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Microelectronic Circuits by Sedra/Smith present a method to analyze series-shunt feedback circuits by separating the 'A circuit' from the 'B circuit' in which h11 and h22 are consolidated into the 'A circuit' along with source and load resistances and h21 is neglected: enter image description here

The open-loop gain A is determined by Vo/Vi:

enter image description here

Everything makes sense until they present the following example circuit:

enter image description here

They redraw the circuit so that the 'A Circuit' (on the left) and the 'B circuit' (on the right) are clearly shown:

enter image description here

I've highlighted the part that I don't understand. Why are R1, R2, and the input source returned to ground? If the parallel combination of R1 and R2 constitutes h11, then these resistors should not be returned to ground as per the first image I posted. What they seem to be doing is this: enter image description here enter image description here

So the question is: what's the justification for connecting h11 and Vi to ground when no such connection exists in the first image I posted?

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  • \$\begingroup\$ Do you have a reference in the book for \$h_{11} = R_1 // R_2\$ or did you find that yourself? I can't seem to reach that result. \$\endgroup\$ – Sven B May 22 '18 at 9:47
  • \$\begingroup\$ The feedback network is comprised of R1 and R2. The quantity h11 is the impedance seen looking into the input with the output shorted, so it is trivial to show that h11 = R1 || R2. \$\endgroup\$ – pr871 May 22 '18 at 11:39
  • \$\begingroup\$ It isn't trivial to me. The input ends up on the gate of Q1, which has infinite impedance at DC. \$\endgroup\$ – Sven B May 22 '18 at 11:42
  • \$\begingroup\$ Sorry, I should have explained that h11 corresponds to the feedback network only (that is, the feedback network separated from the rest of the circuit). This feedback network is comprised of R1 and R2 only. It's highlighted in the blue box in the example circuit. \$\endgroup\$ – pr871 May 22 '18 at 11:45
  • \$\begingroup\$ In the first image, h11 and h22 belong to the feedback network (the B circuit). However, they are consolidated into the A circuit for the purpose of analysis. \$\endgroup\$ – pr871 May 22 '18 at 11:47
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I started from the complete small-signal equivalent circuit and checked what they are doing.

schematic

simulate this circuit – Schematic created using CircuitLab

Replacing the feedback network with the hybrid parameters yields:

schematic

simulate this circuit

Where \$h_{11} = R_1 // R_2\$, \$h_{12} = \frac{R_1}{R_1+R_2}\$ and \$h_{22} = R_1 + R_2\$ as you probably have found.

So I believe your analysis is spot on. I strongly believe this is an error in the reference schematic.

There is also one other thing that points to an error. In order to make any series-shunt feedback work, you need circuit 'A' to have a differential input, and not a single-ended input as shown in the diagram. Else you cannot connect it in series with something else.

[EDIT]

I got my hands on the 6th edition of the book, and noticed that they are doing it correctly in Figure 10.16. They connect the negative terminal to ground in Example 10.4, 10.5 for series-shunt feedback.

What I also found was that our actual 'circuit A' as we found it, is not a two-port and so it cannot be represented by h-parameters. In order to be a two-port, the current flowing into the positive input terminal needs to be equal and opposite to the current flowing into the negative terminal. Since the positive input terminal is floating, there should be no current flowing from the negative input terminal, ever. I believe the author 'fixed' it shorting the grounds of input and output, turning it into a two-port. However, at the same time the author also tied the negative input and negative output together, which is obviously not possible for series-shunt feedback as you pointed out.

While this action does turn the amplifier into a two-port, it will not lead to a correct representation of the actual circuit. Nevertheless, it may be an okay approximation if the negative input terminal doesn't conduct much current (if it doesn't, we can neglect that current and treat the input as a port).

Applying Kirchoff's Current Law to the following region in blue shows that the current through \$R_{D1}\$ needs to be equal to the current through the negative input terminal (sum of all the currents into the blue region need to add up to 0).

Kirchoff's Current Law

So if the current through \$R_{D1}\$ can be limited in some way, we may be relatively OK.

If we assume that the gain of 'circuit A' is large, then we can also assume that the inputs are pretty much virtually shorted. So to estimate the current flowing from the negative input, we can apply a voltage \$V\$ to both positive and negative input and see what current we get through \$R_{D1}\$ (and so through the negative input).

I found (hope I'm correct)

\$i_- = V\cdot \frac{g_{01}}{g_{01}(h_{11} + R_{D1}) + g_{m1}h_{11} + 1}\$

It seems that the approximation holds for large \$g_{m1}\$, large \$h_{11}\$, large \$R_{D1}\$ and small \$g_{01}\$ (and small \$V\$). This is in line with the typical design of this kind of circuit.

I'm not sure if the author of the book realized this. I personally think they compared their results with simulations, and saw that they matched. But purely theoretically I think there is something off.

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  • \$\begingroup\$ With regard to your last comment, I believe the 'A circuit' does have a differential input (one input at the MOSFET gate and the other at the MOSFET source). Once h11 is moved from the 'B circuit' to the 'A circuit', it connects to the MOSFET source (as you have drawn), and the 'A circuit' terminal is moved to encompass h11. \$\endgroup\$ – pr871 May 22 '18 at 13:56
  • \$\begingroup\$ The open-loop gain, A, is determined by Vo/Vi in the last diagram you provided. The problem is that, unless you connect h11 to ground (which the authors do and I don't understand why), the output is cut-off from the input and so A = 0, which is obviously incorrect. However, if this is an error as you suspect, they make the same 'error' numerous times throughout the chapter, and since this is a popular textbook in its 7th edition, I tend to feel that they have a legitimate reason for connecting h11 to ground. \$\endgroup\$ – pr871 May 22 '18 at 13:59
  • \$\begingroup\$ I was referring to the + and - in the circuit diagram for the 'A circuit'. The minus is clearly meant for the AC ground symbol, but this means the input is not differential. \$\endgroup\$ – Sven B May 22 '18 at 13:59
  • \$\begingroup\$ Yes I agree, this is the whole source of my confusion. But as I said, I have a hard time believing that they're screwing this up. \$\endgroup\$ – pr871 May 22 '18 at 14:03
  • \$\begingroup\$ One possibility is that they are grounding the negative input just for the sake of analogy to the original circuit and for calculating the transfer function of the 'A circuit'. \$\endgroup\$ – Sven B May 22 '18 at 14:08

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