Engineering H-parameter model for non-inverting amplifier

[eyeweyew]

Homework Statement

Results from simple nodal analysis for non-inverting amplifier are not the same as results after converting the feedback portion to h-parameters circuit

Relevant Equations

Nodal analysis and h-parameter model

For a standard non-inverting amplifier as below:

With nodal analysis, I got $$\frac {v_o} {v_i}=\frac {R_EZ_o+aZ_iR_E+aZ_iR_f} {R_ER_f + R_EZ_i + R_EZ_o + R_fZ_i + Z_iZ_o + R_EaZ_i}$$

However, with the feedback network of the amplifier converted to a h-parameter network as below:

With nodal analysis on this converted circuit, I got $$\frac {v_o} {v_i}=\frac {aZ_i(R_E + R_f)^2} {R_ER_f^2 + R_E^2R_f + R_E^2Z_i + R_f^2Z_i + 2R_ER_fZ_i + R_ER_fZ_o + R_EZ_iZ_o + R_fZ_iZ_o + R_E^2aZ_i + R_ER_faZ_i}$$

I checked with Matlab to make sure the two results are different. My question is, are they suppose to produce different results? If so, why are they different? My prof is showing this so I expected they should be equivalent and produce same results.

 

[eyeweyew]

Nevermind, the results matched when I include h₂₁.

 

[The Electrician]

Homework Statement: Results from simple nodal analysis for non-inverting amplifier are not the same as results after converting the feedback portion to h-parameters circuit
Relevant Equations: Nodal analysis and h-parameter model

For a standard non-inverting amplifier as below:
View attachment 341330
With nodal analysis, I got $$\frac {v_o} {v_i}=\frac {R_EZ_o+aZ_iR_E+aZ_iR_f} {R_ER_f + R_EZ_i + R_EZ_o + R_fZ_i + Z_iZ_o + R_EaZ_i}$$

However, with the feedback network of the amplifier converted to a h-parameter network as below:

View attachment 341337

With nodal analysis on this converted circuit, I got $$\frac {v_o} {v_i}=\frac {aZ_i(R_E + R_f)^2} {R_ER_f^2 + R_E^2R_f + R_E^2Z_i + R_f^2Z_i + 2R_ER_fZ_i + R_ER_fZ_o + R_EZ_iZ_o + R_fZ_iZ_o + R_E^2aZ_i + R_ER_faZ_i}$$

I checked with Matlab to make sure the two results are different. My question is, are they suppose to produce different results? If so, why are they different? My prof is showing this so I expected they should be equivalent and produce same results.

Why have you chosen the orientation of your h parameter two port so the signal flow is from left to right?
Don't you want the feedback to flow from the opamp output back to the minus input, which would be right to left?

 

[eyeweyew]

Why have you chosen the orientation of your h parameter two port so the signal flow is from left to right?
Don't you want the feedback to flow from the opamp output back to the minus input, which would be right to left?

The final feedback would be fv_o where f=R_E/(R_E+R_f) so it is from right to left.

 

[The Electrician]

The final feedback would be fv_o where f=R_E/(R_E+R_f) so it is from right to left.

Reference this: https://en.wikipedia.org/wiki/Two-port_network

What I'm getting at is that you are using the output port of the h-parameter two-port you created as the input you're feeding the opamp output into, and taking signal fed into the minus input of the opamp from the input of your two-port. Fortunately the h-parameter two-port equivalent of the two resistor feedback network is bilateral, so it works out for you.

Ordinarily the thing to do is to use two-ports in the conventional manner, feeding your signal into the input port, and taking the output from the output port.

If you created a h-parameter two-port with Re in shunt at the output as I show in the lower right of the image below, it will give you the same result as you got with your choice of orientation, but you would be applying the output of the opamp to the input port and taking the output at the output port.