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I understand that the equivalent circuits describe the behavior of amplifier for signals of low amplitude that allow us to assume that the circuit behaves linearly. My questions are:

  1. Why are all the DC voltage and current sources that aren't varying with time zeroed out? I don't understand the statement- "As far as the signal is concerned, all DC sources have no effect on operation".

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As in the above figure, the small signal equivalent model shows that the resistance tied between drain and VDD(which is shorted) to be in parallel with the the current source gmVgs. However the voltage across resistance and the voltage VDS(Vo) are not quite the same. How is this accounted in the above model?

Is there any other way to derive expressions for gain and I/O impedance? I tried to draw an equivalent circuit with all the dc sources present.Note that the voltage across RD is not VDS(as it should not be). Is it correct?

enter image description here

Applying KVL to output loop:

VDD - gmVgsRD - Vout = 0

VDD/Vgs - gmRD = Vout/Vgs

Voltage gain = Av = Vout/Vgs = VDD - gmRD.

But the expression for gain derived by shorting VDD is -gmRD.

Where am I going wrong?

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  • \$\begingroup\$ I have just drawn the circuit as it is by including all the dc voltage sources. What is wrong in that? \$\endgroup\$ – Aditya Patil Jul 11 '15 at 7:46
  • \$\begingroup\$ Small-signal analysis is like doing superposition and ignoring the direct voltages and currents. So just turn all the direct voltage sources down to zero and replace them with short circuits. Your equivalent circuit is wrong because the transconductance is defined as: \$g_m = \Delta I_d/\Delta V_{gs} = i_d/v_{gs}\$, i.e. change in output current divided by change in input voltage. \$\endgroup\$ – Chu Jul 11 '15 at 7:53
  • \$\begingroup\$ Yeah. That's what is shown in the circuit right? \$\endgroup\$ – Aditya Patil Jul 11 '15 at 8:27
  • \$\begingroup\$ No, your circuit shows: \$i_d = g_m (v_{gs} + bias)\$. It should be \$i_d = g_m v_{gs}\$ \$\endgroup\$ – Chu Jul 11 '15 at 8:31
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    \$\begingroup\$ Exactly, and it's the inclusion of direct voltages that makes it wrong. \$\endgroup\$ – Chu Jul 11 '15 at 8:33
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The true answer to your question unfortunately involves some bits of advanced calculus. Small signal models are derived from a first-order multi-variable Taylor expansion of the true non-linear equations describing the actual circuit behavior. This process is called circuit linearization.

Let's consider a very simple example with only one independent variable. Assume you have a non-linear V-I relationship for a two-terminal component that can be expressed in some mathematical way, for example \$i = i(v) \$, where \$i(v)\$ represents the math relationship (a function). Regular (i.e. one-dimension) Taylor expansion of that relation around an arbitrary point \$V_0\$, gives:

$$ i = i(V_0) + \dfrac{di}{dv}\bigg|_{V_0} \cdot (v-V_0) + R = i(V_0) + \dfrac{di}{dv}\bigg|_{V_0} \cdot \Delta v + R $$

where \$R\$ is an error term which depends on all the higher powers of \$\Delta v = v - V_0\$. The linearization consists in neglecting the higher order terms (R) and describe the component with the linearized equation:

$$ i = i(V_0) + \dfrac{di}{dv}\bigg|_{V_0} \cdot \Delta v $$

This is useful, i.e. gives small errors, only if the variations are small (for a given definition of small). That's where the small signal hypothesis is used.

Keep well in mind that the linearization is done around a point, i.e. around some arbitrarily chosen value of the independent variable V (that would be your quiescent point, in practice, i.e. your DC component). As you can see, the Taylor expansion requires to compute the derivative of \$i\$ and compute it at the same quiescent point \$V_0\$, giving rise to what in EE term is a differential circuit parameter \$\frac{di}{dv}\big|_{V_0}\$. Let's call it \$g\$ (it is a conductance and it is differential, so the lowercase g). Moreover, \$g\$ depends on the specific quiescent point chosen, so if we are really picky we should write \$g(V_0)\$.

Note, also, that \$i(V_0)\$ is the quiescent current, i.e. the current corresponding to the quiescent voltage. Hence we can call it simply \$I_0\$. Then we can rewrite the above linearized equation like this:

$$ i = I_0 + g \cdot \Delta v \qquad\Leftrightarrow\qquad i - I_0 = g \cdot \Delta v \qquad\Leftrightarrow\qquad \Delta i = g \cdot \Delta v $$

where I defined \$\Delta i = i - I_0\$.

This latter equation describes how variations in the current relate to the corresponding variations in the voltage across the component. It is a simple linear relationships, where DC components are "embedded" in the variations and in the computation of the differential parameter g. If you translate this equation in a circuit element you'll find a simple resistor with a conductance g.

To answer your question directly: there is no trace of DC components in the linearized (i.e. small signals) equation, that's why they don't appear in the circuit.

The same procedure can be carried out with components with more terminals, but this requires handling more quantities and the Taylor expansion becomes unwieldy (it is multi-variable and partial derivatives pop out). The concept is the same, though.

Small signal models are nothing more than the circuit equivalent of the differential parameters obtained by linearizing the multi-variable non-linear model (equations) of the components you're dealing with.

To summarize:

  • You choose a quiescent point (DC operating point): that's \$V_0\$
  • You compute the dependent quantities at DC (DC analysis): you find \$I_0\$
  • You linearize your circuit around that point using the DC OP data: you find \$g\$
  • You solve the circuit for small variations (aka AC analysis) using only the differential (i.e. small-signal) model \$g\$.
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  • \$\begingroup\$ Thanks! I do understand how small signal models are derived and the underlying approximations. We fix a bias point and assume that the signal excursions are small enough for the circuit to be linear. If we are to draw a complete model, why not include the DC sources? As I've asked above, the equivalent model shows RD having same voltage across it as Vout. Isn't it wrong? \$\endgroup\$ – Aditya Patil Jul 11 '15 at 11:23
  • \$\begingroup\$ Thank you so much!Great explanation. It would be a great help if you can answer this. electronics.stackexchange.com/questions/177745/… \$\endgroup\$ – Aditya Patil Jul 11 '15 at 11:32
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    \$\begingroup\$ Aditya Patil, assume a slightly improved model with finite rds (small signal dynamic resistance between D and S). In this case, the DC source would drive a current through rds which simply is wrong. Hence, in your model, you cannot apply the known classical rules for circuit analysis. \$\endgroup\$ – LvW Jul 11 '15 at 11:36
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When calculating the gain of an amplifying stage, do the DC voltages appear in the gain formula ? NO! However, the value of some of the parameters (like the transconductance gm) DEPEND on DC parameters.

Therefore, it is the purpose of the the so-called "small-signal" equivalent circuit diagram to visualize the interaction between all the relevant parameters - thus, allowing to derive the desired formulas (for gain, input- and output impedances,..).

But note that such a an eqivalent circuit diagram is valid for small-signals only because transistor parameters like gm are dynamic/differential values only. Moreover, it is valid for a fixed DC operating point only (but the DC values do not appear in th diagram).

Answering your last question: For our calculations (gain, input impedace,..) we do not need these small-signal eqivalent diagrams at all. But they can help to derive equations and to UNDERSTAND the interrelations between some parts of the circuit (wanted, unwanted feedback effects). These diagrams are a visualization of the known formulas which were derived from transistor physics - together with known rules for circuit analysis (Ohms and Kirchhoffs laws). And all these formulas can be derived also from the actual (real) circuit diagram also.

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  • \$\begingroup\$ "When calculating the gain of an amplifying stage, do the DC voltages appear in the gain formula ? NO!" They don't appear because while deriving the expressions, DC voltages are ignored. That's exactly my question. Why!? \$\endgroup\$ – Aditya Patil Jul 11 '15 at 8:28
  • \$\begingroup\$ No - they are not ignored! They determine the VALUES of some transistor parametrs. And because gain, input and output impedances are also small-signal values, only small-signal parameters of the circuit must be taken into account. And the small-signal internal (dynamic, differential) source resistance of supply voltages is ZERO ! \$\endgroup\$ – LvW Jul 11 '15 at 8:46
  • \$\begingroup\$ I don't see why I should not get the same result by including DC sources as well. Could you please explain me? \$\endgroup\$ – Aditya Patil Jul 11 '15 at 8:50
  • \$\begingroup\$ I think, the last three lines of your first contributions give the answer. You are asking: "Where am I going wrong" because your result is wrong - taking the DC bias into consideration. The transconductance gm is DETERMINED by the DC bias voltage, but you must not consider this DC value twice while including it again in the value of Vgs (as you did). \$\endgroup\$ – LvW Jul 11 '15 at 9:08
  • \$\begingroup\$ Or - as another example - think of the BJT equivalent diagram with a finite small-signal input resistance rbe (h11, yie). Does the DC voltage at the base node drive a DC current through rbe? No - of course not! Because rbe applies to dynamic/differential values only! \$\endgroup\$ – LvW Jul 11 '15 at 9:12

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