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I have the following circuit with an ideal opamp and \$R_1=R_2=...=R\$

with \$U_{out}=-2U_{in}\$. If I now naivly redefine the ground like this

the behaviour of the circuit changes to \$U_{out}=-U_{in}\$

Why can't I redefine the ground like this?

(I'm aware that those circuits aren't super useful. This is just an example to analyze a circuit. I'm just a little confused that I can't choose an abitrary net as ground for this schematic.)

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    \$\begingroup\$ What are you trying to achieve? \$\endgroup\$
    – Chu
    Commented Aug 19, 2020 at 16:40
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    \$\begingroup\$ In the first case the current that comes from/goes into the output of the op-amp also goes through R1. In the second case it doesn't. \$\endgroup\$ Commented Aug 19, 2020 at 16:42
  • \$\begingroup\$ I though that I could drive the transfer function by moving the ground potential and then using the equation if the inverting amplifier. \$\endgroup\$
    – someonr
    Commented Aug 19, 2020 at 16:49
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    \$\begingroup\$ Hm, how much current does flow through R4 in your second circuit? How much voltage drop is then over R4? \$\endgroup\$ Commented Aug 19, 2020 at 16:57
  • \$\begingroup\$ If you ground the pos terminal in the second circuit, the voltage will be zero. That means zero current from gnd2 as all current goes to gnd \$\endgroup\$
    – Voltage Spike
    Commented Aug 19, 2020 at 19:39

2 Answers 2

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When you quote absolute voltages like \$U_{IN}\$, you are really referring to the difference between the potential at some node you have labelled as \$U_{IN}\$ and some other node you have labelled as "0V", or ground.

By moving your 0V ground reference point from one place to another in a circuit, all you've done is add or subtract something to all the "absolute" values you've quoted, including 0V.

This can cause a relationship like \$U_{OUT} = -2 U_{IN}\$ to seem to change in absolute terms, when actually it didn't in relative terms.

For example, take that very relationship:

$$ U_{OUT} = -2 U_{IN} $$

Imagine you move the ground point to somewhere which is \$\frac{1}{3}U_{IN}\$ higher than where it was. Now all the voltages you quote will be \$\frac{1}{3}U_{IN}\$ lower, so you have to subtract \$\frac{1}{3}U_{IN}\$ from every variable in the equation:

$$ (U_{OUT} - \frac{1}{3}U_{IN}) = -2 (U_{IN} - \frac{1}{3}U_{IN}) $$ $$\begin{aligned} U_{OUT} &= -2 (U_{IN} - \frac{1}{3}U_{IN}) + \frac{1}{3}U_{IN} \newline \newline &= -2U_{IN} + \frac{2}{3}U_{IN} + \frac{1}{3}U_{IN} \newline \newline U_{OUT} &= -U_{IN} \end{aligned}$$

Coincidence? Maybe.

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First circuit

It resembles the classic op-amp differential amplifier but without one resistor. It is interesting that all its components - R1, R2, R3, and Vin1 are connected in series to the op-amp output. So, the same output current flows through them.

Note the two op-amp inputs are equal and floating. So, when the input voltage varies, both they vary (a common-mode signal).

The op-amp will try to set zero voltage between its inputs. So the input "real voltage source" (Vin1 + R2) will be virtually shorted and Vin1 = VR2. This current flowing through R2 will be I = Vin1/R2. The same current flows through all resistors... and the output voltage Vout1 is....... Only, what is the point of this?

It would be interesting to understand the mechanism of the H&H "virtual short" between the op-amp inputs because there are no good explanations for this phenomenon. What gives the inputs short is the network of three elements in series - R1, Vout and R3. It behaves like a "piece of wire" with zero resistance because the op-amp output voltage neutralizes the total voltage drop across R1 and R3 (Vout - VR1 - VR3 = 0). Saying it in a more attractive way, the op-amp output acts as a negative "resistor" with resistance -(R1 + R3) that compensates the total positive resistance (R1 + R3) and the resulting total resistance is zero (R1 + R3 - R1 - R3 = 0).

Second circuit

The second circuit is simply an inverting ampifier with R1 = R2 = 1 k. Only there are two strange things - Vin2 is floating and R4 is between the load and ground (in series to the load). Both they do nоt make sense.

The ground

I'm only interested about why I can't choose the ground arbitrary on this schematic.

The ground has two functions (meanings). First, it is a common point for all devices (source, amplifier, load); so it sinks all currents (if negative). Second, it is a reference point when measuring the voltages inside circuits (where you connect the black probe of the voltmeter).

In your circuits, the ground is the middle point of the power supply (or the middle point of two voltage sources in series). Since the output voltage should be negative, the current will flow through the negative supplying voltage source. In the first circuit, the input current will flow through R1 while in the second circuit this current will not flow through this resistor (named R4 here). Now the load current will flow through R4 (if there is a load connected).

Simply speaking, the ground is just one of the power supply terminals (positive, negative or middle). So, when changing its place, you connect this terminal to a different point of the circuit... and this is already a different circuit.

So, the conclusion is: You can arbitrary choose only the reference point for the voltage measurement but not the common return point since this changes the current path and the circuit itself.

Applications

I'm aware that those circuits aren't super useful.

Still there is a reasonable application of the first circuit... It is a "unbalanced differential amplifier" relative to the common-mode signal but this does not matter here because the input source is floating. There is no problem to use it with a differential input voltage what Vin1 is.

What is more interesting is that this amplifier can be single-supplied. When the input voltage increases, both op-amp input voltages increase as well (acting as a common-mode input signal). Thus they stay somewhere between the rails as though the op-amp is biased.

I have seen this trick of an "auto biasing" in the circuit of a photo diode amplifier below in the H&H bestseller (see my answer). The photodiode acts as the "Vin1 + R2" network in the first OP's circuit.

Photodiode amplifier

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  • \$\begingroup\$ So essentielly the problem is that the ground of the opamp is kind of hidden? Real Opamps would have a split power supply with a ground. \$\endgroup\$
    – someonr
    Commented Aug 19, 2020 at 19:17
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    \$\begingroup\$ Op-amps do not have a "ground terminal". Their input, internal and output voltages simply vary between negative and positive rails of the power supply. Think of their internal stages as voltage dividers "stretched" between rails; so their output voltages vary between -Vee and Vcc. They can be referenced to any point inside the power supply (usually, in the middle, as you noted) that is named "ground"... or outside the supply (obtained by something like a voltage divider) that can be named "artificial ground" or "virtual ground". \$\endgroup\$ Commented Aug 19, 2020 at 20:26

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