Amplifier Circuit with Negative Feedback

Hi All

I am asked for the output voltage of the entire circuit (see attached).

The hint I get is to:

Find B, then ACL then VO

Where

B = the Feedback Network (i.e. a voltage divider) and X = the comparison element (compares input signal with output signal).

B is a voltage divider shown as R1 and R2 and it takes the output voltage, divides it, then feeds it into X.

My attempt:

Step 1 - Find B

B = output of feedback network / input applied to feedback network

So the formula for the output of a feedback network is:

Vo = Vi (R2 \ R1 + R2)

Vo = 0.03V (200 \ 9800 + 200) = 0.0006

Then

the input applied to the feedback network is:

Vin x Amplifier Gain = 0.03V x 500 = 15 V

Then

B = 0.0006 \ 15 = 0.00004

Step 2 - Find ACL

ACL = the gain of a closed loop.

AOL = the gain of an open loop (no voltage divider).

ACL = AOL \ 1 + BAOL

So

ACL = 15 \ 1 + 0.00004 x 15 = 15

Step 3 - Find Vo

Vo = voltage output

Okay, I am lost here. I already used the Vo formula in step 1 to get the voltage output of the feedback network (i.e. the voltage divider) so do I need to use it again?

I don't get it really. If Vin is 0.03 V why can't I simply multiply that with the amplifier gain of 500?

I suspect the divider circuit influences how you find the voltage of the entire circuit but I just can't see what formula to apply here for each of the 3 steps.

Any help much appreciated

Maybe this will help
On the basis of schematics we get three equations :
(1) V1=Vin-A
(2) Vout=V1*G
(3) A=Vout*B

G is a open-loop gain of a opamp ( forward gain ) = 500[V/V]
B - feedback factor ( feedback gain ), sometime we use letter "β" or "K" for feedback gain.

B=R2/(R1+R2)=0.02[V/V]

And now we can calculate the close loop gain, substitute 1 to 2
Vout=V1*G=(Vin-A)*G
And 3 for A
Vout=(Vin-Vout*B)*G
Vout=Vin*G-Vout*G*B
Vout+Vout*G*B=Vin*G
Vout=Vin*G/(1+G*B)
Vout/vin=G/(1+G*B)=45V/V
Or dividing all by G we get:
Vout/vin=1 / [ (1/G)+B ]
http://en.wikipedia.org/wiki/Negative_feedback_amplifier#Voltage_amplifiers

Here are several steps that if calculated properly will yield the answer:

Step 1: Take a close look at the input to the negative terminal of the summing junction. This voltage is Vout times the resistor attentuator. This is one important term.

Step 2: Using the other input to the summing junction (the input voltage Vin) and subtracting the voltage calculated in step 1, you will have the voltage at the output of the summing junction.

Step 3: Since the gain of the amplifier is known, it is a straightforward exercise to write the expression for the input of the amplifier as a function of the output Vout.

Step 4: From step 2 you have the voltage at the output of the summing junction and from step 3 you have the voltage that would need to be applied to the amplifier's input to to get Vout.

Step 5: You can set these two expressions equal to each other and solve for Vout since all other terms are given.

hgmjr

Thanks Jony and hgmjr

I need a bit of time to sit down and go through your advice step by step.

I will try to find B, then ACL then VO based on your advice. I need to stick to this approach as that is the hint in the question (i.e. Find B, then ACL then VO)

I will let you know how I go soon.

Jony130 said:

Maybe this will help
On the basis of schematics we get three equations :
(1) V1=Vin-A
(2) Vout=V1*G
(3) A=Vout*B

G is a open-loop gain of a opamp ( forward gain ) = 500[V/V]
B - feedback factor ( feedback gain ), sometime we use letter "β" or "K" for feedback gain.

B=R2/(R1+R2)=0.02[V/V]

And now we can calculate the close loop gain, substitute 1 to 2
Vout=V1*G=(Vin-A)*G
And 3 for A
Vout=(Vin-Vout*B)*G
Vout=Vin*G-Vout*G*B
Vout+Vout*G*B=Vin*G
Vout=Vin*G/(1+G*B)
Vout/vin=G/(1+G*B)=45V/V
Or dividing all by G we get:
Vout/vin=1 / [ (1/G)+B ]
http://en.wikipedia.org/wiki/Negative_feedback_amplifier#Voltage_amplifiers

Click to expand...

Jony
As usual your answer is pure genious. Not just because you know the subject, but because you have a rare talent of being able to teach!

Some material I read is written by some very clever people, but being clever shouldn't be a license to teach. I wish uni's would wake up to this and be careful who they hire. We need real teachers, not aloof intellects!

Anyway, please excuse my frustration here, hope they promote you in this forum soon Jony, because you are a true teacher and an intellect (but stick to the teaching part). Thanks so much for your much needed help.

Okay, now to try and use your teachings to express the solution the way the question has asked i.e. Find B, then ACL, then Vo.

Find B:
B=R2/(R1+R2)=0.02[V/V]

Okay, thats a repeat of your answer Jony, but repeated here to clarify. I missed this version of the voltage divider formula that was expressed in a very subtle way in my textbook (by an intellectual type that can't teach).

Find ACL:
Jony's expression: Vout/vin=G/(1+G*B)=45V/V

Forumla in my textbook: Acl = Aol / 1 + BAol

Where Acl = gain of closed loop
Aol = gain of open loop
B = feedback factor (or gain)

I am not convinced here that Vout/vin actually means Acl as Vout/Vin should mean Aol and not Acl

But ignoring the confusion for now (hopefully Jony will answer this one), I get, using my textbook formula:

Aol = 500
B=0.02
So
Acl = 500 /1 + (0.02 * 500) = 45V

Find Vout:

Jony's formula: Vout=Vin*G/(1+G*B)

Forumla in my textbook. Could not find one! This is where I swear alot...

Anyway, when Jony says G, my textbook would call that Aol. I hate my textbook....

So, moving on, using Jony's formula because I have no other choice:

Vout = 0.03 * 500 / 1 + (500 * 0.02) = 1.36 V

Which is the correct answer! Thanks again Jony!

Thanks for the good word.

Find B:
B=R2/(R1+R2)=0.02[V/V]
Okay, thats a repeat of your answer Jony, but repeated here to clarify. I missed this version of the voltage divider formula that was expressed in a very subtle way in my textbook

Click to expand...

As I said the B or "β" is a feedback factor, the "gain" of a feedback network.
The "gain" of a voltage divider is Vout/Vin


Vout=I*R2 and I=Vin/(R1+R2)

Vout=[ Vin/(R1+R2)] *R2
Vout=Vin*R2/(R1+R2)
Vout/Vin=R2/(R1+R2)

I am not convinced here that Vout/vin actually means Acl as Vout/Vin should mean Aol and not Acl

Click to expand...

In your diagram we have the amplifier with closed loop negative feedback.
So Vout/Vin is a Acl=Au_cl or whatever letter you use.

\(Acl=\frac{Aol }{1+Aol*\beta}\)

And if we remove the feedback network from the circuit, now Vou/Vin is equal Aol.

Negative feedback it's all about the gain stabilization.