No. The problem is not the same.
In the original circuit, there is negative feedback.
In post #18, there is no feedback.

The trick in op amp circuits is to take advantage of infinite gain and reduce the gain to something useful by using negative feedback.

 

No. The problem is not the same.
In the original circuit, there is negative feedback.
In post #18, there is no feedback.

The trick in op amp circuits is to take advantage of infinite gain and reduce the gain to something useful by using negative feedback.

so, the point here, is input voltage = 0 or we say Vp = Vn, because Vp connected to ground so that Vp = 1, Vn not connected to ground so we must calculate. how about this case I connect ground to Vn source like this below

 

Now you have no feedback and the gain is infinite.
What is Vo in this case?

 

so, the point here, is input voltage = 0 or we say Vp = Vn, because Vp connected to ground so that Vp = 1, Vn not connected to ground so we must calculate. how about this case I connect ground to Vn source like this below
View attachment 339199

You mean that output voltage = infinity*(2V-1V) this case = infinity?

 

Now you have no feedback and the gain is infinite.
What is Vo in this case?

You mean that output voltage = infinity*(2V-1V) this case = infinity?

 

You mean that output voltage = infinity*(2V-1V) this case = infinity?

Yes and no. It should be negative infinity. You’ve got the signs wrong and it’s limited by the supply voltage.

 

Yes and no. It should be negative infinity. You’ve got the signs wrong and it’s limited by the supply voltage.

Can you shortly show the wrong signs ? and how is it limited by supply volt.

 

Vo = A (Vp - Vn)

You are being sloppy.

Where is the voltage coming from?
You cannot get 100 V from a 12 V circuit.

 

Vo = A (Vp - Vn)

You are being sloppy.

Where is the voltage coming from?
You cannot get 100 V from a 12 V circuit.

I thought From equation Vout = Gain factor * ( V2-V1) = infinite*(2-1). Is something wrong here? How is it limited by supply voltage as you said? Can you shortly explain?

 

Vo = A (Vp - Vn)

You are not being careful.
A = ∞
Vp = 1 V
Vn = 2 V

As for power supply limits,
suppose this is a real circuit powered by +12 V on the positive supply rail and -12 V on the negative supply rail.

Can you see that the output voltage cannot go beyond ±12 V?

 

Vo = A (Vp - Vn)

You are not being careful.
A = ∞
Vp = 1 V
Vn = 2 V

As for power supply limits,
suppose this is a real circuit powered by +12 V on the positive supply rail and -12 V on the negative supply rail.

Can you see that the output voltage cannot go beyond ±12 V?

oh, it means that Vout depends on A and how much supply voltage put in, so use KVL for the loop with supply voltage , we have the max value of Vout can reach, is that right? like this below

 

While we are at it, here are the properties of an ideal op amp.

input impedance = ∞ (input takes zero current)
output impedance = 0 Ω (output can supply infinite current)
voltage gain = ∞ (output can supply infinite voltage)
bandwidth = ∞ (can operate at any frequency)

As an exercise, select a real op amp and compare the real specifications with those of an ideal op amp.

 

While we are at it, here are the properties of an ideal op amp.

input impedance = ∞ (input takes zero current)
output impedance = 0 Ω (output can supply infinite current)
voltage gain = ∞ (output can supply infinite voltage)
bandwidth = ∞ (can operate at any frequency)

As an exercise, select a real op amp and compare the real specifications with those of an ideal op amp.

yeah, i already simulateD with OP07

 

While we are at it, here are the properties of an ideal op amp.

input impedance = ∞ (input takes zero current)
output impedance = 0 Ω (output can supply infinite current)
voltage gain = ∞ (output can supply infinite voltage)
bandwidth = ∞ (can operate at any frequency)

As an exercise, select a real op amp and compare the real specifications with those of an ideal op amp.

oh, it means that Vout depends on A and how much supply voltage put in, so use KVL for the loop with supply voltage , we have the max value of Vout can reach, is that right? like this below

 

I am not discussing simulation. I am talking about a real device.
Go to the datasheet and write down the specifications given in the datasheet for the parameters I listed, along with the maximum supply voltages.

Get into the habit of looking up and reading real device datasheets.

 

oh, it means that Vout depends on A and how much supply voltage put in, so use KVL for the loop with supply voltage , we have the max value of Vout can reach, is that right? like this below
View attachment 339205

Is this correct btw?

 

is this equation correct?, then Vo= -K, K is gain
View attachment 339181

A voltage can't equal a gain, so you know it is wrong.

The voltage at the inverting input is only 2 V if the voltage at the top of the 2 V source is 0 V.

Since it is an ideal opamp, what is the current in the 10 kΩ resistor?

What does that say about the voltage drop across the 10 kΩ resistor?

What does that say about the voltage at the top of the 2 V source relative to the output voltage?

 

Can you read #8 from Mr @WBahn again, I follow his equation, Vo= K(Vp-Vn), Vp is voltage at non inverting terminal = 1, Vn is voltage at inverting terminal = 2. oh, you mean Vn here not 2V, I thought 2 VDC connect to inverting terminal then it is 2V.

According to this reasoning, if you put 200 1.5 V batteries in series and put the negative terminal of the end one at ground, then the voltage at the positive terminal at the other end is only 1.5 V.

Or, also by this reasoning, if you connect the negative terminal of a 12 V battery to a 10,000 V, relative to ground, then the voltage at the positive terminal of the battery is only 12 V.

 

and in the problem is the same, right? Why Cannot apply the definition of Op amp like the picture I just added? is something different between the problem and the Definition circuit in the picture I just added?

They are NOT the same. In your second picture, the voltage at the negative end of the source connected to each opamp input terminal is tied to ground (0 V). But the negative terminal of the 2 V source in your original circuit is NOT tied to 0 V, it is tied to the output of the opamp via that 10 kΩ resistor.

 

According to this reasoning, if you put 200 1.5 V batteries in series and put the negative terminal of the end one at ground, then the voltage at the positive terminal at the other end is only 1.5 V.

Or, also by this reasoning, if you connect the negative terminal of a 12 V battery to a 10,000 V, relative to ground, then the voltage at the positive terminal of the battery is only 12 V.

Can you give a picture c about what you mean in the bold line?