General Formula for the inputs to an op-amp?

Question

I was wondering if there is a single formula that expresses the voltage to the inputs of an op-amp, lets say an inverting op-amp, in every case?

In the example attached, I understand how the contribution from \$ V_o\$ : \$ \frac{R_1}{R_1+R_f} \cdot V_o \$ as it is just the output voltage multiplied by the feedback. But why is the contribution from \$V_{in}\$ : \$ V_{in} \cdot \frac{R_f}{R_1 + R_f} \$? Why is it not just \$-3V\$? Is there a general formula that states:

\$V_- = V_{0} \cdot \frac{R1}{R1+Rf} + V_{in} \cdot \frac{Rf}{R1+Rf} \$

I understand that it may be on an example to example basis, but I am struggling to understand how you deduce which resistor you are meant to look at.

Answer 1

I'm not sure if this is the kind of thing you are asking for, but it might help. For the following arrangement, there is a general formula for the relationship between \$V_{OUT}\$, \$V_A\$ and \$V_B\$:

^{simulate this circuit – Schematic created using CircuitLab}

$$ V_{OUT} = V_B\left( 1 + \frac{R_2}{R_1} \right) - V_A\frac{R_2}{R_1} $$

This can be interpreted to mean that any variation in \$V_B\$ (while \$V_A\$ remains constant) will be subject to gain \$1+\frac{R_2}{R_1}\$, and any variation in \$V_A\$ (while \$V_B\$ is fixed) will be subject to gain \$-\frac{R_2}{R_1}\$.

By setting \$V_A\$ to zero, or \$V_B\$ to zero, effectively you are producing the following two configurations respectively:

^{simulate this circuit}

No doubt you recognise these as classic non-inverting and inverting amplifiers, and you can see by plugging in \$V_A=0\$ and \$V_B=0\$ into the above general equation, you obtain their well-known gain equations.

On the left, with \$V_A = 0\$ we have the non-inverting relationship:

$$ \frac{V_{OUT}}{V_{B}} = 1 + \frac{R_2}{R_1} $$

On the right, inverting, when \$V_B = 0\$ we have:

$$ \frac{V_{OUT}}{V_{A}} = -\frac{R_2}{R_1} $$

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If you are comfortable with the concept of negative feedback (via the potential divider formed by R1 and R2) causing the op-amp to produce whatever output is necessary to equalise its two inputs, then the general formula (the first equation I wrote above) is easy to derive. Consider this:

^{simulate this circuit}

The op-amp's inverting input potential is \$V_Q\$, and its non-inverting input has potential \$V_B\$. From op-amp behavior with negative feedback, we can state:

$$ V_Q = V_B $$

\$V_Q\$ is some potential that lies between \$V_A\$ and \$V_{OUT}\$, as defined by the potential divider of R1 and R2. It might help to redraw that divider like this:

^{simulate this circuit}

If you do the algebra you'll find:

$$ V_Q = V_A + (V_{OUT} - V_A)\frac{R_1}{R_1 + R_2} $$

From before, we stated that op-amp action with negative feedback will always cause \$V_Q\$ to settle at a value to equal \$V_B\$, so replace \$V_Q\$ with \$V_B\$:

$$ V_B = V_A + (V_{OUT} - V_A)\frac{R_1}{R_1 + R_2} $$

If you re-arrange that to make \$V_{OUT}\$ the subject, you will end up with the same general equation I started with:

$$ V_{OUT} = V_B\left( 1 + \frac{R_2}{R_1} \right) - V_A\frac{R_2}{R_1} $$

Answer 2

There are a few simple rules, also called the golden rules of op-amps.

In general, in a properly working circuit with feedback, the inverting and non-inverting input voltages are equal, and no current flows in or out through the input pins.

That is enough to know before applying KCL or KVL to the rest of the circuit to come up with a formula for every op-amp circuit you encounter.

Even ones with capacitors or inductors as they all are just impedances and then you can find their transfer function even based on input frequency.

So, by applying the golden rules to your example circuit, if your V+ input is -5V, it means you don't even have to calculate anything to know the V- input will be -5V too, for example in a properly working inverting amplifier.

Answer 3

Easily by applying Millmann Theorem at the inverting input : \$V_- = \frac{\frac{V_{in}}{R_1} + \frac{V_o}{R_f}}{\frac{1}{R_1}+\frac{1}{R_f}}\$ and another extra developping the expression you get the desired formula you stated.

why is the contribution from Vin: Vin * R1 / (R1 + Rf)? Why is it not just -3V?

I mean clearly there is a resistor R1 in front of the inverting input, so it takes a way a portion of the input voltage Vin, if it wasnt there yes it would be a -3V contribution from Vin

Answer 4

But why is the contribution from \$V_{in}\$ : \$ V_{in} \cdot \frac{R1}{R1 + Rf} \$? Why is it not just \$-3V\$?

First to note that the expression is not \$ V_{in} \cdot \frac{R1}{R1 + Rf} \$ but \$ V_{in} \cdot \frac{Rf}{R1 + Rf} \$.

The input voltage Vin = -3 V cannot be directly applied to the inverting input since it will override the output voltage Vo acting through the resistor Rf. Vin must act through another resistor (R1). Thus the two resistors R1 and Rf constitute a resistor summer with weighted inputs. Their "gains" are Gin = Rf/(R1 + Rf) for the circuit input and Gout = R1/(R1 + Rf) for the op-amp output. So the voltage of the summing point (op-amp inverting input) is V- = Gin.Vin - Gout.Vout. If the input voltage would be directly applied (R1 = 0), then Gin = 1 and Gout = 0, so V- = Vin.

See more about the “philosophy” behind the two basic op-amp configurations with negative feedback in another answer of mine.