(Note that this question is related to this one, though this question represents my having somewhat better understood the "modification" alluded to in the linked question and wanting to understand why said modification is acceptable from an equivalent circuit perspective.)

Though it is not usually explicitly stated, every analog circuits text I have read (Gray and Hurst, Sedra and Smith, Razavi) proceeds with the (low-frequency, though the analogy extends if we use s-domain analysis) analysis of amplifiers in the following way. In what follows, := means "by definition". I refer to the figure below (taken from Sedra and Smith) for the discussion. Basically, we are interested in taking the general form of Figure (a) and making it into an equivalent circuit of the form in Figure (b).

enter image description here

(1) Compute \$R_{in}:= v_i/i_i\$. Note that this is defined with the \$R_L\$ termination attached on the output.

(2) Compute \$R_{o}\$, defined as the Thevenin resistance looking into the output (of course, this means \$v_{sig}\$ is short, since a Thevenin resistance is defined as having all independent sources off). Sedra and Smith I think make an error here: they define \$R_{o}\$ as the resistance looking into the amplifier output when \$v_i = 0\$ (see the Figure (c) below) but I don't think this can be correct (and Gray and Hurst disagree with them per my linked question) because it will miss that for bilateral circuits, \$R_{sig}\$ will matter for the output resistance. I expand on this below in my justification section.

(3) Determine \$A_{vo}:= (v_o/v_i)|_{i_o = 0}\$. Note that this is defined with the output circuit open-circuited.

(4) It is then claimed, as a "circuit theorem" as it were, that we can represent the actual amplifier circuit of Figure (a) with that of Figure (b) given these values.

Effectively, my question is why this is permitted. I am hoping for an explicit demonstration of the validity of this procedure from established circuit theorems like Thevenin's theorem and two-port theory. Let me emphasize at the outset that the aforementioned procedure is rigorously correct is our amplifier two-port network is unilateral (defined as \$y_{12} = 0\$). The problem is justifying this procedure in the bilateral case. Two-port theory is generally abandoned in favor of the procedure noted above in this bilateral case.

My attempt at justification of the aforementioned procedure:

The circuit of Figure (a) is a one-port circuit. It follows from Thevenin's thorem (of course we assume the amplifier has been linearized) that this one-port can thereby be represented as below, where as per the statement of Thevenin's theorem, \$R_{TH}\$ has been computed as in (2) above (that is, \$R_{TH} \equiv R_o\$) and \$V_{TH}\$ has been computed as the open-circuit voltage of the circuit. Since \$v_{sig}\$ is the only independent source in the circuit, it follows by linearity/superposition that this open circuit output voltage \$V_{TH}\$ is proportional to \$v_{sig}\$, say \$V_{TH} = G_{vo}v_{sig}\$.

enter image description here

This is fully rigorous and no discrepancies have yet arisen. With this stated, we are still not of the form in Fig (b). I emphasize that Figure (b) uses \$A_{vo}v_i\$ for the controlled voltage source, and that this is computed (per (3) above) at open circuit on the output for a given excitation \$v_i\$ on the input to the amplifier. We want to replace the independent \$V_{TH}\ = G_{vo}v_{sig}\$ in the equivalent circuit above with the controlled source of Figure (b) and the given input circuit. Recalling that \$V_{TH}\$ was computed from Figure (a) with the output open circuited (i.e. \$R_L\$ not attached), we might be tempted (indeed, I was) to argue as follows. This open-circuited Figure (a) is still linear and so \$v_i\$ is proportional to the lone excitation, \$v_{sig}\$. The constant of proprtionality obtains by considering KVL applied to Figure (a) when open-circuited, and in this case one has \$v_{sig} = i_iR_{sig} + v_i\$ which can be rearranged to give $$v_i|_{i_o = 0} = v_{sig}\frac{R_{in}}{R_{sig}+R_{in}}$$ where, importantly, \$R_{in} := (v_i/i_i)|_{i_o = 0}\$. From here, if one defines $$A_{vo}:= \frac{R_{sig}+R_{in}}{R_{in}}G_{vo}$$ which is the same thing we would have arrived at if we had have just done, as in the procedure above, \$(v_o/v_i)|_{i_o = 0}\$. It is then equivalent to break out a circuit as in the input circuit of Figure (b) which gives us \$v_i\$ here.

But this is a different definition than in the procedure part (1) above, where \$R_L\$ was part of the circuit in the \$R_{in}\$ calculation, whereas in my justification for Figure (b) here, \$R_{in}\$ was computed with the output circuit opened. For a bilateral amplifier, there is a difference. The justification I have given above is somewhat different than that performed in these classic textbooks. Basically, the discrepancy arises from me having a calculation of \$R_{in}\$ which is different in justifying the manipulations which get us to Figure (b)'s form. Where have I gone wrong? Do you agree with my procedure?

  • \$\begingroup\$ For a bilateral amplifier, there is a difference. Have you tried a specific example using these two methods and shown the difference? \$\endgroup\$
    – internet
    Commented Dec 25, 2023 at 14:14
  • \$\begingroup\$ See for example the input resistance of a source/emitter follower, where \$R_L\$ is in the standard expression for input resistance. The source/emitter follower is of course bilateral. @internet \$\endgroup\$
    – EE18
    Commented Dec 25, 2023 at 14:41
  • 1
    \$\begingroup\$ Sedra and Smith, I believe, make an error here: they define Ro as the resistance looking into the amplifier output when vi=0. No, Ro in this context refers to the output ressistance of the amplifier alone and does not include Rsig. \$\endgroup\$
    – internet
    Commented Jan 3 at 13:19
  • 3
    \$\begingroup\$ To answer the question in the title, I don't think the amplifier model in Figure 7.34 considers bilateral aspects. If you wish to take that into account, include a reverse gain part in the model \$\endgroup\$
    – internet
    Commented Jan 3 at 13:25
  • 1
    \$\begingroup\$ Yes, Fig 7.34b does not and that is the point. The amplifier in Fig 7.34a can in general be bilateral, and yet we represent it as in Fig 7.34b. @internet \$\endgroup\$
    – EE18
    Commented Jan 3 at 13:41

4 Answers 4


I think you might be looking for something that Sedra and Smith are not actually trying to provide. The purpose of Figure 7.34 is to introduce (really, re-introduce, see section 1.5) basic parameters for amplifier models. (The particular circuit in the figure is abandoned almost immediately in favor of transistor models based on controlled current sources.) There is no reason to be particularly rigorous or to place the simplified model into the larger context of two-port network theory, so the authors do not do so. The intent is to describe a simple unilateral voltage amplifier driving a passive load.

I'm not sure why you're looking for a feedback connection between input and output here. The model provided does not include such a connection, so the input and output resistances are, in fact, independent. The circuit in your other question is a T model, so it's not so much a disagreement as a different choice of model.

The reason that \$R_L\$ is shown connected is because that's going to be important later, when common-emitter/source amplifiers are discussed and the input and output resistance are re-derived.

In the hybrid-pi model, \$r_o\$ is specifically used to model channel length modulation and the Early effect, which are drain-source/collector-emitter phenomena.

There are, of course, more complex models that do include connections from output to input. They are discussed much later in the chapters on high-frequency models and feedback.

EDIT: I dug up a copy of Gray et al and studied your questions again. I looked at the common-collector amplifier (I'm not as familiar with common base/gate), they seem to do exactly what you're looking for in section 3.3.6. They find expressions for the input and output resistances and gain that depend on the source and load resistances, then give approximate formulas for the case where β >> 1 and \$R_S\$ and \$R_L\$ are reasonable values. Partial excerpt:

Analysis and Design of Analog Integrated Circuits, section 3.3.6


Slightly further in the section from the above

So in this case they get to something like S&S's 7.34(b) through a deliberate simplification.

But it seems like your question is deeper than that -- why is Figure 7.34(b) used as a standard model for amplifiers, even though it doesn't cover every case? I think the answer to that still comes down to the purpose of the text. Both Sedra and Smith and Gray and Hurst are teaching design. In that context, 7.34(b) is as much a goal as a description. It's how an amplifier is supposed to behave. The basic transistor amplifier circuits are used because they behave approximately like 7.34(b). When that approximation breaks down, further analysis is required, and seen in S&S and (much more extensively) in Gray and Hurst. But even then, it's often convenient to describe more complex behavior as a modification of simple behavior, e.g. in the Miller approximation. Eventually, if you want to move beyond approximations, you turn to simulation, which uses vastly more complex models for transistors anyway.

The models used in textbooks fit into a certain space -- accurate enough to give insight into the operation of real circuits, but simple enough to readily understand and use quickly. The concepts of input resistance, output resistance, and gain fall into this category. Are these values really always independent, as in 7.34(b)? No. Are they approximately independent? Most of the time. Are they a useful mental model? Definitely!

Hopefully this helped. If I have misunderstood and you would like me to look at this some more, please feel free to comment.

EDIT 2: I think I may finally understand the heart of your question. Rephrasing in my own words:

  1. We find a Thevenin equivalent looking into the input port to get an input resistance.
  2. We find a different Thevenin equivalent looking into the output port to get an output bias voltage and an output resistance.
  3. We then smoosh those two Thevenin equivalents together into a two-port equivalent circuit like Figure 7.34(b).
  4. Is the smooshing really a valid transformation (from a circuit theory standpoint), especially given that there's a dependent source in the middle?

My answer to that is no. Gray et al's treatment of the common-collector amplifier states this outright, as seen in the quoted section above. The equivalent resistance \$R_{in}\$, which includes \$R_L\$, does not align with the control voltage as show in 7.34(b). So theoretically, it's not a valid transformation in this case.

It is, of course, still a useful model, even as an approximation. The main way it is useful is in answering the following questions:

  1. How does the amplifier load the source? \$(R_i)\$
  2. What is the limit on the amplifier's output current? \$(R_o)\$
  3. Given reasonable sources and loads, what is the gain of the amplifier? \$(A_v)\$

For these questions, we only look at one part of the amplifier's behavior at a time, so we're not really relying on the "two-sided" nature of the equivalent.

  • \$\begingroup\$ Thank you for your comment. I guess my point is that when these complex models are introduced (i.e. bilateral models) we still represent the given amplifier stage in this way, and I am trying to understand the justification therefor. Even a simple model illustrates this -- the source/emitter follower with \$r_o\$ considered is bilateral, and yet we still represent it as in Fig 7.34b). \$\endgroup\$
    – EE18
    Commented Jan 4 at 15:13
  • \$\begingroup\$ @EE18 All (for a MOSFET) or almost all (for a BJT) of the output current flows from drain/collector to source/emitter. The signal resistance's contribution to the output resistance is is thus zero or negligible. This is discussed in S&S's development of the T model, where they refer to the source absorption theorem (is that what you're looking for?). Perhaps I'm misunderstanding your question? My point is that this is an introductory text for sophomores, not a reference work on modeling. In that context, Fig 7.34b seems to me to be more of a definition or a design goal than a description. \$\endgroup\$
    – Adam Haun
    Commented Jan 4 at 18:09
  • \$\begingroup\$ That's well taken as it pertains to S&S. Let me perhaps instead refer to Gray and Hurst which I think is meant to be much more of a reference (although maybe I'm wrong here too). Even still, they seem tacitly to use this strategy of taking any amplifier stage, computing \$R_{in}, R_{out}, G_m\$, and then representing it as in Fig 7.34b. In the case of a unilateral amplifier this coincides with two-port theory but for a bilateral amplifier it does not, and indeed these definitions include load resistance in \$R_{in}\$ and source resistance in \$R_{out}\$, so it is surely very different... \$\endgroup\$
    – EE18
    Commented Jan 4 at 21:18
  • \$\begingroup\$ ...than two-port theory in this case. \$\endgroup\$
    – EE18
    Commented Jan 4 at 21:18
  • 1
    \$\begingroup\$ No worries at all. I don't think it's a you problem but simply me not totally following (or perhaps not even understanding exactly what I am asking). At any rate, I am going to leave this open but want to thank you for your time and help with this. I am going to ask my professor about this too and will report back if I glean anything! \$\endgroup\$
    – EE18
    Commented Jan 12 at 16:35

I edited this answer to better address the question..... Consider the special case of a network that is terminated at the input and output with simple one ports (that are not interconnected). By choosing appropriate values for the two resistors and the gain parameter of the circuit of figure B, this circuit can have the same input impedance, output impedance, and loaded gain as any two port network. It will not have the same "feedback" parameter, but if the original network is "mostly" unilateral, this may be OK. For bilateral networks, the input and output impedances change as the output and input loadings change. This is obviously not true for the circuit of figure B. Therefore, using this equivalent circuit will only work for the particular sources and loads that were used to derive the equivalent circuit.
The three parameters of the equivalent circuit can be measured from the original loaded circuit, or can be computed if the two port parameters of the original circuit are known. For example, the output impedance of a circuit with a known generator resistance is Zout=z22-z12*z21/(z11+zgen). You can see that the bilateral nature of the original circuit effects the values of the equivalent circuit, even though the equivalent circuit is unilateral.

  • \$\begingroup\$ The trouble is why it's acceptable to miss the feedback term. Note also how the general characterization I refer to here includes the load and source resistances in the input and output resistances, respectively. This is certainly not something we do when calculating two-port parameters. \$\endgroup\$
    – EE18
    Commented Dec 27, 2023 at 21:47


simulate this circuit – Schematic created using CircuitLab

Here is an example of two networks, one which is bilateral and one which is of the form of figure B. Both have the same generator and load, and both have the same input impedance, output impedance, and gain. The equivalent circuit is only valid with the terminations shown.

  • \$\begingroup\$ Thank you for this, this is very helpful. I guess I'm looking for a rigorous proof of why this procedure works though too. \$\endgroup\$
    – EE18
    Commented Dec 29, 2023 at 14:49


An operational amplifier is modeled as a 2-port passive linear network.

In 2-port networks there are 4 variables: Vin, Iin, Vout, Iout

2-port networks give us 2 relationships between the 4 variables and we need to add extra 2 relationships to solve the linear system of 4 variables and 4 equations.

Let's see 3 examples where we add 2 extra relationships:

Example 1:

Vin = sin(t)
Vout = 100 * Iout

A voltage source is connected to input port. A resistor of 100 Ohm is connected to output port.

Example 2:

Vin = 50*In + 10 Iout = 0

A voltage source of 10 V and a series resistor of 50 Ohm is connected to intput port. The output port is left open.

Example 3:

Iin = 5
Iout = 0

A current source of 5 A is connected to input port. The output port is left open.

The first 2 examples do modify the topology of the 2-port network because when you turn off the voltage source there is a wire that connects the 2 input nodes.

And so what?

The circuit designer connects anything he/she wants to the 2-port network. He/she doesn't care about modifying the topology of the the 2-port network

The bottom line is:

If you want to calculate the internal resistance R0 of your operational amplifier, connect to the input and output ports 2 circuits that once turned off they don't modify the topology of your 2-port network (the light blue circuit).

To measure Ro I would:

  1. Add a current source Iout of 1 Ampere to the output
  2. Leave open the input port (*)
  3. Measure the voltage Vout across Iout
  4. Calculate the ratio Ro = Vout/Iout

(*) Rin will receive no current and therefore the control voltage Vi will be zero.

Compute Ro, defined as the Thevenin resistance looking into the output (of course, this means vsig is short, since a Thevenin resistance is defined as having all independent sources off). Sedra and Smith I think make an error here: they define Ro as the resistance looking into the amplifier output when vi=0 (see the Figure (c) below) but I don't think this can be correct (and Gray and Hurst disagree with them per my linked question) because it will miss that for bilateral circuits, Rsig will matter for the output resistance. I expand on this below in my justification section.

Sedra's Ro is the output resistance of the whole circuit, that is Vsig, Rsig and the 2-port amplifier (the light blue circuit).

If you want to evaluate Ro of the 2-port amplifier (the light blue circuit) only, you have to leave the input port open and set Rsig to zero.

More precisely:

You have to use a current source Isig in parallel to the input port.

Isig indeed doesn't modify the topology of the 2-port amplifier when you turn it off and you can safely calculate Ro of the 2-port amplifier.

  • \$\begingroup\$ I'm not sure I understand so perhaps you can expand. How is Sedra's Ro the output resistance of the whole circuit if it ignores the resistance of the signal source, Rsig? \$\endgroup\$
    – EE18
    Commented Dec 29, 2023 at 14:51

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