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Op Amp With Feedback

Have a look at the diagram above. It shows an op-amp with negative feedback. Ignoring the feedback loop for now, the voltage (pressure) coming from the 5V battery, acting on the inverting pin, should amount to 5V.

Now, I understand that the feedback loop is adjusting the voltage on the inverting pin to make it equal to the reference (non-inverting) pin or 3.3V and you can clearly see that the output is producing a negative voltage to achieve this. I also understand that the input pins have a very high impedance. What's confusing me is how it actually does this (considering my rudimentary knowledge of electronics). I considered that the output was sinking some of the current produced (by the battery) to alleviate the pressure on the input pin and bring it down to 3.3V, but clearly that isn't the case (since the 1 ohm resistor is passing 5A). So, I'm struggling to understand how 5V of pressure from the battery results in 3.3V on the input pin. I'd appreciate it if somebody could explain this in an intuitive way.

Much Appreciated.

Edit:

Many thanks for your all your comments and insights. The circuit itself is a distraction. At the heart of this is my attempt to understand op-amp and negative feedback behaviour. I wanted to understand the following:

a) if you could apply an input voltage to the op-amp that was different to the reference voltage (more specifically that you have voltage "pressure" acting on the input pin that is higher than the reference pin). This also assumes that you have some other input other than the feedback loop.

b) if a is true (and assuming that the op-amp adjusts its output to equalize the voltage on both pins) how is this achieved? Clearly the feedback loop plays a part and the only thing I could think of was that the output was sinking current to take the pressure off the input pin and thus explain the voltage drop between the between the higher voltage source and the voltage appearing at the input pin. Incidentally, the question would still stand even if you had a resistor between the source voltage and the feedback loop/input pin junction.

A better circuit example is shown below (taken from http://www.electronics-tutorials.ws/opamp/opamp_2.html):

Inverting Op-amp

But with V1 set to some voltage other than 0V.

Final Edit (Promise!):

From the electronics tutorial article linked above:

"This negative feedback results in the inverting input terminal having a different signal on it than the actual input voltage as it will be the sum of the input voltage plus the negative feedback voltage giving it the label or term of a Summing Point."

This is exactly where my confusion lies. As I understand it voltage is simply pressure exerted measured with respect to some reference point (usually ground). What the paragraph above is suggesting is that at the "summing point" there's a change in voltage (pressure) because of the addition of the feedback voltage. Intuitively, I'm thinking that the output pin "sinks" some current to reduce the voltage at the summing point. But that sinking of current would reduce Iin (since no current flows through the inverting pin). The result would seem to be that Vin drops. But is this the case?

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  • \$\begingroup\$ The resistor is in parallel to the batterey, and for that reason you can remove it from the circuit. It doesn't influence the opamp at all. \$\endgroup\$
    – jippie
    Aug 21, 2014 at 19:54
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    \$\begingroup\$ This circuit makes no sense. You are impressing 5 volts across the 1 ohm resistor so the op amp cannot do anything. You don't appear to have an input signal. What is this circuit supposed to do? \$\endgroup\$
    – Barry
    Aug 21, 2014 at 19:54
  • \$\begingroup\$ I updated my answer to address your edits. \$\endgroup\$ Aug 22, 2014 at 15:06

5 Answers 5

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To gain insight into what is happening, replace the op-amp with an ideal voltage amplifier model (we assume the gain \$A \rightarrow \infty\$):

schematic

simulate this circuit – Schematic created using CircuitLab

Now it's easy to see two important points

  • \$R\$ can only change the current through the 5V source - it has no other effect
  • there is no path for output current thus the output current is zero.

Thus, in this odd circuit, the output voltage adjusts to be 5V less than the voltage applied to the non-inverting terminal which, in this case, implies

$$V_O = -1.7\mathrm V$$

and the resistor is irrelevent to this result.


(Added to address edited and expanded question)

As I understand it voltage is simply current pressure measured with respect to some reference point (usually ground). In this case, we have Iin producing Vin "pressure"

I'm not sure what you mean by the "current pressure" but, in this circuit, it is commonly understood that the voltage \$V_{in}\$ is an independent variable - a given - which means that \$V_{in}\$ isn't 'produced' by \$I_{in}\$ but, rather, produced externally to the circuit.

To make this clear, one can explicitly add the external source to the circuit, e.g.,

schematic

simulate this circuit

Now it's clear that \$I_{in}\$ depends on \$V_{in}\$ but \$V_{in}\$ is fixed by the voltage source, i.e., changing the value of \$R_{in}\$ will change the value of \$I_{in}\$ but not the value of \$V_{in}\$.

Intuitively, I'm thinking that the output pin "sinks" some current to reduce the voltage at the summing point. But that sinking of current would reduce Iin (since no current flows through the inverting pin). The result would seem to be that Vin drops. But is this the case?

The voltage at the output of the ideal op-amp, if negative feedback is present, will be whatever it needs to be so that the inverting input voltage equals the non-inverting input voltage.

Now, this might mean that the output must sink current or it may mean that the output must source current.

In my opinion, the most intuitive, straightforward way to think about this is to apply voltage division.

By voltage division, the voltage at the inverting input is given by

$$V_- = V_{in}\frac{R_F}{R_{in} + R_F} + V_{out}\frac{R_{in}}{R_{in} + R_F}$$

This result is elementary and holds even if the op-amp is removed from the circuit and \$V_{out}\$ is produced by an independent voltage source.

So, at this point, we can ask the question

  • What must \$V_{out}\$ be such that the inverting input voltage, \$V_-\$, equals the non-inverting input voltage, \$ V_+\$?

A little bit of quick algebra yields the answer

$$V_{out} = V_+\left(1 + \frac{R_F}{R_{in}} \right) - V_{in}\frac{R_F}{R_{in}}$$

Thus, if \$V_{out}\$ equals the above, the inverting input voltage will equal the non-inverting input voltage.


just one more thing: in the case where Vout is positive what effect does this have on Iin?

We can straightforwardly write the equation for \$I_{in}\$ as follows:

$$I_{in} = \frac{V_{in} - V_{out}}{R_{in} + R_F}$$

But, under the assumption that \$V_{out}\$ is whatever it needs to be so that the inverting input voltage equals the non-inverting input voltage, we have

$$I_{in} = \frac{V_{in} - V_+}{R_{in}}$$

Carefully note that, under the above assumption (which is the same as assuming an ideal op-amp), \$I_{in}\$ does not depend on \$V_{out}\$ period. This is a consequence of the constraint \$V_- = V_+\$.

In summary, assuming an ideal op-amp, there is no instant in which \$V_- \ne V_+\$.

For physical op-amps, we must add additional circuit elements to model the departure from non-ideal behaviour and that is beyond the scope of this answer.

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  • \$\begingroup\$ I think I'm nearly there with this AC, just one more thing: in the case where Vout is positive what effect does this have on Iin? I think my issue here is with how the current is behaving as the voltage equalization is achieved. I'm guessing that if Vout is positive it will effect Iin, right - acting to slow it down. By the way, I completely understand your equations and I get the balancing of voltages to achieve V+. It's the current behaviour I'm not grokking. I think I need a few strong pale ales! \$\endgroup\$
    – Buck8pe
    Aug 22, 2014 at 19:44
  • \$\begingroup\$ @user50500, I've updated my answer to address your comment. \$\endgroup\$ Aug 22, 2014 at 23:50
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The resistor is in parallel to the battery, and for that reason you can remove it from the circuit. It doesn't influence the opamp at all. You can easily redraw the circuit with the resistor in parallel to the 5V battery without changing anything to the remainder of the circuit.

Rule of thumb: An ideal opamp with negative feedback will try to set both inputs V+ and V- to the same voltage, using its output.

So given that both V+ and V- are at the same potential of 3.3V, set by the only voltage source that is referenced to ground, you can calculate the output voltage. It will be 5 volts lower than the 3.3 input voltage. 3.3 - 5 = -1.7V

As a matter of fact, because all voltages are set by voltage sources, you don't need the opamp either. You can safely remove it, then tie V+ to V- and again, nothing changes to the circuit. Hence Iout = 0A.

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  • \$\begingroup\$ You are absolutely correct but I can't see how a beginner would use such a circuit - I think he's simulated it hence he gets the -1.7V voltage but his simulation makes no sense. \$\endgroup\$
    – Andy aka
    Aug 21, 2014 at 20:04
  • \$\begingroup\$ Everything makes sense except for seeing any application for this circuit. \$\endgroup\$ Aug 21, 2014 at 20:47
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In this configuration you are missing an all-important resistor known as the inverting input resistor. This allows the inverting input pin node to be dragged down to match the non-inverting pin voltage: -

schematic

simulate this circuit – Schematic created using CircuitLab

I'm also assuming that the brown wire in your diagram is meant to be ground because it wouldn't make sense otherwise.

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    \$\begingroup\$ The battery is not referenced to ground in my opinion, Only the 3.3V seems to be ground refrenced. (assumptions I guess). \$\endgroup\$
    – jippie
    Aug 21, 2014 at 19:55
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    \$\begingroup\$ No, look again. The 5V battery is the feedback loop (in parallel with the 1-ohm resistor). \$\endgroup\$
    – Dave Tweed
    Aug 21, 2014 at 19:55
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Introduction

Need for conceptual questions

Rarely are such in-depth questions about a fundamental phenomenon as negative feedback. Indeed, the question was asked more than nine years ago, but such questions never get old. They are set and will be set in the future because this is a concept, and concepts are eternal. I myself have been asking myself this all my life.... but I think I have already managed to answer it, and I can share my philosophy here.

The weakness of formal explanations

What's confusing me is how it actually does this.

There is a well-developed paradigm in textbooks that is repeated over generations. As in some "religion" (or, in more modern terms, AI), it gives us straightforward explanations that are easy to teach by teachers and remember by students. The only problem with these "explanations" is... that actually they do not explain circuit phenomena. They only tell us what was done but not why it was done that way and "how it actually does this".

How do we reveal the basic idea?

I will not go into the OP's specific circuit because it is been done in other answers. It is clear that the errors in it come from the lack of idea of what all this is being done for. That is why I set out to explain it here.

I have several approaches to explain the phenomenon of negative feedback, but this one is my favorite. It is the most original and somehow paradoxical which gives rise to thought.

Conceptual NFB follower

The idea here is to find out which is the most basic negative feedback configuration and then build step by step the more complex configurations. Let's get started then!

Zero-voltage stabilizer

Arrangement: After much thought, I came to the conclusion that the most elementary possible negative feedback configuration is an inverting amplifier whose output is connected to its input... nothing more, just that. In the conceptual schematic below, it is implemented by the abstract voltage-controlled voltage source VCVS acting as an amplifier with a high gain of -100,000.

schematic

simulate this circuit – Schematic created using CircuitLab

Operation: If for some reason the voltage tries to increase, the amplifier will in turn decrease it and restore the initial voltage. The higher the gain, the lower the voltage, and at high enough gain it becomes almost zero (it would be interesting to investigate this, but there is no way here because the voltage is zero). So this arrangement can be called a “zero-voltage stabilizer” or "virtual ground". To really feel "how it actually does this", we can put ourselves in its place by taking a variable voltage source Vvar and adjusting its voltage to zero while looking at the voltmeter.

schematic

simulate this circuit

Just imagine how many analog circuits are based on this simple "output to input" connection! And if the amplifier is non-inverting, then how many digital circuits (latches, registers, memories...) are based on it!

Voltage-disturbed stabilizer

But what is the point of producing a virtual-ground zero voltage in such a complicated way - by negative feedback (monitoring and maintaining it)? Is not it simpler to use the real-ground zero voltage? The point is that this voltage is stable and not affected by disturbances.

Floating voltage source: Let's check it out by inserting a "disturbing" voltage source Vdist in series to the input (between the output and the inverting input). Thus, its voltage is added to the zero input voltage. The "amplifier" responds to this by starting to raise its output voltage in the opposite direction (- Vdist +, + Vout -, going around the loop) so as to subtract it from the disturbing voltage. As a result, Vout = Vdist.

schematic

simulate this circuit

Eureka! We can use the mirror voltage as an output voltage since it is grounded and buffered copy of the input voltage Vdist! So that is the idea of this negative feedback trick - to compensate the "disturbing" input voltage with an equivalent "anti-voltage", and then to use it instead of the original input voltage.

Grounded voltage source: But there is a "small" problem with this arrangement - Vdist is "floating" (ungrounded) input voltage source. Then let's move it between the "amp" input and ground (it is possible here since the VCVS has a "floating" input). Thus it becomes a real grounded input voltage source Vin.

schematic

simulate this circuit

Op-amp NFB follower

Now let's implement the conceptual circuits above with a real op-amp with differential input. The same explanations as above apply here.

Without disturbance (no input voltage)

schematic

simulate this circuit

With floating "disturbing voltage source"

schematic

simulate this circuit

With grounded input voltage source

schematic

simulate this circuit

Op-amp non-inverting amplifier

The circuits above are followers that are disturbed by an additive disturbance (the disturbing voltage is added). If we want them to amplify, we can insert another but multiplicative disturbance. This means to attenuate the output voltage and subtract a part of it from the input voltage. In the circuit below, it is implemented by the R1-R2 voltage divider.

schematic

simulate this circuit

The op-amp is forced to further increase its output voltage (R1 + R2)/R1 times to compensate for the disturbance caused by the voltage divider. The bottom line is that a non-inverting amplifier is a deliberately disturbed voltage follower.

Op-amp inverting amplifier

Above we came up with the idea of "disturbing" the op-amp by adding/subtracting the "disturbing" input voltage to/from its input voltage in a series manner. And can not we do it in parallel? Let's try it.

Directly (no resistors): However, the op-amp output is also connected in parallel with the input. So a conflict occurs between the two perfect voltage sources (Vin and Vout), and it is not clear who will win in this "tug of war". Note that because the elements are "ideal", a huge current flows between the voltage sources.

schematic

simulate this circuit

So, I'm struggling to understand how 5V of pressure from the battery results in 3.3V on the input pin.

Through resistors: To mitigate the conflict, we can connect the voltage sources Vin and Vout through resistors R1 and R2 to the op-amp input. They actually form a resistor summer with weighted inputs. Its output voltage (in the midpoint between resistors) is between the two voltages. That is why "5V of pressure from the battery results in 3.3V on the input pin".

schematic

simulate this circuit

Intuitively, I'm thinking that the output pin "sinks" some current to reduce the voltage at the summing point.

Vin and Vout are connected through the resistor network R1-R2 so a current flows from the higher voltage to the lower voltage. Vin1-R1 and Vout-R2 can be considered as cascaded current source and sink (the former produces a current and the latter absorbs the current).

Summary

  • The most elementary possible negative feedback configuration is an inverting amplifier whose output is connected to its input.

  • The input voltage source disturbs the amplifier trying to change its zero input voltage.

  • This disturbance is additive (a voltage that is added or subtracted).

  • The output voltage is with the same polarity (regarding ground) to be subtracted from the input voltage in the loop.

  • The amplifier responds to the disturbance by producing a "mirror copy" of the input voltage disturbance.

  • The "mirror voltage" is used as a buffered and grounded output voltage.

  • It can be multiplied by disturbing the op-amp with another but multiplicative disturbance β (attenuation).

  • The amplifier reacts to the new disturbance by a "mirror amplified copy".

  • Another way to disturb the amplifier is in parallel through resistors that mitigate the conflict between the voltage sources.

  • In this case, the output voltage is with opposite polarity (regarding ground) to be subtracted from the input voltage.

  • Vin and β are useful disturbances; all others are unwanted and must be suppressed.

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Your circuit is missing two absolutely vital resistors. The circuit should be something like:

schematic

simulate this circuit – Schematic created using CircuitLab (The op-amp needs a power supply, of course)

R1 allows a voltage drop between the +5V source and the - input, so that the negative feedback through R2 can be effective.

R3 is the output load - your drawing seems to show the output shorted to the negative side of both voltage sources.

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    \$\begingroup\$ No, like Andyaka, you've misread the diagram. The 5V battery is referenced to the output, while the 3.3V supply is referenced to the (hidden) ground. \$\endgroup\$
    – Dave Tweed
    Aug 21, 2014 at 21:34

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