I don't see how the 2nd theorem is true for an ideal operational amplifier. enter image description here

This was from the Engineering Circuit Analysis 8th Edition textbook by Hayt.

If this is correct can you somebody help me understand why? I'm looking at this model they gave me and still can see why the voltage across is zero.

enter image description here

Ri approaches infinity and Ro is zero.

  • 4
    \$\begingroup\$ This is the definition of an ideal op amp. Real op amps are not ideal, but approximate it. The statement that the voltage is zero is only valid when you have negative feedback. \$\endgroup\$
    – Hearth
    Commented Nov 1, 2018 at 0:21
  • \$\begingroup\$ Yes, however for an ideal Op Amp as the resistance approaches infinity the input terminals act more and more like a open circuit. Open circuits can have any voltage across them right? This was my way of think about it. \$\endgroup\$
    – Nick Yarn
    Commented Nov 1, 2018 at 0:23
  • \$\begingroup\$ That's what I thought, however in the book it never states negative feedback for this rule. \$\endgroup\$
    – Nick Yarn
    Commented Nov 1, 2018 at 0:25
  • 4
    \$\begingroup\$ No text book is ever ideal... \$\endgroup\$
    – user105652
    Commented Nov 1, 2018 at 0:26
  • \$\begingroup\$ zero current through a resistor will create zero voltage across the resistor \$\endgroup\$
    – jsotola
    Commented Nov 1, 2018 at 0:27

3 Answers 3


The voltage gain of an ideal opamp is infinity, so if there was any voltage difference between the inputs there would be infinite voltage at the output.

Ok, so this was probably a mistake right? They may have forgot to say for a negative feedback configuration.

The ideal opamp only results in a valid (converging) circuit if there is negative feedback. If there isn't, the output voltage is infinite, then infinite voltage across infinite resistance results in indeterminate current, etc.

  • \$\begingroup\$ Yes, but I'm asking about the voltage across the input terminals \$\endgroup\$
    – Nick Yarn
    Commented Nov 1, 2018 at 0:34
  • \$\begingroup\$ Not correct. The voltage at the output cannot exceed the supply rails. Also an open loop gain is seldom over 100,000-which is not infinite, but enough to fix the output at or close to one of the supply rails. \$\endgroup\$
    – user105652
    Commented Nov 1, 2018 at 0:34
  • 3
    \$\begingroup\$ Ideal opamps don’t have supply rails. \$\endgroup\$
    – τεκ
    Commented Nov 1, 2018 at 0:51
  • \$\begingroup\$ @NickYarn If the voltage across the input terminals was not zero, the output would have infinite voltage, which would make the circuit invalid. \$\endgroup\$
    – τεκ
    Commented Nov 1, 2018 at 0:54
  • 3
    \$\begingroup\$ So, the whole thing about this "ideal op-amp" is that it's an ideal -- as in a Platonic ideal, that floats around in whatever imaginary space Platonic ideals exist in, rubbing elbows with ideal chairs and ideal rocks and ideal spheres. It is not a real thing. It's a tool for doing the first step in solving vexing op-amp problems, and the second step starts with saying "of course, I don't have an ideal op-amp". \$\endgroup\$
    – TimWescott
    Commented Nov 1, 2018 at 1:08

Its because for most OPAMPS \$R_i=\infty\$ There are various misconceptions with ideal opamp concepts here. You need to take these things with a pinch of salt.

Firstly, Opamps were realized in earlier days with BJTs, Implying that \$R_i\neq\infty\$ for such OPAMPs Most modern OPAMPs use MOSFETs, for which this assumption is fairly true.

Secondly, The voltage difference across the input terminals for a OPAMP is zero ( is not true in general).

Take the following example,


simulate this circuit – Schematic created using CircuitLab

The Voltage difference between the two terminals of the OPAMP can be anything ( based on the external circuit).

When they say that the input difference between the two input terminals is zero. It is under the assumption that the OPAMP is being operated with negative feedback ( and you are not forcing both the voltages simultaneously, you are letting one of the terminals to decide for itself ).

To explain further, Observe the circuit below.


simulate this circuit

In the above circuit, the OPAMP is operating in negative feedback.

  1. We are forcing one terminal Vp = 1
  2. The other terminal Vm is being set by the negative feedback itself.

Now assume that the OPAMP has a gain of A ( which is very high ). And also assume that \$R_i=\infty\$

We can write the following equations

\$ (0-V_m)/R=i \$ --------- (1)

\$ (Vm-V_{out})/R=i \$ --------- (2)

Equating these two we can say that

\$ V_{out}=2V_m \$ --------- (3)

We also know that OPAMP being a differential amplifier follows the following equation

\$ V_{out}=A(V_p-V_m) \$ --------- (4)

Now if you look at (3) it says \$ V_{out}=2V_m \$

putting this in (4), we get

\$ 2V_m=A(V_p-V_m) \$

On rearranging

\$ \frac{2V_m}{A}=(V_p-V_m) \$

\$ \frac{2V_m}{A}+V_m=V_p \$

\$ (\frac{2}{A}+1)V_m=V_p \$

\$ V_m=\frac{V_p}{(\frac{2}{A}+1)} \$

For any practical OPAMP, the gain A is in the order of \$ 10^3 - 10^6 \$

If you look at the last equation and put this value of A.

You will realize that,

\$ V_m = V_p \$

This is happening due to the beauty of negative feedback. OPAMP as such has nothing to do with this.

As long as the gain of the OPAMP is very high and it is connected in a negative feedback configuration, you can assume that forcing one node ( either \$ V_p \$ and leaving the other node to fix itself by negative feedback ) will always ensure that

\$ V_m = V_p \$

  • \$\begingroup\$ Great! So an ideal Op Amp in an open loop configuration can have any voltage across the input terminals. \$\endgroup\$
    – Nick Yarn
    Commented Nov 1, 2018 at 1:30
  • \$\begingroup\$ yes it can.. However the output wont be useful. Every Opamps output voltage is limited by the Power supply of the OPAMP. If you were to give a input difference of 1v and if the gain of the OPAMP is 1000, you would expect a voltage of 1000 but in reality it can never go beyond the power supply of the OPAMP \$\endgroup\$ Commented Nov 1, 2018 at 1:44

Forget about rule 2 for a second. It's useful for doing mathematical calculations but it's important to understand how the op amp works first.

An op amp is basically the same as a comparator, if the positive input is higher than the negative input it turns on, and if it's lower it turns off.

Now look at the most basic op amp circuit, the buffer:

op amp buffer diagram

Lets say VIN is 1v. What's going to happen is that the output will go high until the (-) input reaches 1v then the output will go low. It's going to oscillate around 1v, and because the speed isn't infinite, it's basically going to get "stuck" at 1v.

That sounds unstable for a comparator right? It just wants to be on or off, not between. Well the difference between an op amp and a comparator is that op amps are designed to work in this manner, and will hold a stable output voltage.

This balance point will occur at the point where the op amp is "switching" on and off. And that occurs where + and - are equal. Therefore when you're doing math, you can start with the assumption that + and - will be equal, and figure out what value of V_out would be required for that to happen.

This is the basis for the statement "there is no voltage difference between the input terminals", as you can see it's only true if you set up the circuit in a certain way.

  • \$\begingroup\$ So when they use the term Operational Amplifier they are referring to a negative feed back configuration right? \$\endgroup\$
    – Nick Yarn
    Commented Nov 1, 2018 at 1:40
  • \$\begingroup\$ Not exactly. Op Amps are devices designed to be used in a negative feedback configuration. You can use an op amp without negative feedback and its still an op amp. Just like how you can hook up a voltage regulator backwards and it's still a voltage regulator, even though the "rules" of voltage regulators will no longer apply. \$\endgroup\$
    – Drew
    Commented Nov 1, 2018 at 2:58

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