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\begin{center}
  {\LARGE Trabalho 2. Theoretical Values}\\
  {\large P.\ Stallinga}\\
  {\tiny 30.11.2004}\\
  \rule{10cm}{0.5mm}
\end{center}

\vspace{1cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                1 Common Emittor                              %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
  {\Large \textsc{1: Common Emitter}}
\end{center}

\begin{figure}[b]
\begin{flushright}
  \vspace*{0.7cm} {\small Typeset with {\LaTeX}
    2\raisebox{-0.3ex}{$\varepsilon$}} \vspace*{-0.7cm}
\end{flushright}
\end{figure}

\noindent Assume: $\beta$ = 200 ($\alpha$ = 1), V$_{\rm A}$ = 120
V, $C_\mu$ = 8 pF, $C_\pi$ = 25 pF\\

\mbox{\scalebox{0.9}{\includegraphics{1.eps}}}\hfill\fbox{\parbox[b]{3cm}{
\noindent
\fbox{Actual values:}\\
$R_{\rm S}$ = 1 k$\Omega$\\
$C_{\rm S}$ = 10 $\mu$F\\
$R_{\rm B1}$ = 10 k$\Omega$\\
$R_{\rm B2}$ = 20 k$\Omega$\\
$R_{\rm C}$ = 4.7 k$\Omega$\\
$C_{\rm L}$ = 10 $\mu$F\\
$R_{\rm E}$ = 3.3 k$\Omega$\\
$C_{\rm E}$ = 100 $\mu$F\\
$R_{\rm L}$ = 10 k$\Omega$\\
$V_{\rm CC}$ = +10 V

}}\\

\noindent
\fbox{Bias conditions:}\\
$V_{\rm B}$ = +10 V $\cdot$ 10 k$\Omega$ / (20 k$\Omega$ + 10
k$\Omega$) = 3.33 V\\
$V_{\rm E}$ = $V_{\rm B}$ $-$ 0.7 V = 2.63 V\\
$I_{\rm E}$ = 2.63 V / 3.3 k$\Omega$ = 0.797 mA\\
$V_{\rm C}$ = 10 $-$ 0.797 $10^{-3}$ * 4700 = 6.25 V\\
$r_{\rm e}$ = 26 mV / 0.798 mA = 32.58 $\Omega$\\
$r_\pi = (\beta +1)r_{\rm e} = 6.5449\;{\rm k}\Omega$\\
$r_{\rm o}$ = 120 V / 0.798 mA = 150 k$\Omega$\\
$g_{\rm m}$ = 1/$r_{\rm e}$ = 30.7 mS\\


\noindent
\fbox{Gain:}\\
The gain is defined by the standard formula for a CE amplifier:
\begin{eqnarray}
\nonumber
A_{\rm v} = - \frac{R_{\rm C}}{R_{\rm E} + r_{\rm e}}
\end{eqnarray}
In the intermediate frequency regime all the condensators (of the
LPFs) can be considered short-circuited. Thus, the resistor $R_{\rm
  E}$ is bypassed and the gain increases to
\begin{eqnarray}
\nonumber
A_{\rm v} = - \frac{R_{\rm C}}{r_{\rm e}} = - \frac{4700}{32.58} = -144.26
\end{eqnarray}
To calculate the input resistance of the transistor and the stage we
also consider $R_{\rm E}$ bypassed:
\begin{eqnarray}
\nonumber
r_{\rm in}({\rm transistor})= (\beta + 1)r_{\rm e} = 201\cdot
32.58\;\Omega = 6.549\;{\rm k}\Omega
\end{eqnarray}
\begin{eqnarray}
\nonumber
r_{\rm in}({\rm stage}) &=& R_{\rm B1} \| R_{\rm B2} \| r_{\rm in}\\
\nonumber
&=& 10\;{\rm k}\Omega \| 20\;{\rm k}\Omega \| 6.549\;{\rm k}\Omega =
3.304\;{\rm k}\Omega
\end{eqnarray}
\begin{eqnarray}
\nonumber
\frac{v_{\rm o}}{v_{\rm i}} &=& A_{\rm v}\left[ \frac{r_{\rm in}{\rm (stage)}}
{r_{\rm S} + r_{\rm in}{\rm (stage)} } \right] \left[ \frac{R_{\rm L}}
{R_{\rm L} + R_{\rm C} } \right]\\
\nonumber
&=& -144.26 \left( \frac{3.304}{1+3.304} \right) \left( \frac{10}
{10 + 4.7} \right) \\
\nonumber
&=& -144.26\cdot 0.7677\cdot 0.6803 = -75.33
\end{eqnarray}\\

\parpic[r]{ \scalebox{0.9}{\includegraphics{t2_2.eps}}}
\noindent
\fbox{lower cut-off frequency:}\\
Using the short-circuit-time-constants method (all other C's are
short circuited)\\
\textbf{1:} The condensator ($C_{\rm S}$) at the input is part of
a high-pass filter (HPF). To calculate the cut-off frequency of
this filter we have to calculate the RC time and hence we have to
determine the resistance this condensator sees. Figure 2
summarizes this. The part after the collector can be neglected,
because it is much larger. The resistor $R_{\rm E}$ is bypassed by
the capacitor $C_{\rm E}$.
\begin{eqnarray}
\nonumber
R &=& R_{\rm S} + R_{\rm B1} \| R_{\rm B2} \| (\beta + 1) r_{\rm e}\\
\nonumber
  &=& 1{\rm k}\Omega + 10\;{\rm k}\Omega \| 20\;{\rm k}\Omega \|
  6.549\;{\rm k}\Omega\\
\nonumber
  &=& 4.304\;{\rm k}\Omega\\
\nonumber
C_{\rm S} &=& 10\;\mu {\rm F}\\
\nonumber
\tau_S &=& RC = 43.04\;{\rm ms}\\
\nonumber f_{LS} &=& \frac{1}{2\pi \tau_S} = 3.7\;{\rm Hz}
\end{eqnarray}

\parpic[r]{ \scalebox{0.9}{\includegraphics{t2_3.eps}}}
\noindent
\textbf{2:} For the condensator at the load we have:
\begin{eqnarray}
\nonumber
R = R_{\rm L} + R_{\rm C} = 10\;{\rm k}\Omega + 4.7\;{\rm k}\Omega =
14.7\;{\rm k}\Omega
\end{eqnarray}
(note: the resistance after the collector is again neglected).  With
the capacitance $C = C_{\rm L}$ = 10 $\mu$F we have
\begin{eqnarray}
\nonumber
\tau_L &=& RC = 147\;{\rm ms}\\
\nonumber f_{LL} &=& \frac{1}{2\pi \tau_L} = 1.08\;{\rm Hz}
\end{eqnarray}

\parpic[r]{ \scalebox{0.9}{\includegraphics{t2_4.eps}}}
\noindent
\textbf{3:} Finally, for the emittor condensator we have (everything
at the collector ignored):

\vspace{1cm}

\begin{eqnarray}
\nonumber
R &=& R_{\rm E} \| (r_{\rm e} + \frac{R_{\rm B1}}{\beta +1} \| \frac{R_{\rm B2}}{\beta +1}
\| \frac{R_{\rm S}}{\beta +1} )\\
\nonumber
&=& 3300\;\Omega \| (32.58\;\Omega + 49.75\;\Omega \| 99.50\;\Omega \|
4.975\;\Omega )
= 36.5\;\Omega\\
\nonumber
C &=& C_{\rm E} = 100\;\mu{\rm F}\\
\nonumber
\tau_E &=& RC = 3.65\;{\rm ms}\\
\nonumber f_{LE} &=& \frac{1}{2\pi \tau_E} = 43\;{\rm Hz}
\end{eqnarray}
The cut-off frequency ($f_L$) is determined by the higher of the
three $f_{LS}$, $f_{LL}$ and $f_{LE}$, namely by condensator at
the emittor, $f_L = f_{LE}$. More exact, $f_L = f_{LS} + f_{LL} +
f_{LE}$ = 3.7 Hz + 1.08 Hz + 43 Hz = 48 Hz. Below this frequency
the attenuation is 20 dB / decade. The phase slowly shifts
$90^{\rm o}$ and is $45^{\rm o}$ at $f_L$. When the frequency is
further lowered it hits the second cut-off frequency
(approximately $f_{LS}$) from where the attenuation will be 40 dB
/ decade, etc. Each filter introduces an attenuation of 20 dB /
decade and a phase shift of $90^{\rm o}$.\\

\vspace{1cm}

\noindent
\fbox{higher cut-off frequency:}\\
Using the open-circuit-time-constants method (all other high
frequency C's are open circuit).\\
The higher cut-off frequencies are determined by the shunt
capacitances that connect the input or the output to ground. We
have $C_\pi$ (between base and emittor) and $C_\mu$ (between base
and collector). Note that in this configuration the latter is
connected between the input and the output (with negative gain)
and is therefore susceptible to feedback effects (resulting in
so-called Miller capacitance,
$C_{\rm M}$).\\
For the higher cut-off frequencies it is most convenient to decompose
the system into an input filter and an output filter. This is allowed
because the resistance between the input and the output is so high
that there is virtually no contact between these points. At the input
we have:
\begin{eqnarray}
\nonumber
R &=& R_{\rm S} \| R_{\rm B1} \| R_{\rm B2} \| (\beta +1)r_{\rm e}\\
\nonumber
&=& 1\;{\rm k}\Omega \| 10\;{\rm k}\Omega \| 20\;{\rm k}\Omega \|
6.549\;{\rm k}\Omega\\
\nonumber
&=& 767.6\;\Omega\\
\label{eq:C}
C &=& C_\pi + C_{\rm M} = C_\pi + C_\mu (1 - A_{\rm v})\\
\nonumber
&=& 25\;{\rm pF} + 8\;{\rm pF}(1+98.14)\\
\nonumber
&=&818.12\;{\rm pF}\\
\nonumber
\tau_i &=& RC = 627.99\;{\rm ns}\\
\nonumber
f_{Hi} &=& \frac{1}{2\pi \tau_i} = 253\;{\rm kHz}
\end{eqnarray}
Note that for the amplification $A_{\rm v}$ we have to use the
voltage amplification factor between the collector and the base
because the condesator $C_\mu$ is connected between those points.
It does not include the voltage devider at the source:
\begin{eqnarray}
\nonumber
A_{\rm v} = -\frac{R_{\rm C} \| R_{\rm L}} {r_{\rm e}} = -98.14
\end{eqnarray}
At the output we have:
\begin{eqnarray}
\nonumber
R &=& r_{\rm c} \| R_{\rm C} \| R_{\rm L}\\
\nonumber
&=& 150\;{\rm k}\Omega \| 4.7\;{\rm k}\Omega \| 10\;{\rm k}\Omega\\
\nonumber
&=& 3.131\;{\rm k}\Omega\\
\nonumber
C &=& C_\mu(1 - 1/A_{\rm v})\\
\nonumber
 &=& 8\;{\rm pF}\cdot 1.0102\\
\nonumber
 &=& 8.08\;{\rm pF}\\
\nonumber
\tau_o &=& RC = 25.3\;{\rm ns}\\
\nonumber
f_{Ho} &=& \frac{1}{2\pi \tau_o} = 6.3\;{\rm MHz}
\end{eqnarray}
The cut-off frequency is therefore determined by the filter at the
input. Using the open-circuit-time-constants method $\tau = \tau_i
+ \tau_o$ = 627.99 ns + 25.3 ns = 655 ns and $f_H = 1/2\pi\tau$ =
243 kHz.

Finally, the internal cut-off frequency of the transistor is
determined by (see p. 437 of Bogart)
\begin{eqnarray}
\nonumber
f_{\rm T} &=& \frac{1}{2\pi r_{\rm e}(C_\pi + C_\mu)} =
\frac{1}{2\pi
  (32.58\;\Omega)(12\;{\rm pF})} = 407\;{\rm MHz}
\end{eqnarray}
(According to the specifications (datasheets), for the 2N2222 the
unitary current amplification frequency is $f_{\rm T}$ = 250 MHz.)
This cut-off frequency is not important for the circuit. The gain
and cut-off frequencies are summarized in the Bode plot on the
next page.

\vspace{2ex}

\noindent
\fbox{phase diagram:}\\
The overall gain of the amplifier at mid-band frequencies is
negative. This is equal to a phase shift of 180$^{\rm o}$. The
filter at the source with $f_{LS}$ introduces a phase shift of
+90$^{\rm o}$ (see p.\ 33 of Sedra) at low frequencies. In fact,
every HPF introduces a phase shift of 90$^{\rm o}$. On the other
side of the spectrum, every LPF introduces a phase shift of
$-90^{\rm o}$ at high frequencies.  The figures show the voltage
gain (in dB) at the load and the phase
relative to the source voltage:\\
\begin{center}
  \scalebox{0.8}{\includegraphics{t2_5a.eps}}\\
  \scalebox{0.8}{\includegraphics{t2_5b.eps}}
\end{center}

\noindent
\fbox{Without $C_E$:}\\
When the shunt capacitance $C_{\rm E}$ is pulled, the gain of the
amplifier reduces to
\begin{eqnarray}
\nonumber
 A_{\rm v} = - \frac{R_{\rm C}}{r_{\rm e}+R_{\rm E}}
=-\frac{4700}{32.58+3300} = -1.410
\end{eqnarray}
The voltage gain between the base and the collector (to which
$C_\mu$ is connected; we will use this value to calculate the
Miller capacitance) when a load resistance is connected is
\begin{eqnarray}
\nonumber
A_{\rm v} = -1.410\cdot\frac{R_{\rm L}}{R_{\rm L}+R_{\rm C}} =
-1.410\cdot\frac{10}{10+4.7} = -0.959
\end{eqnarray}
The voltage divider at the input:
\begin{eqnarray}
\nonumber
r_{\rm in} = (\beta + 1)(r_{\rm e}+R_{\rm E}) - 669.8\;{\rm k}\Omega\\
\nonumber
(r_{\rm in})_{\rm stage} = r_{\rm in}\| R_{\rm B1}\|R_{\rm B2} =
669.8\;{\rm k}\Omega\|10{\rm k}\Omega\|20{\rm k}\Omega = 6.6\;{\rm k}\Omega\\
\nonumber
\frac{v_{\rm i}}{v_{\rm S}} = \frac{(r_{\rm in})_{\rm stage}}
 {(r_{\rm in})_{\rm stage}+R_{\rm S}} = \frac{6.6}{6.6+1} = 0.868
\end{eqnarray}
The ratio of the voltage at the output and the voltage at the input
then becomes (taking into account the voltage dividers at the input
and the output):
\begin{eqnarray}
\nonumber
\frac{v_{\rm o}}{v_{\rm i}} = -0.832
\end{eqnarray}
The filter at the emittor is removed and the highest cut-off frequency
is then $f_{\rm 1}$(S) whose value is also slightly changed by the
elimination of the shunt capacitance:
\begin{eqnarray}
\nonumber
R &=& R_{\rm S} + R_{\rm B1} \| R_{\rm B2} \| (\beta +1)(r_{\rm
  e}+R_{\rm E})\\
\nonumber
&=& 7.601\;{\rm k}\Omega\\
\nonumber
C &=& 10\;\mu{\rm F}\\
\nonumber
\tau_S &=& RC = 76.01\;{\rm ms}\\
\nonumber
f_{LS} &=& \frac{1}{2\pi \tau_S} = 2.09\;{\rm Hz}\\
\nonumber f_{LL} &=& f_{LL} = 1.08\;{\rm Hz}
\end{eqnarray}
For the higher cut-off frequency.\\
1: At the entrance of the transistor:
\begin{eqnarray}
\nonumber
R &=& R_{\rm S}\|R_{\rm B1}\|R_{\rm B2}\|(\beta +1)(R_{\rm E}+r_{\rm
  e})\\
\nonumber
&=& 1{\rm k}\Omega\| 10{\rm k}\Omega\| 20{\rm k}\Omega\| 669.8 {\rm k}\Omega = 868\;\Omega\\
\nonumber C &=& C_\pi + (1-A_{\rm v})C_\mu = 25\;{\rm pF} +
(1+0.959)\cdot 8\;{\rm pF} = 40.7\;{\rm pF}\\
\nonumber
\tau_i &=& RC = 35.3\;{\rm ns}\\
\nonumber f_{Hi} &=& \frac{1}{2\pi \tau_i} = 4.5\;{\rm MHz}
\end{eqnarray}
2: At the exit of the transistor:
\begin{eqnarray}
\nonumber
R &=& R_{\rm L}\|R_{\rm C} = 3.131\;{\rm k}\Omega\\
\nonumber
C &=& (1-1/A_{\rm v})C_\mu = (1+1/0.959)\cdot 8\;{\rm pF} = 16.3\;{\rm pF}\\
\nonumber
\tau_o &=& RC = 51.0\;{\rm ns}\\
\nonumber f_{Ho} &=& \frac{1}{2\pi \tau_o} = 3.12\;{\rm MHz}
\end{eqnarray}
3: internal filter:
\begin{eqnarray}
\nonumber f_T ' = f_T = 407\;{\rm MHz}
\end{eqnarray}
The cut-off frequency is determined by the exit of the transistor.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                2 Cascode                                     %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newpage
\begin{center}
  {\Large \textsc{2: Cascode}}
\end{center}

\mbox{\scalebox{0.9}{\includegraphics{2.eps}}}\hfill\fbox{\parbox[b]{3cm}{
\noindent
\fbox{Actual values:}\\
$R_{\rm S}$ = 1 k$\Omega$\\
$C_{\rm S}$ = 10 $\mu$F\\
$R_{\rm B1}$ = 8.2 k$\Omega$\\
$R_{\rm B2}$ = 8.2 k$\Omega$\\
$R_{\rm B3}$ = 20 k$\Omega$\\
$R_{\rm C}$ = 4.7 k$\Omega$\\
$C_{\rm L}$ = 10 $\mu$F\\
$R_{\rm E}$ = 3.3 k$\Omega$\\
$C_{\rm E}$ = 100 $\mu$F\\
$R_{\rm L}$ = 10 k$\Omega$\\
$V_{\rm CC}$ = 15 V\\
$C_{\rm B}$ = 10 $\mu$F
}}\\

\noindent
\fbox{Bias conditions:}\\
$V_{\rm B1}$ = +15 V $\cdot$ 8.2 k$\Omega$ / (20 k$\Omega$ + 8.2
k$\Omega$
+ 8.2 k$\Omega$) = 3.38 V\\
$V_{\rm E1}$ = $V_{\rm B} -$ 0.7 V = 2.68 V\\
$I_{\rm E1}$ = 2.68 V / 3.3 k$\Omega$ = 0.8119 mA (3.3 k$\Omega$
choosen so that
$I_{\rm E1}$ will be approximately 0.8 mA)\\
$I_{\rm C2}$ = $I_{\rm E2}$ = $I_{\rm C1}$ = $I_{\rm E1}$
($\alpha$=1)\\
$V_{\rm C2} = +15 - R_{\rm C}I_{\rm C2} =$ 11.18 V\\
$r_{\rm e}$ = 26 mV / 0.8119 mA = 32.024 $\Omega$\\
$g_{\rm m} = 1/r_{\rm e}$ = 31.2 mS\\
$r_\pi = (\beta +1)r_{\rm e} = 6.437\;{\rm k}\Omega$\\
$r_{\rm o}$ = 120 V / 0.8119 mA = 150 k$\Omega$\\
The output resistance of Q1 is $r_o$. The output resistance of Q2
is much higher (reasoning is similar to current source of TP1)

\noindent
\fbox{Gain:}\\
\begin{eqnarray}
\nonumber
(r_{\rm in})_{\rm stage} &=& (\beta +1)r_{\rm e1} \| R_{\rm B1} \| R_{\rm B2}\\
\nonumber
&=& 6.437\;{\rm k}\Omega \| 8.2\;{\rm k}\Omega \| 8.2\;{\rm k}\Omega\\
\nonumber
&=& 2.505\;{\rm k}\Omega\\
\nonumber
(r_{\rm o})_{\rm stage} &=& R_{\rm C} \| r_{\rm out,2} = R_{\rm C}\\
\nonumber
&=& 4.7\;{\rm k}\Omega\\
\nonumber A_{\rm v}^1 &=& -\frac{r_{\rm out,1} \| r_{\rm
e2}}{r_{\rm e1}} =
-\frac{r_{\rm e2}}{r_{\rm e1}} = -1\\
\nonumber A_{\rm v}^2 &=& \frac{R_{\rm C} \| r_{\rm out,2}}{r_{\rm
e2} \| r_{\rm c1}}
= \frac{R_{\rm C}}{r_{\rm e2}} = \frac{4700}{32.024} = +147.76\\
\nonumber
A_{\rm v} &=& A_{\rm v}^1\cdot A_{\rm v}^2 = -1\cdot 147.76 = -147.76\\
\nonumber
\frac{v_{\rm L}}{v_{\rm S}} &=& \frac{r_{\rm in}}{r_{\rm in} + r_{\rm S}}
\cdot A_{\rm v} \cdot \frac{r_{\rm L}}{r_{\rm L}+r_{\rm o}}\\
\nonumber
&=& \left( \frac{2505}{1000+2505} \right)\left( -146.76 \right)
\left( \frac{10000}{10000+4700} \right)\\
\nonumber
&=& 0.7147\cdot (-146.764) \cdot 0.6803\\
\nonumber
&=& -71.36
\end{eqnarray}

\noindent
At low frequencies nothing much changes, though there is an extra
filter caused by the condensator $C_{\rm B}$:
\begin{eqnarray}
\nonumber
C &=& C_{\rm S} = 10\;\mu{\rm F}\\
\nonumber
R &=& R_{\rm S} + R_{\rm B1}\| R_{\rm B2}\|(\beta +1)r_{\rm e} = 1 +
8.2\| 8.2\| 6.437 = 3.505\;{\rm k}\Omega\\
\nonumber
\tau_S &=& RC = 35.05\;{\rm ms}\\
\nonumber
f_{LS} &=& 4.54\;{\rm Hz}\\[2ex]
%\end{eqnarray}
%\begin{eqnarray}
\nonumber
C &=& C_{\rm B} = 10\;\mu{\rm F}\\
\nonumber
R &=& R_{\rm B3}\|(R_{\rm B2} + R_{\rm S}\|R_{\rm B1}\|(\beta
 +1)r_{\rm e}) = 20\| (8.2 + 1\|8.2\| 6.437) = 6.199\;{\rm k}\Omega\\
\nonumber
\tau_B &=& RC = 61.99\;{\rm ms}\\
\nonumber
f_{LB} &=& 2.57\;{\rm Hz}\\[2ex]
% \end{eqnarray}
% \begin{eqnarray}
\nonumber
C &=& C_{\rm E} = 100\;\mu{\rm F}\\
\nonumber
R &=& R_{\rm E}\|\left( r_{\rm e} + \frac{R_{\rm B1}\| R_{\rm B2}\| R_{\rm
    S}}{\beta +1}\right)  = 3.3\| \left( 0.032024 + \frac{8.2\| 8.2\|
  1}{201}\right) = 35.63\;\Omega\\
\nonumber
\tau_E &=& RC = 3.563\;{\rm ms}\\
\nonumber
f_{LE} &=& 44.67\;{\rm Hz}\\[2ex]
% \end{eqnarray}
% \begin{eqnarray}
\nonumber
C &=& C_{\rm L} = 10\;\mu{\rm F}\\
\nonumber
R &=& R_{\rm C}+R_{\rm L} = 4.7 + 10 = 14.7\;{\rm k}\Omega\\
\nonumber
\tau_L &=& RC = 147\;{\rm ms}\\
\nonumber f_{LL} &=& 1.08\;{\rm Hz}
\end{eqnarray}
The cut-off frequency is the largest of the four: 45 Hz. Or using
the short-circuit-time-constants theory: $f_L = \sum_i f_i$ = 53
Hz.

\noindent The advantage is in the high frequencies, namely the
virtual elimination of the Miller capacitance because the
capacitance $C_\mu$ at the entrance of the stage is connected to
two points with a voltage gain close to $-1$ as described above.
The Miller capacitance therefore reduces to
\begin{eqnarray}
\nonumber C_{\rm M} = C_\mu \left( 1 - A_{\rm v} \right) =
 8\;{\rm pF}\left(  1 - (-1) \right) = 16\;{\rm pF}
\end{eqnarray}
and the total capacitance becomes
\begin{eqnarray}
\nonumber C = C_\pi + C_{\rm M} = 25\;{\rm pF} + 16\;{\rm pF} =
41\;{\rm pF}
\end{eqnarray}
The resistance is approximately the same as in Equation~\ref{eq:C}.
\begin{eqnarray}
\nonumber
R &=& R_{\rm S}\| R_{\rm B1}\| R_{\rm B2}\| (\beta +1)r_{\rm e} = 1\|
8.2\| 8.2\| 6.537 = 714.7\;\Omega\\
\nonumber
\tau_i = RC &=& 29.3\;{\rm ns}\\
\nonumber f_{Hi} &=& 5.43\;{\rm MHz}
\end{eqnarray}
At the output the capacitor $C_\mu$ ow has one leg connected to
ground ($V_{\rm B2}$) and the Miller effect disappears (1-0 = 1)
\begin{eqnarray}
\nonumber
R &=& R_{\rm C}\|R_{\rm L} = 10 \| 4.7 = 3.197\;{\rm k}\Omega\\
\nonumber
C &=& C_\mu = 8\;{\rm pF}\\
\nonumber
\tau &=& RC = 25.6\;{\rm ns}\\
\nonumber
f_{Ho} &=& 6.2\;{\rm MHz}
\end{eqnarray}
In-between the two transistors we have another filter:
\begin{eqnarray}
\nonumber C &=& C_{\pi 2} +  (1-(-1/1))C_{\mu 1} = 41\;{\rm
  pF}\\
\nonumber
R &=& r_{\rm e} = 32.58\;\Omega\\
\nonumber
\tau &=& RC = 1.34\;{\rm ns}\\
\nonumber f_{Hm} &=& 119\;{\rm MHz}
\end{eqnarray}
Using the theory of open-circuit time constants we find a total
time constant of
\begin{eqnarray}
\nonumber \tau = 29.3\;{\rm ns} + 25.6\;{\rm ns} + 1.34\;{\rm ns}
= 56.2\;{\rm ns}\\
\nonumber f_{\rm H} = 1/2\pi\tau = 2.8\;{\rm MHz}
\end{eqnarray}
Conclusion: there is a significant increase in high-frequency
response by approximately one decade, basically caused by the
elimination of the Miller effect at the entrance.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                3 Differential Pair                           %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newpage
\begin{center}
  {\Large 3: \textsc{Differential Pair with Current Source}}
\end{center}
\noindent

\mbox{\scalebox{0.6}{\includegraphics{3.eps}}}\hfill\fbox{\parbox[b]{3cm}{
\noindent
\fbox{Actual values:}\\
$R_{\rm C}$ = 3.3 k$\Omega$\\
$R_{\rm S}$ = 1 k$\Omega$\\
$I_0$ = 3 mA\\
}}\\

\noindent
\fbox{bias:}\\
$I_0$ = 3 mA, $I_{\rm E}$ = 1.5 mA. $r_{\rm e}$ = 26 mV / 1.5 mA =
17.33 $\Omega$. $r_\pi$ = 201 x 17.3 $\Omega$ = 3.48 k$\Omega$.
$V_{o2}$ = $V_{o1}$ = 10 V - 1.5 mA x 3.3 k$\Omega$ = 5.05 V.
$r_{\rm in}$ = 2 x $r_\pi$ = 7.0 k$\Omega$.\\[2ex]
\fbox{gain:}\\
\begin{eqnarray}
\nonumber
A_{\rm v}^{\rm l} &=& -\frac{\alpha R_{\rm C}}{2r_{\rm e}} = -95.4\\
\nonumber
A_{\rm v}^{\rm r} &=& +95.4\\
\nonumber
\frac{v_{\rm o}}{v_{\rm s}} &=& \frac{r_{\rm in}}{r_{\rm in}+R_{\rm
    S}}A_{\rm v}
= \frac{(\beta + 1)2r_{\rm e}}{(\beta +1)2r_{\rm e}+R_{\rm S}} A_{\rm
  v} = 0.8742\cdot 95.2 = +83.4
\end{eqnarray}

\noindent
\fbox{high-frequencies:}\\
The high-frequency response is not improved compared to a
single-stage amplifier. At the input exists $C_\pi$ = 25 pF and
$C_\mu$ = 8 pF. Both suffer from the Miller effect and are
therefore multiplied by the respective voltage gains between the
two points that are bridged by the capacitance. From Trabalho 1 we
know that in differential mode, the gain between base and
collector of the left transistor (Q1) is given by
\begin{eqnarray}
\nonumber
A_{\rm v} &=& -\frac{\alpha R_{\rm C}}{2r_{\rm e}}\\
\nonumber
&=& -\frac{1\cdot 3300}{2\cdot 17.33}\\
\nonumber
&=& -95.2
\end{eqnarray}
The gain felt by $C_\pi$ is exactly 0.5. The total capacitance at
the input is therefore
\begin{eqnarray}
\nonumber C &=& (1-0.5)C_\pi + C_{\rm M} = 0.5C_\pi + C_\mu\left(
1 - A_{\rm v}
\right)\\
\nonumber &=& 12.5\;{\rm pF} + 8\;{\rm pF}\left( 1 - (-95.4)
\right) = 783.7\;{\rm pF}
\end{eqnarray}
The resistance-to-ground at the input is
\begin{eqnarray}
\nonumber
R &=& R_{\rm S} \| (\beta+1)2r_{\rm e}\\
\nonumber
&=& 1\;k\Omega \| 201\cdot 17.33\;\Omega = 777\;\Omega\\
\nonumber
\tau_i &=& RC = 608.9\;{\rm ns}\\
\nonumber f_{Hi} &=& \frac{1}{2\pi\tau_i} = 261\;{\rm kHz}
\end{eqnarray}
At the output (collector of right transistor Q2) we have $C_\mu$ =
8 pF to the base. Without any Miller effects, because the gain
between emittor and base is zero. The total capacitance is
therefore
\begin{eqnarray}
\nonumber C = C_\mu = 8\;{\rm pF}
\end{eqnarray}
At the output the resistance is $R_{\rm C} \| r_{\rm out,2}
\approx R_{\rm
  C}$ = 3.3 k$\Omega$. Therefore
\begin{eqnarray}
\nonumber
\tau_o &=& RC = 26.4\;{\rm ns}\\
\nonumber f_{Ho} &=& \frac{1}{2\pi\tau_o} = 6.0\;{\rm MHz}
\end{eqnarray}
Note: with a load resistance of 10 k$\Omega$, the equivalent
resistance at the output would be 2.48 k$\Omega$ and the frequency
would further increase to 8.0 MHz.\\
The internal cut-off frequency remains at
\begin{eqnarray}
\nonumber f_T = f_T = 407\;{\rm MHz}
\end{eqnarray}
\\[2ex]

\noindent
\fbox{low frequencies:}\\
Since there are no coupling capacitors at the entrance or the exit,
there is no theoretical lower cut-off frequency. If we \emph{do}
introduce
them (at the source and the load only) we can say the following:\\
Since there is no need for any bypass condensator at the emittor
we can expect an improvement of the lower frequency response,
since in the single-stage amplifiers the lower-frequency response
is limited by this condensator (see first part). There still
remain the filters at the source and at the load. For the source
we have $C_{\rm S}$ = 10 $\mu$F and the resistance is here $R_{\rm
S} + (\beta+1)2r_{\rm e}$ = 4.48 k$\Omega$. The cut-off frequency
is then $f_{LS}$ = 3.6 Hz. At the load the resistance is $R_{\rm
L} + R_{\rm C} \| r_{\rm o} \approx$ 4.3 k$\Omega$ and with a
load-condensator of 10 $\mu$F this
gives a cut-off frequency of $f_{LL}$ = 3.7 Hz.\\[2ex]

\noindent
In conclusion: For a differential pair we see a significant
improvement of the low-frequency performance compared to a
single-stage amplifier (about one decade), mainly due to the
elimination of the emittor condensator.

\vspace{3cm}
\noindent
\emph{\small \textsc{Further information:}\\
  Chapter 10 of ``Electronic Devices and Circuits'', T.\ F.\ Bogart,
  Jr., 4th edition, Prentice Hall, 1997.\\
  Chapter 7 of ``Microelectronic Circuits'', A.\ S.\ Sedra, K.\ C.\
  Smith, 2nd edition, Oxford University Press, 1982.}
\end{document}
