How Does Non-Standard Feedback Affect Op-Amp Behavior?

[Rudinhoob]

OpAmp "Non-standard" Feedback

Hello,

This is my first post here. Please excuse my poor English and grammar. I will try to make my questions clear.

As I understand, with a negative feedback, where the output connected to the inverting input and in the ideal case, the Vn and Vp will be very close in a sense that Vn tracks the voltage of Vp so that the difference is almost zero.
However, when changing the feedback configuration, for example connecting the Vn input to other parts than the output, or using a positive feedback, how does this affect the ideal rules in analysis?

Thanks.

 

[jim hardy]

That's a hard concept to wrap your brain around when first starting out.

The little integrated circuit we call an "op-amp" is just a high gain amplifier.

When we surround an "op-amp" with a feedback circuit that allows it to hold its inputs equal,
we now have an electronic circuit that behaves in accordance with some mathematical "operation" defined by the feedback network.
So we call it an 'operational amplifier' circuit.
Perhaps we could have done better with the naming comvention, perhaps not.

The circuit you describe , signal applied to +in and Vout connected to -in, performs the math operation Vout = Vin - 0.
More precisely, Vout = Vin(1-1/(amplifier's open loop gain)),
and since open loop gain is around a million that reduces to ~Vin - Vin/(10⁶), whhich rounds to Vin - 0.

Remember - it is the duty of the circuit designer to surround his amplifier with a feedback circuit that allows it to hold its inputs equal.
Else it won't operate.


See if this is any help in your studies:
http://www.analog.com/static/import... Amp Applications Book (PDF)/P2 ChH_final.pdf

This site is a treasure trove:
http://www.philbrickarchive.org/

old jim

 

[Rudinhoob]

When we surround an "op-amp" with a feedback circuit that allows it to hold its inputs equal, we now have an electronic circuit that behaves in accordance with some mathematical "operation" defined by the feedback network.

How can we tell that a certain feedback configuration makes the two input voltages almost equal, and hence assume Vn = Vp?

 

[jim hardy]

How can we tell that a certain feedback configuration makes the two input voltages almost equal, and hence assume Vn = Vp?

What i do is write Kirchoff's voltage law solutions for voltage at each input, then set those equal and solve for Vout.
If Vout doesn't fall between the supply rails, the circuit won't "operate".

 

[alphysicist]

Hello,

Thank you for your question. As you correctly stated, in a standard negative feedback configuration, the output voltage is fed back to the inverting input of the op-amp, resulting in a very small difference between the inverting and non-inverting input voltages. This is known as the virtual short concept and allows for accurate amplification and linear operation of the op-amp.

When using non-standard feedback configurations, such as connecting the Vn input to other parts of the circuit or using positive feedback, the ideal rules for analysis may no longer apply. This is because the virtual short concept may not hold true in these cases and the op-amp may not behave in a linear manner.

In such cases, it is important to carefully analyze the circuit and take into account the effects of the non-standard feedback on the op-amp's behavior. This may require using more complex analysis techniques, such as using the op-amp's transfer function or considering the effects of non-idealities in the op-amp itself.

I hope this helps clarify your question. Please let me know if you have any further inquiries.