a fundamental but tricky question on feedback

get stuck on a fundamental question on feedback, looks simple, but just can not figure out what goes wrong, please refer to the attachment

Re: a fundamental question on feedback

my view is that fist try to find out type of feedback.........

a fundamental question on feedback

It's a special case for feedback analysis, the equation 1 is right, however, when you try to get the equation 2, you used a wrong beta, I am sure that you can get a result just like equation 1.
you can search for some literature for this feedback network, sorry I have no such data at hand.

a fundamental question on feedback

you got a wrong polarity in the figure 1?

a fundamental question on feedback

I agree with abcyin

a fundamental question on feedback

i think my beta is correct, here is the reference from the razavi's book, please take a look. the configuration is for capacitor, but essence is identical as resistor.

I think the first equation is correct.

However, in the equation(2):
The transfer function you got is from "the positive input of the Opamp to the
Output" instead of the transfer function of "Vi to Vo".

The means of these two equations means different things.

maninnet said:

get stuck on a fundamental question on feedback, looks simple, but just can not figure out what goes wrong, please refer to the attachment

Click to expand...

Seeing ur first attachment , i can only say pls refer Ramakant Gaekwad , n u w'll get answer of ur all question. Here what i think is u r confusing with two types of OPAMPs, inverting(v0/vi=-R2/R1) & noninverting(1+R2/R1).

Hello manninet,
This feedback analysis of finding the Beta and putting in the general feedback equation does not work always accurately and is very tricky to apply sometimes. The 1st rule is that the loading effects should be included properly. If there starts coming a feedforward factor through the feedback network which becomes significant to the feedforward through the main amplifier you need to start defining and modelling your feedback network to know what is the actual Beta and what constitutes the feedforward network. This is one reason your Beta is simply not the resistor divider expression you take it to be since there is a feedforward path through the resistor network and so the ideal equation won't be exact because if you remember the Beta block in that feedback model is just unidirectional while in this case it is not. So you need to separate the feedforward component of this resistor divider and include that with the main opamp block and that actually would constitute your A for the circuit. There are a couple of ways you can find the A and Beta for this circuit, one way I have attached below. For a complete understand and to really see the limitations of this type of feedback analysis there are some good books and the best way is to analyse circuits using feedback methods and full blown circuit analysis methods yourself. Try GRay and Meyer, Razavi's feedback analysis in his book raises too many questions that go unanswered if you do it from him. Hope this helps.

The analysis is pretty easy here. Firstly, I do accept that the feedback factor is still R1/R1+R2, with R2 being the feeback resistor. But this is with respect to the virtual ground, since the positive node is connected to ground. Hence assuming that the gain is high enough, the voltage at this node is Vin* (R2/R1+R2) using superposition and Vout = 0. Now if you multiply this with the inverse of the feedback factor, you will get Vout = Vin * (-R2/R1). The minus sign comes because of the negative input put forth.

Hi, maninnet,

You have simply used a wrong equation for the closed-loop gain, instead of the right one: Acl = G A / (1+A b), where G is the input gain (for Aol=0). For a large A, Acl=G/b, not 1/b. G=-R2/(R1+R2), b=R1/(R1+R2), and G/b=-R2/R1. Note also that you have to add a block for G in the functional diagram. Your diagram, without G, is related to the "four types of feedback approach" and does not reflect the original circuit. It rather reflects a circuit, where the original components are lamped into blocks 'a', open-loop gain, and 'f', loop gain. Note that 'a' and 'f' do not necessarily equal 'A', open loop gain, and 'b', return ratio, which you have calculated using "Bode approach" ("return-ratio approach"). To conclude: your mistake is that you have mixed the above two approaches, quite a common mistake.

BTW, the input gain G is missing in many textbooks, and this confuses. Fortunately, a visiting prof from Israel had taught us the G gain in an intro course on electronic circuits.

Regards,
Jasmine

i think your diagram doesnt really represent the circuit.

BETA is R1/(R1+R2), but the input in your diagram is in fact R2/(R1+R2)*Vin, and with these expressions you can get your -R2/R1

i think you should find the transfer function of this circuit, without the "virtual ground " approximation.

your beta is wrong... the output is sampled and fed back to input as current... here Vo is across R2 and hence -(Vo/R2)*R1 is the feedback and hence beta is -R2/R1... the negative sign is due to the polarity of Vo...

A.Anand Srinivasan said:

your beta is wrong... the output is sampled and fed back to input as current... here Vo is across R2 and hence -(Vo/R2)*R1 is the feedback and hence beta is -R2/R1... the negative sign is due to the polarity of Vo...

Click to expand...

This is not true. The error signal is voltage, ve, right? To find the return-ratio, or "feedback" as you mentioned it, one has to suppress vi, forget about virtual ground, and excite the opamp dependent source with a unit signal. This gives the return ratio ve'=vo*R1/(R1+R2)=Aol*1*R1/(R1+R2)=Aol*b. Hence b=+R1/(R1+R2).

Actually this circuit has been analysed both as a voltage-voltage feedback and voltage current feedback circuit, your β would change in both cases.

think why we need negative feedback.
the one reason is to reduce the output to a acceptable value.
but there is also a second reason.

aryajur said:

Actually this circuit has been analysed both as a voltage-voltage feedback and voltage current feedback circuit, your β would change in both cases.

Click to expand...

This shunt-shunt feedback circuit is analyzed in the attached paper (see Fig. 2 and 3). Its a and f, as well as, A (k) and Ab (H=RR/k) partial gains are found, and the two corresponding functional diagrams are suggested (see Fig. 1). Note that in Fig. 1(b), b1 is the input gain G, I mentioned previously. Hope this paper clarifies everything.

Fawad Elahi said:

think why we need negative feedback.
the one reason is to reduce the output to a acceptable value.
but there is also a second reason.

Click to expand...

... and also third, etc.

Hi, all,
I think the analysis is correct.
However, there is a forward gain from Vin to V(-) node.
which is -R2/(R1+R2).
By adding this term all equations are correct.

Vo=-A*V(-)
V(-)=R2/(R1+R2)*Vin+R1/(R1+R2)*Vo
Here R1/(R1+R2) is beta
R2/(R1+R2) is forward gain
Loop gain T is still beta*A

Added after 6 minutes:

If you want to think about more, adding a output resistance at opamp and find the transfer function from intput to output. It is hard to apply feedback and get exactly answer.

answer is
Vo/Vin=[-R2/R1+ro/(R1+R2+ro)]*(T/(1+T))

find a easy way to get exact answer...

Added after 1 minutes:

T=A*R1/(R1+R2+ro)

Hi maninnet,

Your approach is correct; But you have considered feedback as Voltage-Voltage or Voltage-series; where as the the actual f/b in circuit is Voltage-Current or Voltage-Shunt i.e. o/p is sampled and current is fed into I/P. So to analyze this circuit correctly, you have to 1st convert I/P to it's nortan equivalent.

in=vin/R1
Aol= -A*R1 (A=open loop gain of opamp)
f=1/R2

Acl= -A*R1/(1+A*R1/R2)= -R2

vout=in*Acl = vin *-R2/R1

Hence,
Vou/Vin= -R2/R1.

Hope this makes sense.

BTW,

leg1234
V(-)=R2/(R1+R2)*Vin+R1/(R1+R2)*Vo

Click to expand...

Vamsi Mocherla
The analysis is pretty easy here. Firstly, I do accept that the feedback factor is still R1/R1+R2, with R2 being the feeback resistor. But this is with respect to the virtual ground, since the positive node is connected to ground. Hence assuming that the gain is high enough, the voltage at this node is Vin* (R2/R1+R2) using superposition and Vout = 0. Now if you multiply this with the inverse of the feedback factor, you will get Vout = Vin * (-R2/R1). The minus sign comes because of the negative input put forth.

Click to expand...

Both above approaches are similar but flawed. The eqn of V(-) is valid only when open-loop gain of opamp, A=0.

Hope I am able to make point clear.

Regards
ipsc.