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This question already has an answer here:

I never quite understood why negative feedback is more used than positive feedback.

Using feedback makes it possible to adjust the gain, and thereby specify how much the input should be amplified.

But why is negative feedback more commonly used when we want to amplify small voltages from sensor readings such as photodiodes etc.

The output of the op-amp becomes the negative amplified value of the input (input and output has different polarity), which don't make sense why it should be useful..

Hence would i always use positive feedback, but I can see that the internet disagrees with me, so must be wrong somewhere in my "reasoning" so what am I doing wrong?

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marked as duplicate by Olin Lathrop, Sparky256, Scott Seidman, Bence Kaulics, Daniel Grillo Jul 14 '16 at 11:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ Op amps always want to use their open-loop gain. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 23 '16 at 9:24
  • \$\begingroup\$ But in a negative feedback will this not happen.. you could simplify the circuit into a voltage divider, due to the high input resistance, will no current enter the op-amp. \$\endgroup\$ – Carlton Banks Jan 23 '16 at 9:26
  • \$\begingroup\$ And how do you imagine that positive feedback would prevent it? \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 23 '16 at 9:28
  • \$\begingroup\$ The same for positive feedback?.. \$\endgroup\$ – Carlton Banks Jan 23 '16 at 9:28
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    \$\begingroup\$ There is a fundamental misunderstanding of what "feedback" is here... \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 23 '16 at 9:29
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You are confusing 'inverting' with 'negative feedback'.

Open loop

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1: op-amp with open-loop inverting mode.

In Figure 1 the op-amp will amplify the difference between its inputs by the open loop gain. Let's say the open loop gain is 1,000,000 and we apply +1 mV at the '-' input. Since it is higher than the '+' input the output will go to -1 mV x 1000000 = -1000 V. (Obviously on a real op-amp it will stop at the negative supply rail.)

This circuit would not make a good amplifier as the gain would not be controllable, any variation or drift in the input offsets would wreak havoc with the output and any non-linearities of the amplifier would go un-checked. It might, however, make a useable comparitor to detect if the input voltage was above or below, in this case, zero volts.

Negative feedback

Negative feedback occurs when some function of the output of a system, process, or mechanism is fed back in a manner that tends to reduce the fluctuations in the output, whether caused by changes in the input or by other disturbances. Wikipedia.

Applying negative feedback brings the output under control.

schematic

simulate this circuit

Figure 2: inverting op-amp with negative feedback. Figure 3: non-inverting opamp with positive feedback.

Now if we apply negative feedback by putting back R2 the output counters the input. Imagine the operation as a sequence:

  • Vin = 0. Vout = 0.
  • Vin = 1 V. Vout = 0 at this instant. R1 and R2 form a voltage divider and the '-' input is at 0.5 V.
  • Vout starts to swing negative. This pulls the '-' input back towards 0 V.
  • The output settles down at Vout = -1 V.

In this configuration the output will move to whatever voltage makes the '-' input the same as the '+' input. This will occur when \$Vout = -Vin \frac {R2}{R1}\$.

Positive feedback

Figure 3 shows the amplifier with positive feedback.

In this situation the scenario will be as follows:

  • Vin = 0. Vout = 0.
  • Vin = 1mV. Vout still at 0 so '+' input goes to +0.5 mV.
  • Vout starts to rise. The '+' input voltage starts to follow it at about half the rate (because of the voltage divider effect).
  • Vout hits the + supply rail.

This circuit is often used to make a Schmitt trigger to add some hysteresis so that turn-on and turn-off points are different.

Non-inverting with negative feedback

schematic

simulate this circuit

Figure 4. Non-inverting amplifier.

The non-inverting amplifier also uses negative feedback although in a different fashion. I find it most useful to think that the op-amp is 'happy' or stable when both inputs are at the some potential. In this case it's when \$Vout = Vin \frac {R1 + R2}{R1} = Vin (1 + \frac {R2}{R1}) \$.

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  • \$\begingroup\$ Formulas for \$V_{out}\$ for both the inverting/non-inverting are not quite there. \$\endgroup\$ – Captainj2001 Jul 14 '16 at 0:28
  • \$\begingroup\$ I must have been asleep. All OK now? Thanks. \$\endgroup\$ – Transistor Jul 14 '16 at 6:18
  • \$\begingroup\$ Looks good, I would have edited myself but it wouldn't let me because it wasn't enough characters! \$\endgroup\$ – Captainj2001 Jul 14 '16 at 11:34
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I think you're misunderstanding the concepts of positive and negative feedback. Negative feedback sets the gain to a fixed level, while positive feedback lets gain go to infinity.

Negative feedback does not mean that your output is the "negative amplified value of the input". Instead, it means that a part of the output is subtracted from the input.

schematic

simulate this circuit – Schematic created using CircuitLab

In contrast, consider this example for positive feedback (conceptually):

  1. your input signal is of magnitude 1
  2. because you are using positive feedback, your output gets added to your input
  3. adding your output moves your input signal to magnitude 2
  4. because you are using positive feedback, your output gets added to your input
  5. adding your output moves your input signal to magnitude 4
  6. ...

you can see how positive feedback leads to your signal not being amplified by a certain amount, but until it is saturating the op amp.

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  • \$\begingroup\$ Hmm.. ok positive drive the output toward saturation.. Makes sense.. Negative feedback drives it towards zero?, but still doesn't make sense why would one use a controllable negative value? \$\endgroup\$ – Carlton Banks Jan 23 '16 at 9:56
  • \$\begingroup\$ Quote: ...."drives it to zero". No - negative feedback drives the INPUT to a smaller value. For an opamp with a very large open-loop gain the input differential voltage is reduced NEARLY to zero (in practice: some µVolts). \$\endgroup\$ – LvW Jan 23 '16 at 10:15
  • \$\begingroup\$ ...ok.. The way i see the situation i can either choose positive or negative feedback.. How both doesn't seem to do what i want to do. As i see it positive feedback => output always +/- voltage source negative feedback =>output always inverse controlled amplified input. \$\endgroup\$ – Carlton Banks Jan 23 '16 at 10:16
  • \$\begingroup\$ Positive f/b is unstable, negative f/b is normally stable. Negative f/b does not necessarily invert (ie change the sign of ...) the input signal but in any case inversion is not a disadvantage, in fact for many applications it's an advantage. Negative f/b drives the effective op-amp differential input to zero, not the output signal. \$\endgroup\$ – Chu Jan 23 '16 at 12:35
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One of the characteristics of ideal op-amp is Infinite Open loop Gain.

So if you connect any signal to input of op-amp with open loop the gain will more in practical op-amp, hence the output will be more(near to +Vcc)

So to control this high gain mostly preferring -ve feedback rather than increasing gain using positive feedback

And one more added point here is, the opamp having high input impedance. So it won't sink much more current. Mostly sensor output current driven capability will be in mA. It's not that much enough to drive other circuits hence mostly sensors output is fed through op-amp.

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  • \$\begingroup\$ what about the negative output? As i see it positive feedback => output always +/- voltage source negative feedback =>output always inverse controlled amplified input. \$\endgroup\$ – Carlton Banks Jan 23 '16 at 10:15
  • \$\begingroup\$ In inverting amplifier ouput will be -ve. Where else you asking ?. \$\endgroup\$ – Photon001 Jan 23 '16 at 10:23
  • \$\begingroup\$ ok.. I see the benefit of being able to control the gain, but having a polarity inverses input just seems a bit inconvenient.. I mean and example could be a TIA. \$\endgroup\$ – Carlton Banks Jan 23 '16 at 10:26
  • \$\begingroup\$ If -ve bother you mean you have to add one more stage op-amp. then it will become +ve. Its depends on the applications. Suppose if any input signal is -ve then output(inverting op-amp) will be +ve. \$\endgroup\$ – Photon001 Jan 23 '16 at 10:59
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Negative feedback seems to be a problem in your understanding so consider a servo controller like this: -

enter image description here

The 10k pot sets a "demand" and the output transistors drive the motor until the feedback position sensor (attached to the motor) is producing an output voltage that matches the demand. The triangle-shaped thing that looks like an op-amp should be regarded as an error amplifier. Any error between demand and fed-back voltage will be amplified and this will nudge the motor a bit further until that error is very small.

If the error amplifier has more gain, then the motor will move closer towards the desired and demanded set-point i.e. the error becomes smaller.

Now imagine throwing away the motor and feedback position sensor and wiring the output from the push-pull amplifier directly to the error amplifier: -

enter image description here

What you now have is an op-amp and this can be configured in a number of ways: -

enter image description here

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