Small signal equivalent circuits

(a)

An answer:

Assuming voltage-controlled voltage sources as amplifiers or voltage controlled current sources with resistor — whatever floats your goat — you get:

pict

\begin{array}{rcl} \frac{v_{o 1}}{v_{i}} & = a_{v 1} \cdot \frac{r_{in 1}}{r_{in 1} + R_{g}} \\ = 30 \cdot \frac{6 k Ω}{6 k Ω + 600 Ω} \approx 27 \\ \frac{v_{o 2}}{v_{i}} & = a_{v 2} \cdot \frac{r_{in 2}}{r_{in 2} + R_{g}} \\ = 60 \cdot \frac{600 Ω}{600 Ω + 600 Ω} = 30 \end{array}

(b)

An answer:
The equivalent relevant part of the circuit for this question is given below.

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\begin{array}{rcl} H (jω) & = \frac{v_{o}}{v_{i}} \\ = \frac{r_{in}}{r_{in} + R_{g} + \frac{1}{jω C_{C}}} \\ = \frac{jω r_{in} C_{C}}{1 + jω (r_{in} + R_{g}) C_{C}} \\ = \frac{r_{in}}{r_{in} + R_{g}} \cdot \frac{jω (r_{in} + R_{g}) C_{C}}{1 + jω (r_{in} + R_{g}) C_{C}} \end{array}

The cut-off frequency of this first order high pass transfer function is

\begin{array}{rcl} f_{c} & = \frac{1}{2 π C_{C} (r_{in} + R_{g})} \end{array}

Solving for $C_{C}$ gives:

\begin{array}{rcl} C_{C} & = 6 μ F & (r_{i} = 600 Ω) \\ C_{C} & = 1.2 μ F & (r_{i} = 6 k Ω) \end{array}

(c)

An answer:
pict

(a)

An answer:

\begin{array}{rcl} I_{D_{1}} & = \frac{1}{2} K {(V_{GS} - V_{TH})}^{2} \\ V_{GS} = V_{G} - V_{S} \\ V_{G} = \frac{R_{g 2}}{R_{g 1} + R_{g 2}} \cdot V_{DD} \\ V_{S} = 0 \\ \Rightarrow \\ I_{D_{1}} & = \frac{1}{2} K {(\frac{R_{g 2}}{R_{g 1} + R_{g 2}} \cdot V_{DD} - V_{TH})}^{2} \end{array}

(b)

An answer:

A first step can be replacing impedances of pasives with their value at signal frequencies, and setting DC sources to zero.

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After which a second step can be replacing the non-linear components by their SSEC and cleaning up the schematic. This latter usually is unwinding the (virtual) ground node.

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(c)

An answer:

\begin{array}{rcl} a_{v} & = \frac{v_{out}}{v_{in}} \\ v_{out} = - g_{m} \cdot v_{gs} R_{d} \\ v_{gs} = - v_{s} \\ v_{s} = v_{in} \\ \Rightarrow \\ a_{v} & = \frac{- g_{m} \cdot - v_{in}}{v_{in}} R_{d} \\ = g_{m} R_{d} \end{array}

(d)

An answer:
Applying Ohm’s law to the input port and (re)using the input voltage source to drive input port:

\begin{array}{rcl} r_{in} & = \frac{v_{in}}{i_{in}} \\ i_{in} = - g_{m} \cdot v_{gs} \\ v_{gs} = - v_{s} \\ v_{s} = v_{in} \\ \Rightarrow \\ r_{in} & = \frac{1}{g_{m}} \end{array}

(e)

An answer:
Assuming a current source driving the output port:

\begin{array}{rcl} r_{out} & = \frac{v_{out}}{i_{out}} \\ v_{out} = (i_{out} - g_{m} \cdot v_{gs}) \cdot R_{d} \\ v_{gs} = 0 \\ \Rightarrow \\ r_{out} & = R_{d} \end{array}

An answer:
3 cascaded stages would give

{(100)}^{3} = 1 0^{6}

x gain. 3 parallel stages would give

100

x gain.

An answer:
Both

V_{DD}

and the capacitors are modelled by a short-circuit. Failing so most likely will force errors somewhere.

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The derivation of $r_{in}$ is straight forward, leading to:

\begin{array}{rcl} R_{in} = R_{3} + R_{1} ∕ ∕ \frac{αfe}{g_{m}} \end{array}

(a)

An answer:
Non-inverting. This can be shown in multiple ways. You could derive a proper equation for voltage gain and have a look at its sign (note that by setting reactances to shorts or opens there is no phase shift). Another way is to do an if-this-than-that analysis:

\begin{array}{rcl} an increase in v_{in} \Rightarrow \\ a decrease in v_{GS} \Rightarrow \\ a decrease in i_{D_{1}} \Rightarrow \\ a lower voltage drop across R_{d} \Rightarrow \\ an increase of v_{OUT} \end{array}

(b)

An answer:

pict

(c)

An answer:

\begin{array}{rcl} a_{v} & = \frac{v_{out}}{v_{in}} \\ v_{out} = - g_{m} v_{gs} R_{d} \\ v_{gs} = - v_{in} \\ \Rightarrow \\ a_{v} & = g_{m} R_{d} \end{array}

(d)

An answer:

\begin{array}{rcl} r_{in} & = \frac{v_{in}}{i_{in}} \\ i_{in} = - g_{m} v_{gs} \\ v_{gs} = - v_{in} \\ \Rightarrow \\ r_{in} & = \frac{v_{in}}{g_{m} v_{in}} = \frac{1}{g_{m}} \end{array}

(a)

An answer:

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(b)

An answer:

\begin{array}{rcl} a_{v} & = \frac{v_{out}}{v_{in}} \\ v_{out} = - g_{m} v_{be} R_{2} \\ v_{be} = v_{in} - v_{e} \\ v_{e} = g_{m} v_{be} (1 + \frac{1}{αfe}) R_{4} \\ v_{be} = v_{in} - g_{m} v_{be} (1 + \frac{1}{αfe}) R_{4} \\ = \frac{v_{in}}{1 + g_{m} (1 + \frac{1}{αfe}) R_{4}} \\ a_{v} & = - \frac{g_{m} R_{2}}{1 + g_{m} (1 + \frac{1}{αfe}) R_{4}} \end{array}

(a)

An answer:

pict

\begin{array}{rcl} a_{c} & = \frac{v_{out}}{v_{in}} \\ v_{out} = - g_{m} v_{be} R_{2} \\ v_{be} = \frac{(\frac{αfe}{g_{m}} ∕ ∕ R_{1}) v_{in}}{\frac{1}{jω C_{in}} + R_{3} + (\frac{αfe}{g_{m}} ∕ ∕ R_{1})} \\ \Rightarrow \\ a_{v} & = - \frac{g_{m} R_{2} (\frac{αfe}{g_{m}} ∕ ∕ R_{1})}{\frac{1}{jω C_{in}} + R_{3} + (\frac{αfe}{g_{m}} ∕ ∕ R_{1})} \\ = - g_{m} R_{2} \frac{jω (\frac{αfe}{g_{m}} ∕ ∕ R_{1}) C_{in}}{1 + jω (R_{3} + (\frac{αfe}{g_{m}} ∕ ∕ R_{1})) C_{in}} \end{array}

Where the shorthand notation for parallel impedances is used: $A ∕ ∕ B$ is the impedance of $A$ and $B$ in parallel.

(b)

An answer:
The big chunk of the work was done in the previous answer. From that is can readily be derived that the transfer function is first order high-pass. The high-frequency voltage gain is:

\begin{array}{rcl} a_{v} (\infty) & = - g_{m} R_{2} \frac{\frac{αfe}{g_{m}} ∕ ∕ R_{1}}{R_{3} + (\frac{αfe}{g_{m}} ∕ ∕ R_{1})} \end{array}

The pole (and zero) is:

\begin{array}{rcl} ω_{0} & = \frac{1}{(R_{3} + (\frac{αfe}{g_{m}} ∕ ∕ R_{1})) C_{in}} \end{array}

(c)

An answer:

pict

(a): An answer:
(b): An answer:
In the derivation already included is $v_{gs} = v_{in}$ . $\begin{array}{rcl} a_{v} & = \frac{v_{out}}{v_{in}} \\ v_{out} = R_{d} \cdot i_{Rd} \\ = R_{d} \cdot (- g_{m} v_{in} - \frac{v_{out} - v_{in}}{R_{g}}) \\ = \frac{- g_{m} R_{d} \cdot v_{in} + \frac{R_{d}}{R_{g}} v_{in}}{1 + \frac{R_{d}}{R_{g}}} \\ \Rightarrow \\ a_{v} & = \frac{\frac{R_{d}}{R_{g}} - g_{m} R_{d}}{1 + \frac{R_{d}}{R_{g}}} \\ = \frac{R_{d} (1 - g_{m} R_{g})}{R_{g} + R_{d}} \end{array}$
(c): An answer:
$z_{out} = R_{g} ∕ ∕ R_{d}$
(d): An answer:
$z_{out} = \frac{1}{g_{m}} ∕ ∕ R_{d}$

(a)

An answer:

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(b)

An answer:

\begin{array}{rcl} I_{D} & = \frac{1}{2} K {(V_{GS} - V_{T})}^{2} \\ V_{GS} = V_{GG} - I_{D} R_{S} \\ I_{D} & = \frac{1}{2} K {(V_{GG} - V_{T} - I_{D} R_{S})}^{2} \\ = \frac{1}{2} K {(V_{x} - I_{D} R_{S})}^{2}, with V_{x} = V_{GG} - V_{T} \\ = \frac{1}{2} K (V_{x}^{2} - 2 V_{x} I_{D} R_{S} + I_{D}^{2} R_{S}^{2}) \\ \Rightarrow (solve quadratic equation) \\ 0 & = I_{D}^{2} \cdot \frac{1}{2} K R_{S}^{2} - I_{D} (1 + K V_{x} R_{S}) + \frac{1}{2} K V_{x}^{2} \\ I_{D} & = \frac{1 + K V_{x} R_{S} \pm \sqrt{{(1 + K V_{x} R_{S})}^{2} - {(K V_{x} R_{S})}^{2}}}{K R_{S}^{2}} \end{array}

Of these 2 solutions, only one is valid. This valid solution corresponds to $V_{GS} > V_{T}$ .

(c)

An answer:

\begin{array}{l} g_{m} = \sqrt{2 K I_{D}}, substitute I_{D} from (b) \end{array}

(d)

An answer:

\begin{array}{l} v_{out} = - g_{m} v_{in} R_{D} \Rightarrow \frac{v_{out}}{v_{in}} = - g_{m} R_{D} \end{array}

(e)

An answer:
The input resistance is infinite.

(a)

An answer:

\begin{array}{rcl} a_{v} & = - g_{m} R_{C} \\ g_{m} & = 40 \cdot I_{C} \\ I_{C} & = \frac{V_{R_{C}}}{R_{C}} \end{array}

from which it follows that the maximum achievable (absolute) voltage gain is

\begin{array}{rcl} | a_{v} | & = 40 \cdot V_{RC} \end{array}

which is directly determined by (and limited by) the supply voltage. Note that the $40 = q ∕ kT \approx 40$ at room temperature; at low temperatures this factor is higher.

(b)

An answer:
Using the square law:

\begin{array}{rcl} a_{v} & = - g_{m} R_{D} \\ g_{m} & = \sqrt{2 K I_{D}} \\ I_{D} & = \frac{V_{R_{D}}}{R_{D}} \end{array}

from which it follows that the maximum achievable (absolute) voltage gain is

\begin{array}{rcl} | a_{v} | = \sqrt{2 K \frac{V_{R_{D}}}{R_{D}}} \cdot R_{D} = \sqrt{2 K R_{D} V_{R_{D}}} \end{array}

From which it follows that for maximum (absolute) voltage gain, the $K$ and the $R_{D}$ should be as large as possible and consequently the $I_{D}$ should be set very low. There seems to be no limit to these which would indicate that $| a_{v} |$ can be set arbitrarily high.

(c)

An answer:
For the MOS question, you probably used the square law. This square low is not valid for

v_{GS} - V_{T} < kT ∕ q

as then a different operation region is entered: moderate inversion (MI) or even weak inversion (WI). Weak inversion is ignored in this course: there the MOS transistor is modelled as being ”off”. Moderate inversion is not dealt with in this book as it is a transition region between strong inversion (square law) and the (here neglected) weak inversion region. This simplification for the MOS element equations makes derivation (results) for which

v_{GS} - V_{T} < kT ∕ q

far from realistic. Including MI and WI equations would show that MOS amplifiers end up at lower gain than for the bipolar case.

(a): An answer:
A partial answer: you should be able to map this on a corresponding SSEC. In the end, that’s what you did for an MOS transistor and/or for a BJT previously. $\begin{array}{rcl} \frac{\partial i_{G}}{\partial v_{GC}} = 0 \to open \\ \frac{\partial i_{G}}{\partial v_{AC}} = 0 \to open \\ \frac{\partial i_{A}}{\partial v_{GC}} = K \cdot \frac{3}{2} \cdot {(v_{GC} + \frac{v_{AC}}{μ})}^{1 ∕ 2} \to voltage controlled current source \\ \frac{\partial i_{A}}{\partial v_{AC}} = K \cdot \frac{3}{2 μ} \cdot {(v_{GC} + \frac{v_{AC}}{μ})}^{1 ∕ 2} \to resistor \end{array}$

(a): An answer:
A partial answer: you should be able to map this on a corresponding SSEC. In the end, that’s what you did for an MOS transistor and/or for a BJT and/or for the triode tube previously. $\begin{array}{rcl} \frac{\partial i_{G}}{\partial v_{GC}} = 0 \to open \\ \frac{\partial i_{G}}{\partial v_{AC}} = 0 \to open \\ \frac{\partial i_{A}}{\partial v_{GC}} = K \cdot \frac{3}{2} \cdot {(v_{GC} + \frac{v_{AC}}{μ_{c}} + \frac{V_{HIGH}}{μ_{s}})}^{1 ∕ 2} \to voltage controlled current source \\ \frac{\partial i_{A}}{\partial v_{AC}} = K \cdot \frac{3}{2 μ_{c}} \cdot {(v_{GC} + \frac{v_{AC}}{μ_{c}} + \frac{V_{HIGH}}{μ_{s}})}^{1 ∕ 2} \to resistor \end{array}$

(a)
(b)

4 Small signal equivalent circuits