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I am trying to derive the gain of the inverting op-amp stage from feedback theory. I have done it for the non-inverting stage.

I'm having some difficulty - how do I deal with the Vin in the feedback factor for the inverting stage?

enter image description here

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3 Answers 3

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In both cases, non-inverting and inverting, the feedback factor \$\beta\$ is the same, because it is a measure of what fraction of a change in output is returned to the error-amplifier's input. A changing input does not feature in the calculation of \$\beta\$.

I'll try to explain this intuitively, and then I'll demonstrate the idea more rigourously.

I think it's more instructive to start with a general case, which encapsulates both inverting and non-inverting behaviours, and then adapt this general case to one or the other. This is the general op-amp configuration, with two input nodes P and Q, and output node Z:

schematic

simulate this circuit – Schematic created using CircuitLab

You have a single value for \$\beta\$ here, even though there are two inputs able to change. To start with I'll just treat the circuit using the well known input/output relationship, and I'll deal with \$\beta\$ in a moment. Here, the relationship between input potentials \$V_P\$ and \$V_Q\$, and output potential \$V_Z\$ is:

$$ \begin{aligned} V_Z &= \overbrace{+V_P\left(1+\frac{R_2}{R_1}\right)}^\text{non-inverting} \ \overbrace{-V_Q\frac{R_2}{R_1}}^\text{inverting} \end{aligned} $$

If you held \$V_P\$ fixed at some arbitrary potential, the left non-inverting term becomes constant, representing some fixed offset \$V_{OFS}\$:

$$ \begin{aligned} V_Z &= V_{OFS} -V_Q\frac{R_2}{R_1} \end{aligned} $$

If that offset is zero (node P grounded, so \$V_P=0\$), then you have the classic inverting amplifier behaviour:

$$ \begin{aligned} V_Z &= -V_Q\frac{R_2}{R_1} \end{aligned} $$

Holding \$V_Q\$ fixed causes the right-hand "inverting" term to become a constant offset:

$$ \begin{aligned} V_Z &= V_P\left(1+\frac{R_2}{R_1}\right) - V_{OFS} \end{aligned} $$

If node Q is grounded, so that \$V_Q=0\$, that offset term becomes zero, and you have the familiar non-inverting relationship between \$V_P\$ and \$V_Z\$:

$$ \begin{aligned} V_Z &= V_P\left(1+\frac{R_2}{R_1}\right) \end{aligned} $$

That last equation can be expanded:

$$ \begin{aligned} V_Z &= V_P + V_P\frac{R_2}{R_1} \end{aligned} $$

In this form it becomes clearer that the amplifier is still multiplying \$V_P\$ by \$\frac{R_2}{R_1}\$, but that input \$V_P\$ is itself offsetting the output, by amount \$V_P\$. Notice that there is no dependence of \$V_P\$ on \$V_Z\$, and in that sense the lone term \$V_P\$ may not be considered part of any feedback. It's just an offset.

To put it another way, the amplifier's feedback fraction \$\beta\$ is the same is it was in the inverting configuration, but now the "zero reference" level is non-zero, moving up and down as \$V_P\$ changes. This is what gives rise to the extra \$+1\$ in the gain expression \$1+\frac{R_2}{R_1}\$.


That's the intuition, but now for some proper rigour. The typical depiction of a servo system, consisting of summing, gain and feedback elements looks like this:

schematic

simulate this circuit

This is a good block representation of the non-inverting op-amp configuration:

schematic

simulate this circuit

You already showed the correct derivation and application of \$\beta\$ to the relationship between \$V_Z\$ and \$V_P\$, for the non-inverting configuration, but that arrangement assumes that the potential at Q is fixed at \$V_Q=0\$.

If you wish to change the potential at Q, as one can do in the first schematic above, then the block diagram must be altered to account for this variability in \$V_Q\$:

schematic

simulate this circuit

Block F provides feedback as a function of both output \$V_Z\$ and input \$V_Q\$. The entire system behaves like this:

$$ \begin{aligned} V_Z &= A(V_P - V_F) \\ \\ &= A(V_P - F(V_Z, V_Q)) \\ \\ \end{aligned} $$

Instead of a performing a simple division, now the resistor divider formed by R1 and R2 has varying potential \$V_Z\$ at one end, and a varying potential \$V_Q\$ at the other. \$V_F\$ becomes:

$$ \begin{aligned} V_F &= F(V_Z, V_Q) \\ \\ &= V_Q + (V_Z - V_Q)\frac{R_1}{R_1+R_2} \\ \\ &= V_Q\left(1 - \frac{R_1}{R_1+R_2} \right) + V_Z \frac{R_1}{R_1+R_2} \\ \\ \end{aligned} $$

We still define \$\beta\$ as we did for the non-inverting amplifier:

$$ \beta = \frac{R_1}{R_1+R_2} $$

Now we have:

$$ \begin{aligned} V_F &= V_Q\left(1 - \beta \right) + V_Z \beta \\ \\ &= V_Q - \beta V_Q + \beta V_Z \\ \\ \end{aligned} $$

The overall relationship between inputs \$V_P\$, \$V_Q\$ and output \$V_Z\$ is now:

$$ \begin{aligned} V_Z &= A(V_P - V_F) \\ \\ &= A(V_P - V_Q + \beta V_Q - \beta V_Z) \\ \\ \end{aligned} $$

At this point we can modify the summing block to operate upon all those terms individually, which will permit us to make a rather surprising discovery:

schematic

simulate this circuit

You'll notice that as I stated right at the top, feedback is considered only as some fraction \$\beta\$ of change in output \$V_Z\$, irrespective of what's happening at the inputs.

This new block diagram illustrates that \$V_Q\$ and \$\beta V_Q\$, being independent of the current state of output \$V_Z\$, are effectively removed from the feedback loop. In other words, they aren't, ultimately, playing any role in feedback at all. They will still appear in the overall transfer function, just not in the role of feedback.

Continue working to obtain \$V_Z\$ as the subject:

$$ \begin{aligned} V_Z &= A(V_P - V_Q + \beta V_Q - \beta V_Z) \\ \\ V_Z + A\beta V_Z &= A(V_P + V_Q(\beta - 1)) \\ \\ V_Z &= \frac{A(V_P + V_Q(\beta - 1))}{1 + A\beta} \\ \\ &= \frac{V_P + V_Q(\beta - 1)}{\frac{1}{A} + \beta} \\ \\ \end{aligned} $$

Find the limit as \$A \rightarrow \infty\$:

$$ \begin{aligned} V_Z &= \lim_{A \rightarrow \infty}{ \frac{V_P + V_Q(\beta - 1)}{\frac{1}{A} + \beta} } \\ \\ &= \frac{V_P + V_Q(\beta - 1)}{\beta} \\ \\ &= \frac{1}{\beta}V_P + \frac{\beta - 1}{\beta}V_Q \\ \\ &= \frac{1}{\beta}V_P - \left( \frac{1}{\beta} - 1 \right) V_Q \\ \\ \end{aligned} $$

Plug in \$\beta = \frac{R_1}{R_1+R_2}\$:

$$ \begin{aligned} V_Z &= \frac{R_1+R_2}{R_1}V_P - \left( \frac{R_1+R_2}{R_1} - 1 \right) V_Q \\ \\ &= V_P \left( 1 + \frac{R_2}{R_1} \right) - V_Q\frac{R_2}{R_1} \\ \\ \end{aligned} $$

That's the very equation I started with, and you can proceed to tie either P or Q to ground (or any fixed potential), to obtain inverting or non-inverting behaviour, but I hope I've made clear that in either case, the feedback factor \$\beta\$ is the same:

$$ \beta = \frac{R_1}{R_1+R_2} $$


To summarise, then, for this circuit:

schematic

simulate this circuit

Feedback factor \$\beta\$ is:

$$ \beta = \frac{R_1}{R_1+R_2} $$

Output \$V_Z\$ is: $$ \begin{aligned} V_Z &= \frac{1}{\beta}V_P - \left( \frac{1}{\beta} - 1 \right) V_Q \\ \\ &= \left( 1 + \frac{R_2}{R_1} \right)V_P - \frac{R_2}{R_1}V_Q \\ \\ \end{aligned} $$

To obtain an inverting amplifier, tie P to ground, so that \$V_P = 0\$, yielding this relationship:

$$ \begin{aligned} V_Z &= -\left( \frac{1}{\beta} - 1 \right) V_Q \\ \\ &= -\frac{R_2}{R_1}V_Q \\ \\ \end{aligned} $$

To obtain a non-inverting amplifier, tie Q to ground, so that \$V_Q = 0\$, yielding:

$$ \begin{aligned} V_Z &= \frac{1}{\beta}V_P \\ \\ &= \left(1 + \frac{R_2}{R_1} \right) V_P \\ \\ \end{aligned} $$

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For the non-inverting amplifier, we have this situation:

enter image description here

Or we could use a block diagram:

enter image description here

So now we can find the closed-loop gain (\$A_{CL}\$) of a non-inverting amplifier.

$$V_{OUT} = V_B * A_{OL}$$ (1)

$$V_B = V_{IN} - V_C $$(2)

$$V_C = V_{OUT}*β $$ (3)

And now we can calculate the close loop gain, substituting 2 into 1

$$V_{OUT} = V_B * A_{OL} = (V_{IN} - V_C)*A_{OL} = V_{IN}*A_{OL} - V_C*A_{OL}$$

Now we use the third equation:

$$V_{OUT} = V_{IN}*A_{OL} - V_{OUT}*β *A_{OL}$$

$$V_{OUT} + V_{OUT}*β *A_{OL} = V_{IN}*A_{OL} $$

$$V_{OUT}(1 + β *A_{OL}) = V_{IN}*A_{OL}$$

$$V_{OUT} = \frac{V_{IN}A_{OL}}{(1 + β *A_{OL})}$$

And finally the gain

$$A_{CL} = \frac{V_{OUT}}{V_{IN}} = \frac{A_{OL}}{(1 + A_{OL}*β)}$$

Now we can try to do the same thing for inverting amplifier

enter image description here

We can apply the superposition principle thus, we can create this block diagram.

enter image description here

And we can calculate the closed loop gain by inspection because we already know the gain of the right part.

$$A_{CL} = - \frac{R_2}{R_1 + R_2} \times \frac{A_{OL}}{(1 + A_{OL}*β)} $$

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schematic

simulate this circuit – Schematic created using CircuitLab

Because (a) the input voltage does not arrive DIRECTLY to the input opamp and - at the same time - the classical feedback model applies to the differential opamp input, we can apply the superposition theorem, which says that the input voltage at the inverting terminal consists of two parts:

  • Vin*Rf/(Ri+Rf), and
  • Vout*Ri/(Ri+Rf)

Therefore we have one block called "forward transfer function" H_fw=Rf/(Ri+Rf) and the classical feedback function H_f=Ri/(Ri+Rf).

Closed-loop Gain Acl: Applying Black`s formula for negative feedback we have for an ideal opamp (open-loop gain -Aol):

Acl=-H_fw * Aol/(1 + Aol * H_f)=-Hfw/[(1/Aol)+H_f], ....and for Aol very large:

Acl=-H_fw/H_f=-Rf/Ri

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