H(s) in closed loop transfer function incorrect?

Question

I need some help verifying the transfer function of this amplifier circuit stage. The below circuit has a gain of 20 from the result of Rf and Rg.

Below is the nodal analysis equation where G is used to show conductance.

The problem I am having is that I have 2 transfer functions in a feedback loop. G(s) and H(s), the op amp open loop transfer function and the closed loop transfer function. When they combine in feedback loop I get a final transfer function of G(s)/(1+G(s)*H(s))

My output transfer function however seems to have gain bode plot of less than unity!

• Green curve = output transfer function
• Blue curve = G(s) op amp open loop transfer function
• Orange curve = close loop H(s)

Shouldn't the amplifier stage, well, amplify? I can clearly see that the math works out to make it less than unity gain, but how is the output voltage going to be amplified? For example: at DC the math works out to be -26dB, gain of ~1/20. Similarly, the closed loop gain is around +26dB.

When applying H(s) to G(s), in feedback it became net negative. But to apply a real voltage at the input and expect an output, let's say for 1 volt input @DC, Vi=1, Vo=ViTF -> Vo=10.05011=0.05011.

The problem here is I am missing 1/x somewhere to get the proper gain of 20. I thought the transfer function is Vo=TF*Vi not Vo=1/TF *Vi?

Answer 1

I am not familiar with Mathematica.

Your system

G should be of the format \$\frac{V3}{V1}\$. i.e. V1 is input, V3 is output of G(s)

H should be of the format \$\frac{V1}{V3}\$. i.e. V3 is input and V1 is output of H(s).

However, your image seems to show that HofS1 is a function which takes V1 as input and produces V3 as output. I think that this actually represents 1/H(s).

So the line SystemsModelFeedbackConnect(..) is actually doing

\$\frac{G(s)}{1 + G(s)\frac{1}{H(s)}} = \frac{G(s)H(s)}{H(s) + G(s)} \$

So for large values of G(s) (below 10^7 Hz?) you may be effectively plotting H(s) which is supported by the observation that green plot and yellow plot are symmetric about 0 dB.

Answer 2

Use a PID (or a PI controller rather:

Source: https://www.semanticscholar.org/paper/Chapter-Ten-Pid-Control-10.1-Basic-Control/32f76117181bcdd012511fdc0d78c96378a46e72 Figure 10

The P is the gain term, you want this to be 20.

\$ K_p = 20 = \frac{R_2}{R_1}\$

The I term will be where you want the pole to be (you only get one with the PI controller with a -20db/dec rolloff)

\$ K_I = 2\pi f = R_2 C_2 \$

If you really do need the output non-inverted, then use another inverting stage with a gain of 1 after the first one.