Hi,

I have a lot of trouble for understanding AC analysis... Is there someone who can tell me how this transfer function has been found ? Is it possible to have a huge explanation please



Thank you very much !

 

Do you know how to find a voltage gain of a circuit that contains an opamp?

 

An op amp has a high open-loop gain (ideally infinite) so, when there is negative feedback, such as shown in your circuit, then the op amp will always adjust its output to keep the voltage difference between it's (+) and (-) inputs equal to zero.
This means the Vo input current to the (-) input must equals the current from the VCONTROL output to the (-) input.

Given that info, it should be apparent how to write the transfer function for that op amp circuit.

 

Hi, thank you for your answers. Unfortunately I know the rules ... The problem is that i do not understand the small signal analysis If i do not do an small signal analysis there will at least Vref in the equation and Rbiais appearing in the equation ...

 

Hi, thank you for your answers. Unfortunately I know the rules ... The problem is that i do not understand the small signal analysis If i do not do an small signal analysis there will at least Vref in the equation and Rbiais appearing in the equation ...

No there won't. Why do you think that is the case? The large open loop gain of the opamp reduces their effects to insignificance. In order for there to be a change in the circuit Vo must change. As soon as V0 changes then Vcontrol will change. This circuit is called a follower. Whatever happens at Vo happens at Vcontrol; it is a linear circuit after all.

So let Vo = .01*sin ωt. Then Vcontrol = -(Z1/Z2)*.01*sin (ωt±φ). The phase angle φ accounts for the finite delay of the opamp, if you can even measure it. Using a 741 you probably can, but using a modern opamp you probably can't.

 

There is a basic analysis here -

https://ocw.mit.edu/resources/res-6...ring-2013/textbook/MITRES_6-010S13_chap13.pdf

This will give you insight how to model the OpAmp for s plane analysis. Page 565....


Regards, Dana.

 

Hi,

Thank you for your comments. I do not understand how you can tell that Rbias will not appear in the equation. The equation is true if we do a small signal analysis. What do you mean about the open loop gain ? Are you talking about the amplifier gain ? Can you explain how the large open loop gain reduce to insignificance the effect of Vref and Rbiais ? Is this really a follower ? Vout doesn't not follow V control. It is rather than an inverting amplifier.

The equation is something like this Vcontrol = A*Vref - Z1/Z2*Vout with A which is a function of Rbias, Z1 and Z2. It seems pretty logic as regards of the circuit. Nevertheless, it is true that if we do a small signal analysis on this equation we will find that Vcontrol = -Z1/Z2*Vout and so the gain DeltaVcontrol/DeltaVout = -Z1/Z2. A little change on delta Vout will only produce a little change on Vcontrol as you said. But i obtained that conclusion because I wrote the equation...

If I try to understand the phenomenon on the circuit, i have not the same conclusion ... And i know that i'm wrong ! If I apply a little change on Vout, I will have a little change on the inverting input of the amplifier which will depend on Rbiais and then I will amplify the difference between Vref and the inverting input voltage by the amplifier gain. Eventually, Vcontrol will be adjusted thank to the feedback in order to set the difference ((+) - (-)) equal to 0. So Vcontrol will move down for a certain quantity but saying that it does not depends on Vref and Rbiais is not clear for me ...

Thank you,

 

Grinding thru equations you will find the following -

1) At DC the OpAmp G, Aol, will force node at Vinv to equal Vninv of the OpAmp,
so I thru Rbias ~=0. This is commonly referred to as virtual ground or short. Also
referred to as 'servo" behavior of differential amplifiers with high G, the concept of
an amplifier that drives a node to a final value.

Vinv = Inverting input of OpAmp
Vninv = Non-Inverting input of OpAmp

https://www.allaboutcircuits.com/te...tanding-the-virtual-short-in-op-amp-circuits/

2) The OpAmp G, starting at Aol, is a complex gain, dropping as freq increases. This causes
increasing Zout, falling Zin, and (Vninv - Vinv) to rise of network, eg. we lose the virtual short or
ground behavior as our OpAmp runs out of "gas", gain. Also our OpAmp Zout starts looking
inductive, in fact an analysis will allow you to actually compute the L equivalent value. As an
aside when you do that analysis you use PFE on the resulting expression to get at the lumped
element values.

Most OpAmps exhibit either 1 or two pole roll off depending on design purpose. The Two pole
more representative of "fast" OpAmps.

Above you see OpAmp starts out at DC with G = Aol, and then at first pole starts to drop.
This is due to C and gm effects in the OpAmp signal path, dominated by the compensation
scheme used in the OpAmp.

3) The s plane model for a single pole V generator, the G element inside OpAmp, is -

https://inst.eecs.berkeley.edu/~ee1...re05-Non-ideal Op Amps (Finite Bandwidth).pdf

Lots of fun analyzing these circuits. A bit of advice, learn signal flow graph analysis to do this
as it gives visual insight into node and path behavior in the circuits. And saves bucket loads of
time getting at T(s)....


Regards, Dana.

 

Here is an AC analysis where you can see as OpAmp G collapses the "virtual ground/short"
between the two inputs gets compromised.



Green is OpAmp output
Blue AC current through your Rbias
Red the AC voltage developing across Rbias as Vinv loses its grip on AC ground due to loss of G in the OpAmp as freq rises.


Regards, Dana.

 

Thank you very much for all your informations That was very interesting !

 

Please carefully analyze this example:

Case one:



The output voltage for this circuit is given by this equation:

Vcontrol = Vref *(1 + R1/(R2||Rbias)) - Vo*(R1/R2) = 2.5V *21 - 4.9V*10 = 3.5V

Also, notice that the voltage at the inverting input is equal to Vref due to "negative feedback magic" and "virtual short".
And the "virtual short" action is thank to the negative feedback (via R1) and very large opamp internal gain (equals infinity in ideal opamp).

Case two:




This time the control voltage is equal to:

Vcontrol = 52.5V - 10V/V*5.1V = 1.5V

Nothing special is going on here.

But form the AC signal point of view what is the gain?

vcontrol = (3.5V - 1.5V)/(4.9V - 5.1V) = -10 V/V Surprised by the result?

For the AC signal, the circuit gain is equal to - R1/R2. Why? Because not AC current will flow into Rbias resistor.
Beouse due to the virtual short action Rbias current is constant and equal to Vref/Rbias. So, from the AC signal potin of view Rbias is seen as a constant current source. And what we do with the constant current source in AC small-signal analysis? We replace it with a "open circuit". This is why Rbi disappear from the equation.
Or you can take the derivative of a gain formula and see that the constant term ( Vref *(1+ R1/(R2||Rbias)) ) will disappear.

 

Thank you Jony for your explanation ! Really great and simple

 

Is the picture you used in your first post coming from "Switching Power Supplies A–Z" book?

 

Exactly ! What do you think about it ?

 

It's one of the best books I've read about this topic. On the downside, the book does not describe all forward topologies (push-pull, half-bridge, full-bridge, etc). But the ideal book does not exist.
I have first and the second edition on my bookshelf. I also have "Switching Power Supply Design and Optimization second edition" from the same author.

 

I really like it too ! I will take a look at the 2nd book when I will finish the first one Thank you for this advice If you have any other advice It will be a pleasure to hear it ! I have also the "Art of electronics 3rd edition" . I really like it too but the switching power supply part is relatively poor.