Operational Amplifier

LvW

Joined Jun 13, 2013
1,752
Hello to all ,

I am pretty sure that the circuit diagram in the mentioded textbook (reproduced in post#1) contains a typing/drawing error, otherwise the related questions make not much sense (in particular, the last question).
For my opinion, both opamp input terminals must be swapped. In this case, negative feedback would dominate and we have linear operation (amplification with a nice factor).
Juss, it is a good exercise for you to find the gain value for interchanged opamp inputs.
 

WBahn

Joined Mar 31, 2012
29,979
I agree. It would be best to make a notation that the circuit, as drawn, is a comparator. Then indicate that you believe the op-amp inputs are swapped and that you are analyzing the circuit with the op-amp inputs configured the way you believe they were intended to be.
 

LvW

Joined Jun 13, 2013
1,752
Hi Juss,
I have another (rather advanced) question to you (assuming that the inputs are swapped):
What could be the reason for this rarely shown circuit with negative as well as positive feedback?
I can assure you that there is a special reason and that this circuit has some specific advantages if compared with classical opamp based amplifier configuratins.
However, the answer requires something more than only a basic knowledge on opamps.
 

Jony130

Joined Feb 17, 2009
5,487
Hi Juss,
I have another (rather advanced) question to you (assuming that the inputs are swapped):
What could be the reason for this rarely shown circuit with negative as well as positive feedback?
I can assure you that there is a special reason and that this circuit has some specific advantages if compared with classical opamp based amplifier configuratins.
However, the answer requires something more than only a basic knowledge on opamps.
Can you shed some more light on this.
For example why anyone would want to add positive feedback to this circuit



I understand that the gain is now much larger thanks positive feedback, rather than R2/R1.
Vout/Vin = R2*(Rz+R3)/(Rz*R2-R1*R3) = 32
Rz = R4||R5
I also know that Rin will be smaller and Rout larger. But is that all?
To be honest I don't see any advantage.
 

LvW

Joined Jun 13, 2013
1,752
Yes - as mentioned by t_n_k - nobody could explain the circuits operation better than this appl. note.
 

Jony130

Joined Feb 17, 2009
5,487
OK I understand common-mode issue.
But I want to know answer for this
What could be the reason for this rarely shown circuit with negative as well as positive feedback?
I can assure you that there is a special reason and that this circuit has some specific advantages if compared with classical opamp based amplifier configuratins.
What is this "special reason" you are talking about? And I don't want to build VCCS. I just want a voltage amplifier.
 

LvW

Joined Jun 13, 2013
1,752
What is this "special reason" you are talking about? And I don't want to build VCCS. I just want a voltage amplifier.
In some cases it desired to select the loop gain independent on the closed-loop gain. Thus, you can assure or improve stability properties of the opamp. This can be shown by deriving the closed-loop gain formula as well as the loop gain expression (feedback factor).
As an example, applying combined positive and negative feedback you can use uncompensated opamp units even for very small closed-loop gain values.
In any case a reduction in loop gain can improve the step response (no or less overshoot) - and this is possible even for unity gain configurations.
 

LvW

Joined Jun 13, 2013
1,752
Thanks t_n_k,

a very interesting paper. It clearly shows that - in contrast to common belief- positive feedback of an amplifier can (can !!) improve stability properties.
This insight is supported by the fact that - also not known to some beginners - negative feedback degrades dynamic stability. (There is a stability improvement for dc only - stability of the dc operating point).
 
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Thread Starter

Juss

Joined Nov 24, 2013
14
Hello, I am new to this topic and kind of struggling with it so I don't really understand you guys discussion. Sorry about that.

This is my attempt after the help I got previously. Can someone check that if I have done anything wrong please? Thanks you.

(d) Vx=6, Vn = Vx*16/(16+4)+Vout*4/(4+16) = 0.8Vx+0.2Vout = 4.8+0.2Vout = Vp =0.6Vout
4.8=0.4Vout
Vout=12V
(e) Vout saturates at supply voltage = 15V
0.8Vx+0.2*15=0.6*15
0.8Vx=6
Vx=7.5V
 

WBahn

Joined Mar 31, 2012
29,979
No need to apologize. They basically hijacked your thread, which is frowned on but happens all the time and is very easy to do without meaning to do so -- I've done it more than a few times and I'm sure I will do it many more.

Always feel free to ask people to come back to your concerns (it is, after all, YOUR thread) and even to take tangent discussions to a different thread (that you may very well choose to continue reading and participating in). You can ask a mod to split them out, too.

I'll take a look at your work in a little bit. Gotta do something else in the next few minutes first.
 

WBahn

Joined Mar 31, 2012
29,979
Hello, I am new to this topic and kind of struggling with it so I don't really understand you guys discussion. Sorry about that.

This is my attempt after the help I got previously. Can someone check that if I have done anything wrong please? Thanks you.

(d) Vx=6, Vn = Vx*16/(16+4)+Vout*4/(4+16) = 0.8Vx+0.2Vout = 4.8+0.2Vout = Vp =0.6Vout
4.8=0.4Vout
Vout=12V
(e) Vout saturates at supply voltage = 15V
0.8Vx+0.2*15=0.6*15
0.8Vx=6
Vx=7.5V
Assuming that the opamp were connected so that it is stable, then yes this is correct.

The cleaner way to go about it is to determine the gain of the amplifier. Do that for a generic input voltage, Vx, and then you can answer (d) and (e) in one simple line each.

Vp = Vout * (6kΩ/(6kΩ+4kΩ)) = 0.6*Vout
Vn = Vx + (Vout - Vx)*((4kΩ/(4kΩ+16kΩ)) = 0.2*Vout + 0.8*Vx

Vp = Vn

0.6*Vout = 0.2*Vout + 0.8*Vx

0.4*Vout = 0.8*Vx

Gain = Vout/Vx = 0.8/0.4 = 2

Now you can answer part (d)

Vout = Gain * Vx = 2 * 6V = 12V

Now you can answer pard (e)

Vx = Vout/Gain = 15V/2 = 7.5V
 

Thread Starter

Juss

Joined Nov 24, 2013
14
Assuming that the opamp were connected so that it is stable, then yes this is correct.

The cleaner way to go about it is to determine the gain of the amplifier. Do that for a generic input voltage, Vx, and then you can answer (d) and (e) in one simple line each.

Vp = Vout * (6kΩ/(6kΩ+4kΩ)) = 0.6*Vout
Vn = Vx + (Vout - Vx)*((4kΩ/(4kΩ+16kΩ)) = 0.2*Vout + 0.8*Vx

Vp = Vn

0.6*Vout = 0.2*Vout + 0.8*Vx

0.4*Vout = 0.8*Vx

Gain = Vout/Vx = 0.8/0.4 = 2

Now you can answer part (d)

Vout = Gain * Vx = 2 * 6V = 12V

Now you can answer pard (e)

Vx = Vout/Gain = 15V/2 = 7.5V
Thanks. I didn't know that Gain = Vout/Vx
 

LvW

Joined Jun 13, 2013
1,752
Assuming that the opamp were connected so that it is stable, then yes this is correct.
Juss - it is important always to remember this pre-condition given by WBahn.
The assumption Vp=Vn may be used as start of the calculation only if the negative feedback factor dominates over the positive feedback factor (which is the case for your example with swapped input nodes). Otherwise, the opamp is not operated as a linear amplifier.
 
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