Argthtjtr

Trying to figure out the Close Loop System transfer function of an op-amp, however the math isn't working out as it should.

Circuit In question:

^{simulate this circuit – Schematic created using CircuitLab}

Using this as the foundation of calculating the Close loop transfer function:

Where A is the Transfer function (open loop) of the LM1875 which taken from the bode plot obtained here: LM1875 DataSheet. TF should be $$ A = H(s) = frac{31.62}{mathrm{707*10}^{-9}*s+1}$$

Where B is the transfer function of the negative feedback $$ B = G(s) = frac{R2}{R1}+1 = frac{1000kOmega}{1000kOmega}+1 = 2$$

The close loop Equation $$ CL(s) = frac{H(s)}{1+H(s)*G(s)} $$

$$ CL(s) = frac{frac{31.62}{mathrm{707*10}^{-9}*s+1}}{1+(frac{31.62}{mathrm{707*10}^{-9}*s+1})*(2)} $$

$$ CL(s) = frac{frac{31.62}{mathrm{707*10}^{-9}*s+1}}{1+(frac{63.24}{mathrm{707*10}^{-9}*s+1})} $$

$$ CL(s) = frac{frac{31.62}{mathrm{707*10}^{-9}*s+1}}{frac{mathrm{707*10}^{-9}*s+1}{mathrm{707*10}^{-9}*s+1}+(frac{63.24}{mathrm{707*10}^{-9}*s+1})} $$

$$ CL(s) = frac{frac{31.62}{mathrm{707*10}^{-9}*s+1}}{frac{mathrm{707*10}^{-9}*s+1+63.24}{mathrm{707*10}^{-9}*s+1}} $$

$$ require{cancel} CL(s) = frac{31.62}{cancel{mathrm{707*10}^{-9}*s+1}} * frac{cancel{mathrm{707*10}^{-9}*s+1 }}{mathrm{707*10}^{-9}*s+1+63.24} $$

$$ require{cancel} CL(s) = frac{31.62}{mathrm{707*10}^{-9}*s+64.24} $$

Using FVT as $$ Sxrightarrow{}0 $$ :

$$ require{cancel} CL(0) = frac{31.62}{mathrm{707*10}^{-9}*(0)+64.24} = frac{31.62}{64.24} = 0.4922 = DC Gain$$

Where did I go wrong?
I know this is wrong as the DC gain should be around 2V/V

I believe your mistake is to assume
$$ B = G(s) = frac{R2}{R1}+1 = frac{1000kOmega}{1000kOmega}+1 = 2$$
B is actually 1/2, as it is the output voltage divided by 2, that is
$$ B = G(s) = frac{R1}{R1+R2} $$
Where the ratio comes from the $R1,R2$ voltage divider.

With this value of B, you would obtain
$$ CL(0) = frac{31.62}{16.81} = 1.85$$

B is actually this:

Source: https://www.letscontrolit.com/wiki/index.php/DC_Voltage_divider

so B is equal to

$ frac{1000}{1000+1000}=1/2$

if the math is done with 1/2 instead of 2 for B you should get the right answer.

You've got B wrong. The voltage gain (or in this case attenuation) from output to summing node is R1/(R2+R1). The equation you listed is the closed loop gain (at DC at least) for the entire system. see: https://www.electronics-tutorials.ws/opamp/opamp_3.html

I'm also surprised at the transfer function you're using for the op amp. The DC gain should be much higher than 30. If you're going by this image below, note that the gain axis is in dB, so that 30 should be more like 1000.

You have the DC gain right. I was assuming 10db/decade, not 20dB/decade as I should have for voltage gain. 10^1.5=31.6

As others have stated, the fundamental flaw is in the feedback gain which is 1/2, not 2. However, I believe there's also a small error in the closed-loop equation. Let's try to derive it from the circuit above. Note, the feedback is attached to the inverting input of the op amp so there is a differencing block at the input, not summing. First, start at the output which I will label Y, and then follow it through the loop. The input is labeled X.

Output:
$$Y$$
Negative feedback gain:
$$BY$$
Input to the op amp:
$$X-BY$$
Output of the op amp:
$$A(X-BY)$$
Which is also equal to the output:
$$A(X-BY)=Y$$
Now we have the loop equation. Next, derive the closed-loop gain as Output divided by Input:
$$AX - ABY = Y$$
$$AX = Y + ABY$$
$$H = frac{Y}{X} = frac{A}{1+AB}$$
$$H = frac{Y}{X} = frac{A}{AB+1}$$
Let's calculate B first. If we assume that the input impedance of the op amp's inverting input is much, much larger than the impedance of the resistor feedback network, we can ignore any current draw it has from the feedback network. Larger means at least 10 times, but preferable 100 time or more. That would be about 100kΩ which I'm sure is smaller than the value on the datasheet. In that case, the current through both resistors is identical. We calculate the voltage drop of each path over it's total resistance. I will call the midpoint V and the ground point 0 in the following. The feedback gain B will be the output at the midpoint V divided by the input which happens to be Y in this system:
$$frac{V - 0}{R2} = frac{Y - 0}{R1 + R2}$$
$$frac{V}{R2} = frac{Y}{R1 + R2}$$
$$V = frac{R2}{R1 + R2}Y$$
$$B = frac{V}{Y} = frac{R2}{R1 + R2}$$
$$B = frac{1kΩ}{1kΩ + 1kΩ} = frac{1}{2} = 1/2$$
One goal of any good op amp is to have a very, very large open-loop gain which we represent by A. To simplify the equation, we can take the limit as A approached infinity to find the ideal gain of our system:
$$lim_{Atoinfty} H = frac{A}{AB+1} = frac{A}{AB} = frac{1}{B}$$
Since AB >> 1, we can drop the + 1 and with B being 1/2, H is 2 as the open-loop gain approaches infinity. Now, you care about the non-ideal op amp where A is less than infinity. Let's just see what happens if A is only 10:
$$lim_{Ato10} H = frac{A}{AB+1} = frac{10}{10(1/2)-1} = frac{10}{5+1}$$
$$lim_{Ato10} H = frac{10}{6} = 5/3 approx 1.67$$
Our actual gain drops. Put in the number from your datasheet here to find your expected gain which should be close, but slightly smaller than 2. Also, this still assumes an infinite input impedance on the op amp. That will affect the feedback gain B and should be accounted for to get a more accurate calculation. Hopefully, this helps clear up the confusion.