2

I can't find any source online which explains this in a different way from the others. I'd like to think I'm not just being slow, but it seems that I might be just that, regarding the op-amplifier.

enter image description here

So the summing-point constraint claims that \$x\rightarrow 0\$, if we assume the op-amplifier to be ideal. To me, this makes no sense when you consider the characteristics of an ideal op-amplifier.

Suppose that, due to \$V_{in}\$, a positive voltage \$V_{DIFF}\$ appears at the inverting input. Then a large negative output voltage result at the output.

\$V_{OUT}\rightarrow-\infty\$

(Here's where I don't agree with explanations anymore)

A fraction of this is sent back through \$R_F\$, meaning that \$V_{DIFF}\rightarrow 0\$ with time.

I just don't get this, if I were to use numbers to explain why I'm confused, suppose that \$V_{DIFF} = 0.1\,v\$ initially, and then \$V_{OUT}=-10\,000\,v\$. If a fraction of this is sent back, say \$0.1\%\$. Then the voltage \$V_{DIFF}=0.1+0.001\cdot V_{OUT}=-9.9\,V\$

And it continues to spiral out of control. How does it really work? Why does \$V_{DIFF}\rightarrow 0\$?

B. Lee
  • 243
  • 2
  • 6
  • 13

3 Answers3

3

suppose that \$V_{DIFF} = 0.1\,v\$ initially, and then \$V_{OUT}=-10\,000\,v\$. If a fraction of this is sent back, say \$0.1\%\$.

The "fraction" that is sent back is not really a voltage, it's a current.

If \$V_{OUT}\$ is -10,000 V, then there must flow through \$R_F\$ a current of \$\frac{10^4}{R_F}\mathrm{A}\$. Since the input impedance of the op-amp is very high (infinite in the ideal case), the resulting feedback voltage at the inverting input is determined by a voltage divider between \$R_F\$ and \$R_{in}\$. So the feedback voltage is \$V_{OUT}\frac{R_{in}}{R_F+R_{in}}\$. It's likely to be much higher than the 0.1% you assumed in your example.

How to solve for the output voltage including the forward gain and feedback terms, is explained in a previous question.

Community
  • 1
The Photon
  • 126,425
  • 3
  • 159
  • 304
2

B.Lee - for answering your question you need to fully understand the "secrets" of negative feedback. Let me try to explain using a simple example:

Let us analyze what happens after switch-on the power supplies +/- Vs=+/-10V. (The described timely sequence may be somewhat „formalistic“ - however, it helps to understand the feedback concept).

Example: Non-inverting gain stage with desired gain of "+2". That means: Feedback factor k=0.5. Open-loop gain: Aol=1E4.

1.) t=0: Apply at an input voltage Vin=1V. The opamp is not yet working in its linear range (feedback not yet active due to time constants within the circuit) and the output will immediately jump to Vs=+10V.

2.) t>0: The voltage at the inverting terminal will rise to 0.5Vs=5V>Vin=1V. Hence, the voltage at this inverting terminal dominates (is larger) and the output voltage will change in the direction to minus 10V.

3.) However, on its way to -10V the ouput voltage is crossing a positive value which produces at the inverting terminal a feedback voltage of +0.99980004V .

4.) At this very moment (assuming an open-loop gain Aol=1E4) , the opamp is in its linear amplification region because the diff. voltage is Vdiff=Vin-0.99980004=1-0.99980004=0.00019996V. As a result, the output voltage is Vout=Vdiff*Aol=0.00019996*1E4=1.9996001V.

5.) This is a stable equilibrium because: the classical feedback formula for a finite value of Aol also gives the output voltage Vout=Vin*[1E4/(1+0.5E4)]=1.9996001 V.

6.) That means: We have an equilibrium because the output voltage has a value which exactly meets the condition Vout=Vdiff*Aol. Any larger/smaller output voltage causes a small reduction/increase for Vdiff thereby correcting this deviation from the equilibrium.

7.) In this example, the input difference voltage, of course, is NOT zero. It never will be zero - however, the diff. voltage is so small (in our case app. 0.2mV) that it can be neglected (assumed to be zero for calculations) in many cases.

LvW
  • 24,857
  • 2
  • 23
  • 52
1

One way of looking at this is the Op-Amp's output voltage (in this inverting stage) is always kept at a value such that the current through Rin is the same as the current through Rf whilst keeping the junction at virtual earth potential. If the input impedance of the op-amp is very high (it can be ignored) then Vin/Rin must be very close to -Vout/Rf. This really avoids the issue of voltages being higher than power supply limits as the balance is a continuous process (and therefore doesn't spiral out of control)

Drewan Buckley
  • 51
  • 1