# More op-amp fun

Discussion in 'General Electronics Chat' started by Distort10n, Jan 6, 2008.

1. ### Distort10n Thread Starter Active Member

Dec 25, 2006
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Here is a derivation that I am stuck on. You may notice that it is a transfer function of an op-amp. I have seen this solution before, but no derivation so I decided to takle it myself.

My question is how in the world is (1-b) reduced from R2K/(R2 + R1 - R1K)?

See attached.

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2. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,501
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How about a schematic, so we can see what we're working on?

3. ### recca02 Senior Member

Apr 2, 2007
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one clarification required,

4. ### hgmjr Moderator

Jan 28, 2005
9,030
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If the question is how to account for the term (1-b) in the expression on the right side of the initial expression through manipulation of the expression on the left side of the equation then it appears that you have the answer already.

$(1-b)= \frac{R2}{R1+R2} \ \ where\ b = \frac{R1}{(R1+R2)}$

As far as I can tell, you have already sorted this out in your manipulations to the right of the line you have drawn down the center of the page.

hgmjr

5. ### recca02 Senior Member

Apr 2, 2007
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thats exactly why asked whether this was abt the manipulation involved.

6. ### Distort10n Thread Starter Active Member

Dec 25, 2006
429
2
These two expressions are equal. I can derive the equation on the left hand side from the circuit (post later). The explanation goes on to say that the left hand side expression is equal to (1-b)*(k/(1-kb)) when b=(R1/(R1+R2)).

What I have done is derive the transfer function along with the example to arrive at the left hand side portion.

I then can show, for myself, that these are equivalent; however...

I have no idea how in god's high holy name it can be manipulated into the right hand side expression.

That is my question.

7. ### Dave Retired Moderator

Nov 17, 2003
6,960
161
(1 - b) = R2/(R1 + R2)

(1 - b)(R1 + R2) = R2

Multiply out left hand expression:

R1 + R2 - bR1 -bR2 = R2

R1 + R2 - b(R1 + R2) = R2

R1 + R2 = R2 + b(R1 + R2)

R1 = b(R1 + R2)

b = R1/(R1 + R2)

If you write it out as I have written on paper it will be clearer.

Dave

8. ### Distort10n Thread Starter Active Member

Dec 25, 2006
429
2
Maybe my brain is not working, but its seems to me that the equivalent expression is arbitrary:

Oh, BTW R2k/(R2+R1+R1k) = (1-b)(k/(1-kb)) if b = R1/(R2+R1).

Seeing that they are equal given the constraint of b = R1/(R2+R1) is fine, the math is easy. I can derive R2k/(R2+R1+R1k) from such a simple circuit with no problem...easy as pie.

Is this a math trick like "multiplying by 1"; i.e., (Rx/Rx) to help simplify an expression?

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9. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,501
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I would proceed as shown in the attachment.

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10. ### Dave Retired Moderator

Nov 17, 2003
6,960
161
Apologies, I misinterpreted what you were asking. The maths trick you looking at is multiplying the whole expression by 1 which will not change the expression) - however the trick is in your case multiply by (1/(R1+R2))/(1/(R1+R2)), which is something multiplied by itself which is 1.

The Electrician's answer shows how this is done in practice.

Dave