Positive Feedback Op-Amp?

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  • #1
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I'm just wondering, if an op-amp is connected in a positive feedback configuration, should the output not saturate immediately?

Here is an example circuit i made in pspice along with its input/output curve (V12 has amp of 3v).
http://img689.imageshack.us/img689/1286/posfeedback.png [Broken]

As you can see, it is behaving exactly like it would if i connected it in a non inverting, negative feedback configuration. I don't want to draw up a non-inverting -ve feedback circuit but I've simulated that and the input/output behaviour is exactly the same with same gain.

Am i misunderstanding the concept of positive feedback? Thanks.
 
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Answers and Replies

  • #2
vk6kro
Science Advisor
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I would expect that to behave like a Schmitt Trigger and give a square wave out.

You shouldn't be getting an undistorted sinewave out.

Try increasing the input signal to about 11 volts peak to peak or make R8 about 10 K.
 
  • #3
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Yes both those methods work (schmit trigger like output) but my problem is why doesn't the first method work? I would have thought that being in positive feedback configuration, even a small amount of voltage in the input (+ve or -ve) would cause saturation at +- 10V?
 
  • #4
1,762
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The output appears exactly as if the two inputs were reversed. Note that the output is not inverted from the input on the inverting input. Try reversing your input pins an see what happens.
 
  • #5
uart
Science Advisor
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I've seen this type of thing happen before. It seems that some simulation packages simply cannot handle postive feedback circuits correctly. The simulation results are just plain wrong.
 
  • #6
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I'm just wondering, if an op-amp is connected in a positive feedback configuration, should the output not saturate immediately?

Here is an example circuit i made in pspice along with its input/output curve (V12 has amp of 3v).
http://img689.imageshack.us/img689/1286/posfeedback.png [Broken]

As you can see, it is behaving exactly like it would if i connected it in a non inverting, negative feedback configuration. I don't want to draw up a non-inverting -ve feedback circuit but I've simulated that and the input/output behaviour is exactly the same with same gain.

Am i misunderstanding the concept of positive feedback? Thanks.
Yes, you are correct. The positive feedback case would not work. It would likely latch up and drive the output to the rail.

I don't know why the simulation shows it working, but I suspect it is a simulation artifact that results from a overly simplified OPAMP model. You can run into the same issue if you derive the response mathematically using the simplest OPAMP model. You will predict the same gain irrespective of the orientation of the + and - terminals.

Clearly, this is not reality. More realistic models will show issues such as latchup, which is DC instability, or oscillations if the positive feedback occurs at a nonzero frequency.

EDIT: It's been a while since I looked at this kind of thing, but I think you will find that there is an unstable solution that a simulation model may be using. This unstable solution is similar to a ball balancing on a hill. A small perturbation causes the ball to roll down the hill, but mathematically, there is an allowed solution. The simulation may be using a small signal model and it does not know that even a small signal change will cause the ball to roll down to the bottom of the hill (i.e. the voltage should go to the rail).
 
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  • #7
245
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Take a look at the ideal relationship for an op-amp:

[tex]V_{out} = G_{open-loop} \times \left( V_+ - V_- \right)[/tex]

In your circuit, there is a voltage divider between the output and the positive input:

[tex]V_+ = 1/2 V_{out}[/tex]

If we do a little algebra we get the following:

[tex]V_{out} = G_{open-loop} \times \left( 1/2 V_{out} - V_- \right)[/tex]

[tex]V_{out} = \frac{ -G_{open-loop} \times V_-}{1 - (1/2) G_{open-loop}} [/tex]

G is very large so we can simplify:

[tex]V_{out} = 2 V_- [/tex]

That's the ideal version of the op amp. It's identical to what you're getting in your simulation. We could start over and add real components of a real-world op amp like: the offset voltage, the finite input resistance, and the offset currents. The result would, no doubt, look something like this: (I'm not going to do all the algebra)

[tex]V_{out} = G_{open-loop}({some-offset-term})V_- + (2 V_-)[/tex]

G is very large so the output will hit the rails.
 
  • #8
4,239
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Okefenokee. I think you've located the reason that some simulations go wrong, but not the cause.

I could add the same offset term times gain to your equation

[tex]V_{out} = 2 V_- [/tex]

for an op amp properly wired as an amplifier, and get the same result as your last equation.

The operational amplifier, wired as offered in the OP, is metastable at

[tex]V_{out} \cong 2 V_- \ ,[/tex]

in the sense that it will saturate when subjected to a small perturbation, whether G is large or small.
 
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  • #9
1,762
59
[tex]V_{out} = \frac{ -G_{open-loop} \times V_-}{1 - (1/2) G_{open-loop}} [/tex]

G is very large so we can simplify:

[tex]V_{out} = 2 V_- [/tex]
When I work through the above I get

[tex]V_{out} = -2 V_- [/tex]

Also when I enter the circuit into PSpice 9.1 Student Version, Vout goes all the way positive and stays there. The only difference between my circuit and the OP's is that because PSpice 9.1 Student doesn't have a model for LM741, I used an LM324.
 
  • #10
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Thanks for the reply guys. I guess it is pspice modeling problem. I was under the impression that the pspice opamps were accurate models of real world components.

When I work through the above I get

Also when I enter the circuit into PSpice 9.1 Student Version, Vout goes all the way positive and stays there. The only difference between my circuit and the OP's is that because PSpice 9.1 Student doesn't have a model for LM741, I used an LM324.
What is your voltage source amplitude set to? The circuit i drew up is at 3v.
 
  • #11
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59
I set mine to 3 V as well and used two 10 V supplies. Your schematic looked like it came from PSpice too. What version are you using? Have you tried using an LM324 to see if you get the same result?

I have heard many reports of SPICE models not accurately modeling the real world very well but this is the first example I've seen where it's been so obvious. I've used it to model the startup times of oscillators and finding spurious emissions of transmitters and it has always been extremely accurate.

The one thing that bothers me a little about my run is that the signal source is a sine wave so it starts at zero and tends positive. This suggests that the output should have gone negative instead of positive, but there may have been other effects that caused it.
 
  • #12
uart
Science Advisor
2,776
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I set mine to 3 V as well and used two 10 V supplies. Your schematic looked like it came from PSpice too. What version are you using? Have you tried using an LM324 to see if you get the same result?

I have heard many reports of SPICE models not accurately modeling the real world very well but this is the first example I've seen where it's been so obvious. I've used it to model the startup times of oscillators and finding spurious emissions of transmitters and it has always been extremely accurate.

The one thing that bothers me a little about my run is that the signal source is a sine wave so it starts at zero and tends positive. This suggests that the output should have gone negative instead of positive, but there may have been other effects that caused it.
Hi skeptic, it's not the op-amp that's the issue but the type of analysis being used. In the old versions of pspice that I've used it always did this if you used "ac" analysis. It uses a linear model of the opamp which doesn't work for positive feedback. If you tell spice to use transient analysis then it does handle the positive feedback properly.

I'm not sure what package the original poster is using, but usually if you are looking at time domain outputs (as in this case) then you'd also expect it to be doing a transient analysis, so it should get it right. Perhaps this one is doing an "ac" analysis and the gui is just translating that output into a time domain plot.
 
  • #13
1,762
59
Thanks uart, I did use transient analysis.
 
  • #14
16
0
I set mine to 3 V as well and used two 10 V supplies. Your schematic looked like it came from PSpice too. What version are you using? Have you tried using an LM324 to see if you get the same result?

I have heard many reports of SPICE models not accurately modeling the real world very well but this is the first example I've seen where it's been so obvious. I've used it to model the startup times of oscillators and finding spurious emissions of transmitters and it has always been extremely accurate.

The one thing that bothers me a little about my run is that the signal source is a sine wave so it starts at zero and tends positive. This suggests that the output should have gone negative instead of positive, but there may have been other effects that caused it.
Hi Skpetics, thanks for that. I tried the LM324 and it does saturate -ve & stays there after 1.5 periods of following the input. It does seem like a PSPICE modeling problem. My version is 9.2 non student edition.

I am also using transient analysis.
 
  • #15
uart
Science Advisor
2,776
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Interesting. In the very old version of pspice I've got it works correctly (saturates), providing that I use transient analysis, when I use either the LM324 or the ua741.

Here's the model of the ua741 that's included in evaluation library of the version I've got.

Code:
*-----------------------------------------------------------------------------
* connections:   non-inverting input
*                | inverting input
*                | | positive power supply
*                | | | negative power supply
*                | | | | output
*                | | | | |
.subckt uA741    1 2 3 4 5
*
  c1   11 12 8.661E-12
  c2    6  7 30.00E-12
  dc    5 53 dx
  de   54  5 dx
  dlp  90 91 dx
  dln  92 90 dx
  dp    4  3 dx
  egnd 99  0 poly(2) (3,0) (4,0) 0 .5 .5
  fb    7 99 poly(5) vb vc ve vlp vln 0 10.61E6 -10E6 10E6 10E6 -10E6
  ga    6  0 11 12 188.5E-6
  gcm   0  6 10 99 5.961E-9
  iee  10  4 dc 15.16E-6
  hlim 90  0 vlim 1K
  q1   11  2 13 qx
  q2   12  1 14 qx
  r2    6  9 100.0E3
  rc1   3 11 5.305E3
  rc2   3 12 5.305E3
  re1  13 10 1.836E3
  re2  14 10 1.836E3
  ree  10 99 13.19E6
  ro1   8  5 50
  ro2   7 99 100
  rp    3  4 18.16E3
  vb    9  0 dc 0
  vc    3 53 dc 1
  ve   54  4 dc 1
  vlim  7  8 dc 0
  vlp  91  0 dc 40
  vln   0 92 dc 40
.model dx D(Is=800.0E-18 Rs=1)
.model qx NPN(Is=800.0E-18 Bf=93.75)
.ends
*-----------------------------------------------------------------------------
 
  • #16
4,662
5
Try changing your positive feedback resistor R8 from 1k to 10k and see what happens. This should give about +/- 1 volt of Schmitt trigger hysteresis.

Bob S
 

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