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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 '14 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 '14 at 19:54
  • \$\begingroup\$ I updated my answer to address your edits. \$\endgroup\$ – Alfred Centauri Aug 22 '14 at 15:06
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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 '14 at 19:44
  • \$\begingroup\$ @user50500, I've updated my answer to address your comment. \$\endgroup\$ – Alfred Centauri Aug 22 '14 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 '14 at 20:04
  • \$\begingroup\$ Everything makes sense except for seeing any application for this circuit. \$\endgroup\$ – Spehro Pefhany Aug 21 '14 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 '14 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 '14 at 19:55
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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 '14 at 21:34

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