Can someone help me derive the equations for this circuit?

[MRW]

Hello, this is from an Intersil datasheet:

http://i8.tinypic.com/504v9fp.jpg

I have two pages of messy equations, but nothing close to what they
have for EQ. 2 and EQ. 3. They mentioned using superposition, so I did
the same thing, but I think I may have an error somewhere.

For my equations, I just used KCL.

Thanks!

 

[Eeyore]

Hello, this is from an Intersil datasheet:

http://i8.tinypic.com/504v9fp.jpg

I have two pages of messy equations, but nothing close to what they
have for EQ. 2 and EQ. 3. They mentioned using superposition, so I did
the same thing, but I think I may have an error somewhere.

For my equations, I just used KCL.

Potassium chloride ?

Well, Iin+ is (Vin+ - Vcc) / (R5 + R6)

That makes the op-amp + input = Vcc + (Iin+ . R6)

The two op-amp inputs are at the same potential (from feedback action).

That should make it dead easy to calculate Iin- and hence Vout. I'm not going to
do it for you.

R3 is redundant in a well-designed circuit btw.

Graham

 

[MRW]

Potassium chloride ?

Kirchoff's Current Law

Well, Iin+ is (Vin+ - Vcc) / (R5 + R6)

That makes the op-amp + input = Vcc + (Iin+ . R6)

The two op-amp inputs are at the same potential (from feedback action).

That should make it dead easy to calculate Iin- and hence Vout. I'm not going to
do it for you.

R3 is redundant in a well-designed circuit btw.

Graham

Thanks, Graham! I should have looked at it that way earlier. I totally
skipped out the non-inverting part because the application note said
that they used superposition. So, I just grounded the Vin+ input and
took the non-inverting input as just the Vinput+ = Vcc+ * (R5 / (R5 +
R6)). From there, I substituted it into my KCL equation and the mess
started. I couldn't figure out how to get their Vo- equation.

 

[The Phantom]

Hello, this is from an Intersil datasheet:

http://i8.tinypic.com/504v9fp.jpg

I have two pages of messy equations, but nothing close to what they
have for EQ. 2 and EQ. 3. They mentioned using superposition, so I did
the same thing, but I think I may have an error somewhere.

For my equations, I just used KCL.

Thanks!

You shouldn't feel bad that you can't get what they have, because their
equations are wrong. Look at the multiplying factors in parentheses,
(R4+R3)/R3 and R4/R3. If R3 becomes zero, these factors will cause the
gains to become infinite. If R3 is zero, then Vo/Vin- is plainly -R4/R1;
their expression doesn't reduce to this when R3 is zero.

 

[The Phantom]

Hello, this is from an Intersil datasheet:

What datasheet is this from?

 

[Jon Slaughter]

Hello, this is from an Intersil datasheet:

http://i8.tinypic.com/504v9fp.jpg

I have two pages of messy equations, but nothing close to what they
have for EQ. 2 and EQ. 3. They mentioned using superposition, so I did
the same thing, but I think I may have an error somewhere.

For my equations, I just used KCL.

NO!!! thats the wrong way(atleast the hard way).

Your making it way harder than what it is

You have to understand how an op amp works. The trick is to realize that the
op amp has infinite input impedence and that it makes the inputs the same.

What you see is that on the non-inverting side its simply a voltage divider
of Vin.

Think of the op amp sensing the voltage divider but not actually changing
the that part of the circuit(its kinda like a volt meter)

So OPP = (VCC - VInp)*R6/(R5 + R6)

(OPP Means op amp non-inverting side)

This is simply the voltage in the middle of R5 and R6. Its called the
voltage divider rule which you hopefully know.

Now the op amp forces the voltage on the other side, the inverting, to be
OPP. This is the voltage between R3 and R4.


So what you have there is


VCC-
|
R2
|
Vinm ---- R1 ----- + ------ R3 ----- OPP -------- R4 -------- V0
^
Vx

Its a very simple circuit with just resistors in it.

Hopefully you can figure out VO from here?

Start with the current through R4

I = (OPP - V0)/R4

But thats the same current going through R3 and Vx

So

I = (Vx - OPP)/R3


But that current is the current through R1 and R2.

You solve this system and you got your answer.


You can solve this also by realizing you have a delta-wye or whatever its
called in there. Maybe some other tricks too.


Theres no need to use KCL or KFC or anything...

Jon

 

[The Phantom]

Hello, this is from an Intersil datasheet:

http://i8.tinypic.com/504v9fp.jpg

I have two pages of messy equations, but nothing close to what they
have for EQ. 2 and EQ. 3. They mentioned using superposition, so I did
the same thing, but I think I may have an error somewhere.

For my equations, I just used KCL.

Thanks!

The result I get from an analysis is:

http://i7.tinypic.com/4ktz8lf.gif

 

[John Popelish]

NO!!! thats the wrong way(atleast the hard way).

Your making it way harder than what it is

You have to understand how an op amp works. The trick is to realize that the
op amp has infinite input impedence and that it makes the inputs the same.

Actually, it is the assumption of infinite differential
voltage gain that allows the negative feedback to produce
any output by amplifying zero voltage difference between the
two inputs.

It is the assumption of infinite input impedance that allows
you to neglect the influence of the inputs on any divider
(connected to the inputs).

Hope that wasn't too pedantic for you.

Carry on.

 

[MRW]

What datasheet is this from?

Hi. I got it from:
http://www.intersil.com/data/an/an9757.pdf

Thanks!

 

[The Phantom]

Hi. I got it from:
http://www.intersil.com/data/an/an9757.pdf

Thanks!

Have you had any luck deriving the equations?

I think I can see where they made their mistake. Just deriving the
expression for the gain from the Vin- node for example, use the Y-Delta
transformation on the R1,R2,R3 triad. Temporarily ground the Vcc- node and
then the Tee network of R1,R2,R3 will become a Pi network, with a resistor
connected from the Vin- node to ground (which can be ignored because it is
just a load on the voltage source input and has no effect on gain), a
series resistor from Vin- to the - input of the opamp (this is the resistor
which sets gain), and a resistor from the - input of the opamp to ground
(which also will have no effect on the gain from the Vin- input). The
inputs to the + side of the opamp are also grounded for this calculation.

The standard gain expression for an inverting amp configuration with a
feedback resistor, Rf, and an input resistor, Ri, is Av = -Rf/Ri. In this
circuit Rf is R4, and using the Y-Delta transformation, we have
Ri = (R1*R2 + R1*R3 + R2*R3)/R2. This gives a gain from the Vin- node of:

-R2*R4
---------------------
R1*R2 + R1*R3 + R2*R3

Now if you left out the first term in the denominator, you would have:

-R2*R4
-------------
R1*R3 + R2*R3

which could be factored as:

R2 ( R4 )
------- * (- -- )
R1 + R2 ( R3 )

which is what they have, and which is incorrect. A similar error is
present in most of their expressions.