block .

It is the duty of the circuit designer
to surround the opamp
with a circuit that allows the opamp to drive its inputs equal, ie to zero* difference.

(* we say zero, in reality it's Vout/(Open Loop Gain).
Open loop gain is A_Vol on datasheet, 10⁵ is a common value.

So , if output is to be less than Vs, difference at inputs must be less than Vs/A_Vol
For your case
<9/10⁵ = <0.00009 which rounds to zero.

Simplest circuit is unity gain voltage follower
where input signal goes to + input pin
and - input pin is wired directly to output pin
wherever + input pin goes, - input pin will follow as far as power supply allows.
It can only operate within the constraints of power supply (and the common mode limit s of the selected amplifier, given in its datasheet.)
Don't feel alone, i had exact same problem in 1966.

 

[Averagesupernova]

It is the duty of the circuit designer
to surround the opamp
with a circuit that allows the opamp to drive its inputs equal, ie to zero* difference.

This is one of the best statements I have ever seen that sums up (no pun intended) opamp operation. (Again, no pun intended).

 

[sophiecentaur]

It is the duty of the circuit designer
to surround the opamp
with a circuit that allows the opamp to drive its inputs equal, ie to zero* difference.

The 'Virtual Earth' amplifier does this. The circuit feedback and biasing keeps the V_- input at Zero Volts but it isn't actually connected to 'Earth'. The sum of currents in and out of the terminal is kept at zero.

 

[sophiecentaur]

@Faiq: Putting numerical values in at the start of any analysis is not useful.** You need to use Algebra. The basic equations that are used in OpAmp theory are not hard and they will tell you how the circuit actually performs. If you try putting a random selection numbers in, you end up with the output going to + or - supply voltage, which is no use. This Hyperphysics page has relevant equations.
** This is rather like trying to work out simple resistive networks without algebra. You can't tell what any particular component does until you include it in the whole picture. That will nearly always lead you into writing down an algebraic equation. (Except when you have done it so often that the answer is intuitive.)

 

[Faiq]

It takes a little backward thinking to get past this all too common mental stumbling block .

It is the duty of the circuit designer
to surround the opamp
with a circuit that allows the opamp to drive its inputs equal, ie to zero* difference.

This simplifies many of the confusions yet raises a new question. How can a circuit predict the required voltage and produce it?

 

[sophiecentaur]

This simplifies many of the confusions yet raises a new question. How can a circuit predict the required voltage and produce it?

You are talking about what happens in the short period after switch on. Fact is that is complicated because of the different time delays in different electronics components. It is only necessary to write down the steady state situation (after things have settled down**). The circuit doesn't actually need to "predict" anything (and this is a popular misconception and the cause of a lot of confusion with beginners). Currents flow through components until a steady state is reached. This, conceptually, is similar to a mechanical system of masses connected together with springs and with a bit of friction involved. The arrangement reaches a state of equilibrium with each spring taking up a length appropriate to the forces on it. It doesn't need to 'know' anything at all about the rest of the set up.
** There are feedback circuits which will never actually settle down (unstable oscillators). They contain reactive elements (capacitors and inductors) and the volts and current around the circuit keep chasing each other and never actually catch up to reach equilibrium. There is a well known saying that "Amplifiers will oscillate and oscillators will amplify - just to be awkward". Stick with the ideal static circuits to start with.

 

[LvW]

How can a circuit predict the required voltage and produce it?

Didn`t you read (or didn`t you understand) my post#3 ? It explains why and when the system has reached an equilibrium state.

 

[jim hardy]

How can a circuit predict the required voltage and produce it?

It doesn't have to predict anything.
All the circuit has to do is detect imbalance and drive toward balance.

Mother Nature loves a balance. So do operational amplifier circuit designers. I presume you're studying to become one.

I'd recommend you buy a breadboard , some LM324's, a resistor assortment , 9v battery and a voltmeter . Then you can build and troubleshoot circuits.
We learn faster by 'doing' than by 'reading about doing'.

Print yourself a copy of this document
https://www.ti.com/ww/en/bobpease/assets/AN-31.pdf
and study the clever circuits

if you get really interested , these guys "Wrote the Book" .
http://web.mit.edu/6.101/www/reference/op_amps_everyone.pdf
it's 400+ pages , so
Print and keep a hardcopy in a binder or buy a copy it'll be a lifelong resource.