5 Small-signal equivalent circuits - amplifiers

Q 5.1

(a)

An answer:
As a first guess, let’s neglect the base current, derive the value for $R_{B1}$ for that assumption and after that determine whether this assumption was valid. From the exercise we can readily get a number of currents and voltages:

$$\begin{array}{rcll}& & V_B=\frac{R_{B2}}{R_{B1}+R_{B2}}\cdot V_{\mathit{CC}}\to & \\ & & R_{B1}=\frac{R_{B2}\cdot (V_{\mathit{CC}}-V_B)}{V_B}& \end{array}$$

$$\begin{array}{rcll}& I_E=1\mathit{mA}& & \\ & R_E=500\mathrm{\Omega }& & \\ & V_E=0.5V& & \\ & V_{\mathit{BE}}=0.6& & \\ & V_B=1.1V& & \end{array}$$

$$\begin{array}{rcll}& R_{B1}& =\frac{R_{B2}\cdot (V_{\mathit{CC}}-V_B)}{V_B}& \\ & & =\frac{220k\mathrm{\Omega }\cdot (5-1.1)}{1.1}& \\ & & =780k\mathrm{\Omega }& \end{array}$$

Neglecting base current may be valid but this must be checked. ONLY if the base current $I_B$ is much smaller than the current in $R_{B1}$ the assumption was/is valid.

$$\begin{array}{rcll}& I_B& =\frac{I_E}{\mathit{\alpha fe}+1}=10\mathit{\mu A}& \\ & I_{B2}& =\frac{V_B}{R_{B2}}& \\ & & =\frac{1.1V}{220k\mathrm{\Omega }}& \\ & & =5\mathit{\mu A}=\frac{1}{2}{I}_B& \end{array}$$

As $I_B$ is not much smaller than $I_{\mathit{RB}2}$, the base current cannot be ignored. Therefore a more elaborate derivation much be done taking the impact of $I_B$ into account!

$$\begin{array}{rcll}& R_{B1}& =\frac{V_{R_{B1}}}{I_{B1}}& \\ & & I_{R_{B1}}=I_B+I_{R_{B2}}& \\ & & V_{R_{B1}}=V_{\mathit{CC}}-V_B& \\ & & I_{R_{B2}}=\frac{V_B}{R_{B2}}& \\ & R_{B1}& =\frac{3.9}{0.015\cdot 10^{-3}}=260k\mathrm{\Omega }& \end{array}$$

(b)

An answer:
There is (negative) feedback that serves to stabilize the bias settings. This can conveniently be explained using a type of flow diagram:

$$\begin{array}{rcll}& I_C& =I_{C0}(T)\cdot e^{\frac{q{V}_{\mathit{BE}}}{\mathit{KT}}}& \\ & & T\uparrow \to & \\ & & I_C\uparrow \to & \\ & & V_E\uparrow \mathit{and}{V}_B\downarrow \to & \\ & & V_{\mathit{BE}}\downarrow \to & \\ & & I_C\downarrow & \end{array}$$

whereby the $\uparrow$ indicates an increase and the $\downarrow$ indicates a decrease in value.

(c)

An answer:
$$\begin{array}{rcll}& g_m& \equiv \frac{\partial i_C}{\partial v_{\mathit{BE}}}& \\ & & =\frac{q}{\mathit{kT}}{I}_C& \\ & & \approx 40\cdot I_C& \\ & & =40\cdot 10^{-3}=0.04A∕V& \\ & \frac{\mathit{\alpha fe}}{g_m}& =\frac{99}{0.04}=2475\mathrm{\Omega }& \end{array}$$

(d)

An answer:

(e)

An answer:

$$\begin{array}{rcll}& v_{\mathit{out}1}& =-i_c\cdot R_C& \\ & v_{\mathit{out}2}& =i_e\cdot R_E& \\ & & =\frac{\mathit{\alpha fe}+1}{\mathit{\alpha fe}}{i}_c\cdot R_E& \\ & \mathit{with}& \mathit{\alpha fe}>>1\to & \\ & & I_C\approx I_E\to & \\ & & R_C\approx R_E=500\mathrm{\Omega }& \end{array}$$

(f)

An answer:

$$\begin{array}{rcll}& v_{\mathit{out}1}& =-i_c\cdot R_C& \\ & v_{\mathit{out}2}& =i_e\cdot R_E=\frac{\mathit{\alpha fe}+1}{\mathit{\alpha fe}}{i}_c\cdot R_E& \\ & v_{\mathit{out}2}& =-v_{\mathit{out}1}\cdot \frac{\mathit{\alpha fe}+1}{\mathit{\alpha fe}}\mathrm{\text{ (signals are in antiphase)}}& \end{array}$$

(g)

An answer:
The (unloaded) output voltage $v_{\mathit{out}2}$ due to $v_{\mathit{in}}$ is:

$$\begin{array}{rcll}& v_{\mathit{out}2}& =R_E\cdot I_{R_E}& \\ & & =R_E\left(\frac{v_{\mathit{be}}}{\mathit{\alpha fe}∕g_m}+g_m{v}_{\mathit{be}}\right)& \\ & & =g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)v_{\mathit{be}}& \\ & & v_{\mathit{be}}=v_b-v_e& \\ & & b_{\mathit{be}}=v_{\mathit{in}}-v_{\mathit{out}1}& \\ & v_{\mathit{out}2}& =g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)\left(v_{\mathit{in}}-v_{\mathit{out}1}\right)& \\ & & =\frac{g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}{1+g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}\cdot v_{\mathit{in}}& \end{array}$$

The next step is to calculate the short circuit output current due to $v_{\mathit{in}}$ - for this we need a new small signal equivalent circuit:

$$\begin{array}{rcll}& i_{\mathit{out}2}& =\frac{v_{\mathit{in}}}{\mathit{\alpha fe}∕g_m}+g_m{v}_{\mathit{be}}& \\ & & =v_{\mathit{in}}\left(\frac{g_m}{\mathit{\alpha fe}}+g_m\right)& \\ & & =v_{\mathit{in}}\left(\frac{1}{\mathit{\alpha fe}}+1\right)\cdot g_m& \end{array}$$

Note that for this SSEC, $v_{\mathit{be}}=v_{\mathit{in}}$ because $R_E$ appears shorted

$$\begin{array}{rcll}& r_{\mathit{out}}& =\frac{v_{\mathit{out}1}}{i_{\mathit{out}1}}& \\ & & =\frac{\frac{g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}{1+g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}\cdot v_{\mathit{in}}}{\left(\frac{1}{\mathit{\alpha fe}}+1\right)g_m{v}_{\mathit{in}}}& \\ & & =\frac{R_E}{1+g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}& \end{array}$$

(h)

An answer:

$$\begin{array}{rcll}& i_{\mathit{in}}& =\frac{v_{\mathit{in}}}{R_{B_1}∕∕R_{B_2}}+\frac{v_{\mathit{be}}}{\mathit{\alpha fe}∕g_m}& \\ & & v_{\mathit{be}}=v_{\mathit{in}}-v_{\mathit{out}1}& \\ \to i_{\mathit{in}}& =\frac{v_{\mathit{in}}}{R_{B_1}∕∕R_{B_2}}+\frac{v_{\mathit{in}}-\frac{g_m{R}_E\left(\frac{1}{\alpha \mathit{fe}}+1\right)}{1+g_m{R}_E\left(\frac{1}{\alpha \mathit{fe}}+1\right)}\cdot v_{\mathit{in}}}{\mathit{\alpha fe}∕g_m}& & \\ & =\frac{v_{\mathit{in}}}{R_{B_1}∕∕R_{B_2}}+v_{\mathit{in}}\cdot \frac{g_m}{\mathit{\alpha fe}}\cdot \left(1-\frac{g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}{1+g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}\right)& & \\ & =\frac{v_{\mathit{in}}}{R_{B_1}∕∕R_{B_2}}+v_{\mathit{in}}\cdot \frac{g_m}{\mathit{\alpha fe}}\cdot \frac{1}{1+g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}& & \\ & =\frac{v_{\mathit{in}}}{R_{B_1}∕∕R_{B_2}}+\frac{v_{\mathit{in}}}{\frac{\mathit{\alpha fe}}{g_m}+R_E(1+\mathit{\alpha fe})}& & \\ \to r_{\mathit{in}}& =R_{B_1}∕∕R_{B_2}∕∕\left(\frac{\mathit{\alpha fe}}{g_m}+R_E(1+\mathit{\alpha fe})\right)& & \end{array}$$

Reusing a previous equation for $v_{\mathit{out}1}(v_{\mathit{in}})$ can shorten the derivation (only) a little. This leads to

$$\begin{array}{rcll}& i_{\mathit{in}}& =\frac{v_{\mathit{in}}}{R_{B_1}∕∕R_{B_2}}+\frac{v_{\mathit{in}}-\frac{g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}{1+g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}\cdot v_{\mathit{in}}}{\mathit{\alpha fe}∕g_m}& \\ & & =\frac{v_{\mathit{in}}}{R_{B_1}∕∕R_{B_2}}+v_{\mathit{in}}\cdot \frac{g_m}{\mathit{\alpha fe}}\cdot \left(1-\frac{g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}{1+g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}\right)& \\ & & =\frac{v_{\mathit{in}}}{R_{B_1}∕∕R_{B_2}}+v_{\mathit{in}}\cdot \frac{g_m}{\mathit{\alpha fe}}\cdot \frac{1}{1+g_m{R}_E\left(\frac{1}{\mathit{\alpha fe}}+1\right)}& \\ & & =\frac{v_{\mathit{in}}}{R_{B_1}∕∕R_{B_2}}+\frac{v_{\mathit{in}}}{\frac{\mathit{\alpha fe}}{g_m}+R_E(1+\mathit{\alpha fe})}& \\ & r_{\mathit{in}}& =R_{B_1}∕∕R_{B_2}∕∕\left(\frac{\mathit{\alpha fe}}{g_m}+R_E(1+\mathit{\alpha fe})\right)& \end{array}$$

Q 5.2

(a)

An answer:
Bias = DC $\to$, $C=\mathit{open}$, $L=\mathit{short}$ and assume $V_{\mathit{BE}}\approx 0.6V$

$$\begin{array}{rcll}& I_C& =\mathit{\alpha fe}{I}_B& \\ & & I_B=\frac{V_{\mathit{CC}}-V_B}{R_B}& \\ & & V_B=\left(1+\frac{1}{\mathit{\alpha fe}}\right)I_C{R}_E+0.6V& \\ & & I_B=\frac{V_{\mathit{CC}}-\left(1+\frac{1}{\mathit{\alpha fe}}\right)I_C{R}_E-0.6V}{R_B}& \\ & I_C& =\mathit{\alpha fe}\cdot \frac{V_{\mathit{CC}}-\left(1+\frac{1}{\mathit{\alpha fe}}\right)I_C{R}_E-0.6V}{R_B}& \end{array}$$

This requires some rewriting: separation of variables to get a closed expression for $I_C$:

$$\begin{array}{rcll}& & I_C\left(1+\frac{(1+\mathit{\alpha fe})R_E}{R_B}\right)=\frac{\mathit{\alpha fe}(V_{\mathit{CC}}-0.6V)}{R_B}⟹& \\ & I_C& =\frac{\mathit{\alpha fe}(V_{\mathit{CC}}-0.6V)}{R_B\left(1+\frac{(1+\mathit{\alpha fe})R_E}{R_B}\right)}& \\ & & =\frac{\mathit{\alpha fe}(V_{\mathit{CC}}-0.6V)}{R_B+(1+\mathit{\alpha fe})R_E}& \end{array}$$

(b)

An answer:
The voltage drop across $L_1$ is zero (short at DC), so $V_C=V_{\mathit{CC}}$. Or you can see it directly from the equivalent circuit for operation at DC in the previous answer.

(c)

An answer:
A common emitter circuit (CEC)

(d)

An answer:
You can ignore the output impedance of the transistor as long as it is significantly higher than other impedances at the relevant node (collector). In this case that means that $Z_{L1}<<Z_{\mathit{out},\mathit{BJT}}$ for signal frequencies.

(e)

An answer:
Set DC sources to 0 (in this circuit that is only $V_{\mathit{CC}}$). Assume that the values of reactances for which no values are specified are large. Note that in this exercise you have to retain $C_{\mathit{in}}$ and $C_1$! This allows to replace them by shorts or opens: $L_1\to$open, $C_E\to$short, $C_{\mathit{out}}\to$short.

SSEC of the BJT (including B/C/E node identifiers)

Indicate $v_{\mathit{in}}$ and $v_{\mathit{out}}$ and redraw. Again: you should (see the description of the circuit) not replace $C_{\mathit{in}}$, $C_1$ by shorts!

(f)

An answer:
In this answer, the output port is driven from an independent voltage source driving the output port. The other independent sources need to be set to zero. The corresponding small signal equivalent circuit is given below. This will be used to derive $z_{\mathit{out}}=\frac{v_{\mathit{out}}}{i_{\mathit{out}}}$.

$$\begin{array}{rcll}& z_{\mathit{out}}& =\frac{v_{\mathit{out}}}{i_{\mathit{out}}}& \\ & & i_{\mathit{out}}=i_{L1}+g_m{v}_{\mathit{be}}& \\ & & v_{\mathit{be}}=0& \\ & & i_{L2}=\frac{v_{\mathit{out}}}{\mathit{j\omega }{L}_1}& \\ & z_{\mathit{out}}& =\frac{v_{\mathit{out}}}{i_{\mathit{out}}}=\mathit{j\omega }{L}_1& \end{array}$$

(g)

An answer:
Using the SSEC from question e) to calculate the voltage gain $a_v=\frac{v_{\mathit{out}}}{v_{\mathit{in}}}$ can be done as follows.

This SSEC can be simplified by capturing $C_1$ and $\mathit{\alpha fe}∕g_m$ into one impedance

$$\begin{array}{rcll}{Z}_B& =\frac{\frac{\mathit{\alpha fe}}{g_m}\cdot \frac{1}{\mathit{j\omega }{C}_1}}{\frac{\mathit{\alpha fe}}{g_m}+\frac{1}{\mathit{j\omega }{C}_1}}& & \\ & =\frac{\frac{\mathit{\alpha fe}}{g_m}}{1+\mathit{j\omega }{C}_1\frac{\mathit{\alpha fe}}{g_m}}& & \end{array}$$

A derivation can be done as shown below:

$$\begin{array}{rcll}& a_v& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \\ & & v_{\mathit{out}}=i_{L1}\mathit{j\omega }{L}_1& \\ & & i_{L1}=-g_m{v}_{\mathit{be}}& \\ & & v_{\mathit{be}}=v_B& \\ & & v_B=\frac{Z_B}{Z_B+Z_{C_{\mathit{in}}}}\cdot v_{\mathit{in}}& \end{array}$$

Now back substitution leads to the/an answer:

$$\begin{array}{rcll}& & v_B=\frac{Z_B}{Z_B+Z_{C_{\mathit{in}}}}\cdot v_{\mathit{in}}& \\ & & =\frac{\frac{\frac{\mathit{\alpha fe}}{g_m}}{1+\mathit{j\omega }{C}_1\frac{\mathit{\alpha fe}}{g_m}}}{\frac{\frac{\mathit{\alpha fe}}{g_m}}{1+\mathit{j\omega }{C}_1\frac{\mathit{\alpha fe}}{g_m}}+\frac{1}{\mathit{j\omega }{C}_{\mathit{in}}}}\cdot v_{\mathit{in}}& \\ & & =\frac{\mathit{j\omega }{C}_{\mathit{in}}\frac{\mathit{\alpha fe}}{g_m}}{1+\mathit{j\omega }\frac{\mathit{\alpha fe}}{g_m}(C_1+C_{\mathit{in}})}\cdot v_{\mathit{in}}& \end{array}$$

$$\begin{array}{rcll}& v_{\mathit{out}}& =-g_m{v}_B\cdot \mathit{j\omega }{L}_1& \\ & & =\mathit{j\omega }{L}_1{g}_m\frac{\mathit{j\omega }{C}_{\mathit{in}}\frac{\mathit{\alpha fe}}{g_m}}{1+\mathit{j\omega }\frac{\mathit{\alpha fe}}{g_m}(C_1+C_{\mathit{in}})}\cdot v_{\mathit{in}}& \\ & a_v& =-\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \\ & & =\frac{{(\mathit{j\omega })}^2\mathit{\alpha fe}{L}_1{C}_{\mathit{in}}}{1+\mathit{j\omega }\frac{\mathit{\alpha fe}}{g_m}(C_1+C_{\mathit{in}})}& \\ & & =-\frac{{(\mathit{j\omega })}^2\mathit{\alpha fe}{L}_1{C}_{\mathit{in}}}{1+\mathit{j\omega }\frac{\mathit{\alpha fe}}{g_m}(C_1+C_{\mathit{in}})}& \end{array}$$

Q 5.3

(a)

An answer:

$$\begin{array}{rcll}& R_E& =\frac{V_E}{I_E}& \\ & & V_E=V_B-V_{\mathit{BE}}& \\ & & I_E=\frac{\mathit{\alpha fe}+1}{\mathit{\alpha fe}}{I}_C& \end{array}$$

$V_B$ is the voltage at the base node, which generally cannot be calculated merely assuming $R_{B1}$ and $R_{B2}$ form a resistive divider and hence applying the formula for that. This can easily be seen in the circuit schematic depicting the voltages and currents at and around $R_{B1}$ and $R_{B2}$.

From this $V_B$ can be derived. Formulating KCL/KVL:

$$\begin{array}{rcll}& \frac{V_B}{R_{B2}}+\frac{V_B-V_{\mathit{CC}}}{R_{B1}}-I_C∕\mathit{\alpha fe}=0& & \end{array}$$

leading to

$$\begin{array}{rcll}& R_{B1}{V}_B+R_{B2}{V}_B-R_{B2}{V}_{\mathit{CC}}+R_{B1}{R}_{B2}{I}_B=0⟹& & \\ & V_B=\frac{R_{B2}{V}_{\mathit{CC}}-R_{B1}{R}_{B2}{I}_B}{R_{B1}+R_{B2}}& & \end{array}$$

Or applying superposition you get the same equation in a different appearance:

$$V_B=V_{\mathit{CC}}\frac{R_{B2}}{R_{B1}+R_{B2}}-I_B\frac{R_{B1}{R}_{B2}}{R_{B1}+R_{B2}}$$

This yields:

$$R_E=\frac{V_{\mathit{CC}}\frac{R_{B2}}{R_{B1}+R_{B2}}+I_B\frac{R_{B1}{R}_{B2}}{R_{B1}+R_{B2}}-V_{\mathit{BE}}}{\frac{\mathit{\alpha fe}+1}{\mathit{\alpha fe}}{I}_C}$$

This yields numerically

$$\begin{array}{rcll}& & V_B=4V& \\ & & V_{\mathit{BE}}=0.7V& \\ & & R_E=330\mathrm{\Omega }& \end{array}$$

(b)

An answer:
When the signals are sufficiently small so that we can approximate the non-linear transistor characteristics by a linear function.

(c)

An answer:

Which can be redrawn (simplified) into

To calculate the output resistance, e.g set $v_{\mathit{in}}=0$ and force a voltage at the output $v_{\mathit{out},\mathit{forced}}$. Then calculate the current delivered by $v_{\mathit{out},\mathit{forced}}$ and apply Ohm’s law.

$$\begin{array}{rcll}& r_{\mathit{out}}& =\frac{v_{\mathit{out},\mathit{forced}}}{i_{\mathit{out}}}& \\ & & i_{\mathit{out}}=\frac{v_{\mathit{out},\mathit{forced}}}{R_E}+\frac{v_{\mathit{out},\mathit{forced}}}{\mathit{\alpha fe}∕g_m}-g_m\cdot v_{\mathit{be}}& \\ & & i_{\mathit{out}}=\left(\frac{1}{R_E}+\frac{g_m}{\mathit{\alpha fe}}+g_m\right)\cdot v_{\mathit{out},\mathit{forced}}& \\ & r_{\mathit{out}}& =R_E∕∕\frac{\mathit{\alpha fe}}{g_m}∕∕\frac{1}{g_m}& \end{array}$$

XXXXXXXXXXXXXXXX

Q 5.4

(a)

An answer:
The gate current (for this model) is zero which does not require any (equivalent) small signal component. The drain current depends only on the gate-source voltage which can be modelled with a voltage controlled current source delivering $i_d=g_m\cdot v_{\mathit{gs}}$.

(b)

An answer:

Which can be simplified into:

This small signal equivalent shows that the gate is connected to ground AND that both the input signal port AND the output port have this ground in common.

(c)

An answer:

$$\begin{array}{rcll}& r_{\mathit{in}}& =\frac{v_{\mathit{in}}}{i_{\mathit{in}}}& \\ & & i_{\mathit{in}}=\frac{v_{\mathit{in}}}{R_S}-g_m\cdot v_{\mathit{gs}}& \\ & & v_{\mathit{gs}}=v_{\mathit{in}}& \\ & & i_{\mathit{in}}=\left(\frac{1}{R_S}+g_m\right)\cdot v_{\mathit{in}}& \\ & r_{\mathit{in}}& =R_S∕∕\frac{1}{g_m}& \end{array}$$

(d)

An answer:
The derivation is pretty much the same as for question (c), $r_{\mathit{out}}=R_D$.

Q 5.5

(a)

(b)

An answer:

(c)

An answer:

$$\begin{array}{rcll}& a_v& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \\ & & v_{\mathit{out}}=-g_m{v}_{\mathit{be}}\cdot (R_C∕∕R_{\mathit{in},\mathit{NEXT}})& \\ & & =+g_m{v}_{\mathit{in}}\cdot (R_C∕∕R_{\mathit{in},\mathit{NEXT}})& \\ & a_v& =g_m\cdot (R_C∕∕R_{\mathit{in},\mathit{NEXT}})& \\ & & =g_m\cdot \frac{R_C\cdot R_{\mathit{in},\mathit{NEXT}}}{R_C+R_{\mathit{in},\mathit{NEXT}}}& \end{array}$$

(d)

An answer:
From the previously derived small signal equivalent circuit:

$$\begin{array}{rcll}& r_{\mathit{in}}& =\frac{v_{\mathit{in}}}{i_{\mathit{in}}}& \\ & & =R_E∕∕\frac{\mathit{\alpha fe}}{g_m}∕∕\frac{1}{g_m}\approx \frac{1}{g_m}& \end{array}$$

(e)

An answer:
Recycling the answer to the previous question:

$$\begin{array}{rcll}\frac{1}{g_m}& =250\mathrm{\Omega }\to g_m=0.004[A∕V]\to I_C=0.1\mathit{mA}& & \end{array}$$

(f)

An answer:
From

$$a_V=g_m\cdot (R_C∕∕R_{\mathit{in},\mathit{NEXT}})$$

it can be derived that

$$\begin{array}{rcll}& R_C& =\frac{a_v\cdot R_{\mathit{in},\mathit{NEXT}}}{\mathit{gm}{R}_{\mathit{in}}-a_v}& \\ & & =40k\mathrm{\Omega }& \end{array}$$

(g)

An answer:

(h)

An answer:

$$\begin{array}{rcll}& a_v& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \\ & & v_{\mathit{out}}=-g_m{R}_C{v}_{\mathit{be}}& \\ & & v_{\mathit{be}}=v_b-v_e& \\ & & v_b=v_{\mathit{in}}& \\ & & v_e=\left(\frac{g_m{v}_{\mathit{be}}}{\mathit{\alpha fe}}+g_m{v}_{\mathit{be}}\right)\cdot R_{E_1}=v_{\mathit{be}}\cdot g_m{R}_{E_1}\left(1+\frac{1}{\mathit{\alpha fe}}\right)& \end{array}$$

Working this out yields:

$$\begin{array}{rcll}& & v_{\mathit{be}}=v_{\mathit{in}}-v_{\mathit{be}}\cdot g_m\cdot R_{E_1}\left(1+\frac{1}{\mathit{\alpha fe}}\right)\to & \\ & & v_{\mathit{be}}\left(1+g_m\cdot R_{E_1}\left(1+\frac{1}{\mathit{\alpha fe}}\right)\right)=v_{\mathit{in}}\to & \\ & & v_{\mathit{be}}=\frac{v_{\mathit{in}}}{\left(1+g_m\cdot R_{E_1}\left(1+\frac{1}{\mathit{\alpha fe}}\right)\right)}& \\ & a_v& =\frac{-g_m\cdot R_C}{1+g_m\cdot R_{E_1}\left(1+\frac{1}{\mathit{\alpha fe}}\right)}& \end{array}$$

(i)

An answer:

$$r_{\mathit{in}}=R_{B1}∕∕R_{B2}∕∕\left(\frac{\mathit{\alpha fe}}{g_m}+(1+\mathit{\alpha fe})\cdot R_{E1}\right)$$

(j)

An answer:

$$\begin{array}{rcll}& r_{\mathit{in}}& =400k\mathrm{\Omega }∕∕100k\mathrm{\Omega }∕∕\left(\frac{\mathit{\alpha fe}}{g_m}+(1+\mathit{\alpha fe})\cdot R_{E1}\right)& \\ & & \approx 80k\mathrm{\Omega }∕∕60k\mathrm{\Omega }\approx 35k\mathrm{\Omega }& \end{array}$$

$r_{\mathit{in}}$ is smaller than the assumed $40k\mathrm{\Omega }$ so the gain of the first stage will be a little lower than the numerical value derived earlier.

Q 5.6

(a)

An answer:

(b)

An answer:

$$\begin{array}{rcll}& a_V& =\frac{v_{\mathit{out}}}{v_{\mathit{mic}}}& \\ & & v_{\mathit{out}}=-g_m\cdot v_{\mathit{gs}}\cdot (R_D∕∕r_{\mathit{in},\mathit{next}})& \\ & & v_{\mathit{gs}}=-v_{\mathit{mic}}& \\ & & v_{\mathit{out}}=+g_m\cdot v_{\mathit{mic}}\cdot (R_D∕∕r_{\mathit{in},\mathit{next}})& \\ & a_V& =\frac{v_{\mathit{out}}}{v_{\mathit{mic}}}=g_m\cdot (R_D∕∕r_{\mathit{in},\mathit{next}})& \end{array}$$

(c)

An answer:

$$r_{\mathit{in}}=R_S∕∕\frac{1}{g_m}\approx \frac{1}{g_m}$$

(d)

An answer:

Recycling the results of the previous question and combining that with one of the equations for $g_m$ of an MOS transistor:

$$\begin{array}{rcll}& g_m& =\sqrt{2K{I}_D}=K(V_{\mathit{GS}}-V_T)=\frac{2I_D}{V_{\mathit{GS}}-V_T}\to & \\ & I_D& =\frac{g_m^2}{2K}& \\ & & g_m=\frac{1}{250\mathrm{\Omega }}& \\ & & K=40\mathit{mA}∕V^2\to & \\ & I_D& =0.2\mathit{mA}& \end{array}$$

(e)

An answer:

(f)

An answer:

$$\begin{array}{rcll}& a_v& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \\ & & v_{\mathit{out}}=-g_m{v}_{\mathit{gs}}{R}_D& \\ & & v_{\mathit{gs}}=v_{\mathit{in}}-V_S& \\ & & v_S=g_m{v}_{\mathit{gs}}{R}_{S1}& \\ & & \dots & \\ & a_v& =-\frac{g_m{R}_D}{1+g_m{R}_{S1}}& \end{array}$$

(g)

An answer:
A derivation can be: $$\begin{array}{rcll}& R_{S2}& =\frac{V_S}{I_S}-R_{S1}& \\ & & I_S=I_D& \\ & & V_S=V_G-V_{\mathit{GS}}& \\ & & V_{\mathit{GS}}=V_T+\sqrt{\frac{2I_D}{K}}& \\ & & V_G=\frac{R_{G2}}{R_{G1}+R_{G2}}\cdot V_{\mathit{DD}}& \end{array}$$

For which back substitution yields:

$$\begin{array}{rcll}& & V_S=\frac{R_{G2}}{R_{G1}+R_{G2}}\cdot V_{\mathit{DD}}-V_T-\sqrt{\frac{2I_D}{K}}& \\ & R_{S2}& =\frac{\frac{R_{G2}}{R_{G1}+R_{G2}}\cdot V_{\mathit{DD}}-V_T-\sqrt{\frac{2I_D}{K}}}{I_D}-R_{S1}& \end{array}$$

(h)

An answer:
Some numerical values are: $$\begin{array}{rcll}& g_m=\sqrt{2k{I}_D}=0.02A∕V& & \\ & V_{\mathit{GS}}=1.5V& & \\ & V_G=6V& & \\ & V_S=4.5V& & \\ & R_{S2}=650\mathrm{\Omega }& & \end{array}$$

(i)

An answer:

$$a_V=-\frac{g_m{R}_D}{1+g_m{R}_{S1}}=-4$$

Q 5.7

(a)

An answer:
You cannot ignore the base current as an $\mathit{\alpha fe}$ is explicitly specified.

$$\begin{array}{rcll}& I_C& =\mathit{\alpha fe}{I}_B& \\ & & I_B=\frac{V_C-V_B}{R_B}& \\ & & V_B=0.6& \\ & & V_C=V_{\mathit{CC}}-R_C\cdot (I_C+I_B)& \\ & & I_B=\frac{V_{\mathit{CC}}-R_C\cdot (I_C+I_B)-0.6}{R_B}& \\ & & I_B=\frac{V_{\mathit{CC}}-0.6}{R_B+(\mathit{\alpha fe}+1)R_C}& \\ & I_C& =\mathit{\alpha fe}\cdot \frac{V_{\mathit{CC}}-0.6}{R_B+(\mathit{\alpha fe}+1)R_C}& \end{array}$$

(b)

An answer:

(c)

An answer:

$$\begin{array}{rcll}& a_V& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \\ & & v_{\mathit{out}}=R_C\cdot (-g_m{v}_{\mathit{be}}-\frac{v_{\mathit{out}}-v_b}{R_B})& \\ & & v_e=0& \\ & & v_b=v_{\mathit{in}}& \\ & & v_{\mathit{out}}=R_C\cdot (-g_m{v}_{\mathit{in}}-\frac{v_{\mathit{out}}-v_{\mathit{in}}}{R_B})& \\ & & v_{\mathit{out}}=v_{\mathit{in}}\cdot \frac{(-g_m{R}_C+\frac{R_C}{R_B})}{1+\frac{R_C}{R_B}}& \\ & a_V& =-R_C\cdot \frac{-g_m+\frac{1}{R_B}}{1+\frac{R_C}{R_B}}& \end{array}$$

If $R_C<<R_B$ this could be simplified to $a_v\approx -g_m{R}_C$

(d)

An answer:

$$\begin{array}{rcll}& R_B& =\frac{V_C-V_B}{I_B}& \\ & & V_C=V_{\mathit{CC}}-R_C{I}_C& \\ & & V_B\approx 0.6& \\ & & I_B=\frac{I_C}{\mathit{\alpha fe}}& \\ & R_B& =\frac{V_{\mathit{CC}}-R_C{I}_C-0.6}{\frac{I_C}{\mathit{\alpha fe}}}& \\ & & =40k\mathrm{\Omega }& \end{array}$$

(e)

An answer:
If the transistor is replaced and $\mathit{\alpha fe}=40$ the collector current and voltage gain will both decrease,but by far not by the amount by which $\mathit{\alpha fe}$ has decreased. Rewriting the previously derived equation directly yields $I_C\approx 0.8\mathit{mA}$.

So $I_C$ decreases with $20\%$ as does the small signal parameter $g_m$.

Q 5.8

(a)

An answer:

$$\begin{array}{rcll}& R_E& =\frac{V_{E2}}{I_{\mathit{RE}}}& \\ & & I_{\mathit{RE}}=(1+\frac{1}{\mathit{\alpha fe}})I_{C2}-I_{B1}& \\ & R_E& =9k\mathrm{\Omega }& \end{array}$$

$$\begin{array}{rcll}& R_{C_1}& =\frac{V_{\mathit{CC}}-V_{B_E,Q2}-V_{E,Q2}}{I_{C,Q_1}-I_{B,Q2}}& \\ & & =15k\mathrm{\Omega }& \end{array}$$

$$\begin{array}{rcll}& R_B& =\frac{V_{R_B}}{I_{R_B}}& \\ & & =360k\mathrm{\Omega }& \end{array}$$

(b)

An answer:

$$\begin{array}{lll}\hfill R_{\mathit{in},Q2}=\frac{\mathit{\alpha fe}}{g_{m2}}=\frac{120}{0.004}=30k\mathrm{\Omega }& & \hfill \end{array}$$

(c)

An answer:

The first stage gives a loaded voltage gain $a_{v1}=g_{m1}\cdot \left(R_{C_1}∕∕\frac{\mathit{\alpha fe}}{g_{m2}}\right)$

The second stage gives a voltage gain $a_{v2}=g_{m2}\cdot R_{C_2}$

$$\begin{array}{llll}\hfill a_v& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \hfill & \\ \hfill & =g_{m1}\cdot \left(R_{C_1}∕∕\frac{\mathit{\alpha fe}}{g_{m2}}\right)\cdot g_{m2}\cdot R_{C_2}& \hfill & \\ \hfill & =\frac{g_{m1}\cdot R_{C_1}\cdot R_{C_2}\cdot \mathit{\alpha fe}}{R_{C_1}+\frac{\mathit{\alpha fe}}{g_{m2}}}& \hfill & \end{array}$$

(d)

An answer:
Recycle the previously derived equation, rewrite for $R_{C2}$:

$$\begin{array}{llll}\hfill & R_{C2}=2.5k\mathrm{\Omega }& \hfill & \end{array}$$

Q 5.9

(a)

(b)

An answer:
For this specific circuit, a non-Sudoku-like derivation is shown below. The base current CANNOT be neglected: that is explicitly stated. $$\begin{array}{rcll}{R}_1& =& \frac{V_B}{I_{R1}}& \\ & & V_B=0.6+R_{3b}\cdot I_C(1+1∕\mathit{\alpha fe})& \\ & & I_{R1}=I_2-I_C∕\mathit{\alpha fe}& \\ R_1& =& \frac{0.6+R_{3b}\cdot I_C(1+1∕\mathit{\alpha fe})}{I_2-I_C∕\mathit{\alpha fe}}& \end{array}$$

(c)

(d)

(e)

Q 5.10

(a)

(b)

(c)

(d)

Q 5.11

(a)

An answer:

After replacing caps by shorts, inductors by opens, and replacing DC sources by their SS-equivalents:

After replacing the non-linear device(s) by their SSEC:

After cleaning up the winding ground wire we get the easiest SSEC:

(b)

An answer:
From the small signal derivations should be performed. Note that assuming driving the output port, all other independent sources are set to zero, here $v_{\mathit{in}}=0$; the other proper way is to use $v_{\mathit{in}}$ and calculate the unloaded output voltage and the output current if the output is (small signal wise) shorted. $$\begin{array}{rcll}& z_{\mathit{out}}& =\frac{v_{\mathit{out}}}{i_{\mathit{out}}}& \\ & & i_{\mathit{out}}=i_{R4}+\mathit{gm}\cdot v_{\mathit{be}}& \\ & & i_{R4}=\frac{v_{\mathit{out}}}{R_4}& \\ & & v_{\mathit{be}}=0& \\ & z_{\mathit{out}}& =R_4& \end{array}$$

(c)

(d)

An answer:
For input impedance, the sign of the $v_{\mathit{in}}$ is irrelevant, according to Ohm. Because the base (in this SSEC, for this circuit) is NOT shorted to SS-ground, the small signal base voltage is NOT zero. This is different from the simplified situation that is frequently used. Tying the base to ground is wrong for this circuit. A proper derivation (one of many proper ones) is: $$\begin{array}{rcll}& z_{\mathit{in}}& =\frac{v_{\mathit{in}}}{i_{\mathit{in}}}& \\ & & i_{\mathit{in}}=i_{R3}+i_{\mathit{\alpha fe}∕\mathit{gm}}-\mathit{gm}\cdot v_{\mathit{be}}& \\ & & i_{R3}=\frac{v_{\mathit{in}}}{R_3}& \\ & & i_{\mathit{\alpha fe}∕\mathit{gm}}=-\frac{\mathit{gm}}{\mathit{\alpha fe}}{v}_{\mathit{be}}& \\ & & v_{\mathit{be}}=-\frac{\mathit{\alpha fe}∕\mathit{gm}}{\mathit{\alpha fe}∕\mathit{gm}+R_1∕∕R2}{v}_{\mathit{in}}& \\ & & i_{\mathit{in}}=\frac{v_{\mathit{in}}}{R_3}+\frac{\mathit{gm}}{\mathit{\alpha fe}}\frac{\mathit{\alpha fe}∕\mathit{gm}}{\mathit{\alpha fe}∕\mathit{gm}+R_1∕∕R2}{v}_{\mathit{in}}+\mathit{gm}\frac{\mathit{\alpha fe}∕\mathit{gm}}{\mathit{\alpha fe}∕\mathit{gm}+R_1∕∕R2}{v}_{\mathit{in}}& \\ & z_{\mathit{in}}& =R_3∕∕\frac{\mathit{\alpha fe}∕\mathit{gm}+R_1∕∕R2}{\mathit{\alpha fe}∕\mathit{gm}}\left(\frac{1}{\mathit{gm}}+\frac{\mathit{\alpha fe}}{\mathit{gm}}\right)& \end{array}$$

(e)

An answer:
A derivation (one of many) is: $$\begin{array}{rcll}& a_v& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \\ & v_{\mathit{out}}& =R_4\cdot -g_m\cdot v_{\mathit{be}}& \\ & & v_{\mathit{be}}=v_b-v_e& \\ & & v_e=-v_{\mathit{in}}& \\ & & v_b=\frac{R_1∕∕R2}{\mathit{\alpha fe}∕\mathit{gm}+R_1∕∕R2}{v}_e& \\ & & v_{\mathit{be}}=\frac{\mathit{\alpha fe}∕\mathit{gm}}{\mathit{\alpha fe}∕\mathit{gm}+R_1∕∕R2}{v}_{\mathit{in}}& \\ & v_{\mathit{out}}& =-R_4\cdot g_m\cdot \frac{\mathit{\alpha fe}∕\mathit{gm}}{\mathit{\alpha fe}∕\mathit{gm}+R_1∕∕R2}{v}_{\mathit{in}}& \\ & a_v& =-R_4\cdot g_m\cdot \frac{\mathit{\alpha fe}∕\mathit{gm}}{\mathit{\alpha fe}∕\mathit{gm}+R_1∕∕R2}& \end{array}$$

(f)