# OpAmp used as adder

Let's say this operational amplifier, wired as adder, is given.

Since the ideal OpAmp has an infinite input impedance, there won't be any current flow in the (-)-input. How can I calculate the potential at the (-)-input ?

I know how to derive $U_A = -R\:(U_1/r_1 + U_2/r_2)$ but consider this: $U_- = U_A-R \:I_K = U_A - R\: (I_1 + I_2) = U_A - R\: (U_1/r_1 + U_2/r_2) = 0$

So, according to my calculations, we basically have a ground there. But how can the OpAmp amplify a potential difference if $\Delta U = U_+ - U_- = 0$ ? Where does it get its information?

• 0 * infinity is not definable as a single value (it has infinite possible values) - The concept of 'ideal op amp' like any other ideal component allows for easy modelling and simplifying calculations based upon certain assumptions such as the 'open loop gain approaches infinity' ( i.e. gets very large) the difference in voltage between the two inputs gets very small (i.e. tends towards zero) so at the limits we get infinite gain , giving 0V difference. In practice (a non ideal op amp) doesn't quite live up to these values but 'its near enough for pit work' as we say in't North. – JIm Dearden May 13 '16 at 19:15
• The circuit being an (inverting) adder has probably nothing to do with the question. The question where does it get its information would be the same for a simple inverting amplifier, wouldn't it? – Petrus May 23 '16 at 20:43

But how can the OpAmp amplify a potential difference if $\Delta U = U_ + −U_−=0$?

It can't. Your suspicions are correct.

In the theoretical ideal op-amp the open-loop gain is infinite so the op-amp will adjust the output until the output voltage is the desired output (say, 2 V) and the $V_-$ input will then be $\frac {2}{\infty} = 0~V$.

Meanwhile, back in the real world, op-amps gain is less than $\infty$ but is still high - maybe 1,000,000. In this case to get our 2 V output the $V_-$ input will then be $\frac {2}{1M} = 2~\mu V$.

Zero? Maybe not, but "virtually"!

It gets its "information", as you call it, by virtue of the virtual ground which the op amp has to maintain. If either U1 or U2 changes, then the current through either R1 and R2 will change. That current, since the input current to the op amp is zero, must go through R which will cause the output, Ua, to change.

• Okay, so what do I need my OpAmp for? I mean, if I cut it out in the picture above, nothing would change except a missing amplification which I could add behind R . – Christian May 13 '16 at 18:27

The terminals of an ideal op amp are of infinite impedance, this means current does not flow into them or out of them. So in this diagram you would get current flow between all of the resistors. In a ideal op amp the output drives the inputs to be the same potential, but only if it is in negative feedback. The "virtual ground" or shorting the terminals is only an easy way to analyze the closed loop negative feedback of the amplifier without having to consider the signals in the time or frequency domain.

Its also worth noteing that an ideal opamp has infinite bandwidth, infinite input impedance and infinite range. A real opamp does not. A real op amp takes time to respond, the output will not track the response from the inputs instantly and will be delayed. A real opamp does not have infinitive impedance, that means some current will flow into the terminals, it is typically int the range of uA's to nA's (or even smaller) but could affect your calculation if your resistors were very large.

You don't have a ground. You have a VIRTUAL ground that is at the same potential as ground. The current going through R1 and R2 have to go somewhere (a la Kirchoff's Current Law), and it goes through R, generating a potential difference between V- and Vout.

Let's derive it, but for the sake of simplicity, lets leave out U2 and R2.

First, the op amp itself has very large gain A, and $$U_a = (V_+ - V_-) \times A$$ That's by DEFINITION! Now, $V_+=0$, as it is grounded, so $$\frac{U_a}{A}= -V_-$$

Now, ideal op amps have near zero current going into the input terminals. Let's assume zero current, and apply Kirchoff's Current Law at the negative input terminal.

$$\frac{U_1 - V_-}{r_1}= \frac{V_- - U_a}{R}$$

Substitute in for $V_-$

$$\frac{U_1 - \frac{-U_a}{A}}{r_1}= \frac{\frac{-U_a}{A} - U_a}{R}$$

Now, take the limit as A goes to infinity, and you get:

$$U_a = -\frac{R}{r_1}U_1,$$ which is the equation for the inverting amplifier.

• okay, so I have a virtual ground. How does the OpAmp get its information if $\Delta U = 0$ ? – Christian May 13 '16 at 18:16
• It's not that complicated, but you didn't really ask the question in a way conducive the get that particular answer. I suggest looking at electronics.stackexchange.com/questions/131182/… – Scott Seidman May 13 '16 at 18:29
• In short, you are multiplying an infinitesimally small difference (zero in the ideal) by a very large gain (infinite in the ideal). – Scott Seidman May 13 '16 at 18:31

Real world amplifiers it's impossible to have perfectly 0 difference between two terminals.

The model from infinite gain and infinite input impedance can be explained by the simplified internal operational amplifier circuit with two stages below.

This exact same circuit could be created with nearly any kind of transistor, BJTs are just an example of one type.

Ignore Q3 and focus on Q1 and Q2. The theory of operation for them (small signal analysis) is that you have a constant current source that tries to maintain its current flow by adjusting the voltage. If one of the inputs is slightly higher or lower than the other it will affect the flow of current causing the voltage to increase so that the other amplifier will compensate for the adjusted current flow change so that i1 is CONSTANT.

Q3 is a second stage for this applifier increasing the gain even further.

These two stages combine to be (Stage1Gain)*(Stage2Gain) == LARGE NUMBER

this is where we get the theoretically "Infinite" gain from.

Additionally, the zero current flow from the terminals comes from the usage of transistors that allow virtually ZERO (imagine mosfet or jfet transistors) current to flow through them.