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% author of this document: Peter Stallinga

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\begin{document}

\begin{center}
  {\LARGE Trabalho 1. Theoretical Values}\\
  {\large P.\ Stallinga}\\
  {\tiny 10.04.2000}\\
  \rule{10cm}{0.5mm}
\end{center}

\vspace{1cm}

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\parpic[l]{ \scalebox{0.9}{\includegraphics{t1_1.eps}}}
\noindent
Assume: $\beta$ = 200 ($\alpha$=1), V$_{\rm A}$ = 200 V\\
$R_{\rm C}$ = 3.9 k$\Omega$, $R_{\rm S}$ = 10 k$\Omega$\\
labelling: Q1 and Q2: NPN transistors of differential pair. Q3: NPN
transistor of the current source. Q4 and Q5: PNP transistors of current mirror.\\

\noindent
\fbox{1a:}\\
Connect $v_{\rm i1}$ and $v_{\rm i2}$ to ground\\
$V_{\rm E1} = V_{\rm E2} = V_{\rm B} - 0.7\;{\rm V} = 0.7\;{\rm V}$\\
$I_{\rm S} = (15 - 0.7) {\rm V} / 10\;{\rm k}\Omega = 1.43\;{\rm mA}$\\
$I_1 = I_2 = I_{\rm S}/2 = 0.715\;{\rm mA}$\\
This gives $r_{\rm e}$ = ($kT/e$)/$I_{\rm E}$ = 26 mV / 0.715 mA =
36.36~$\Omega$\\
$r_\pi = (\beta +1)r_{\rm e}$ = 7.308 k$\Omega$\\
$V_{\rm o1} = V_{\rm o2} = 15 {\rm V} - $3.9~ k$\Omega\cdot
0.715$ mA = 12.2115 V\\[2ex]

\noindent
\fbox{1b:}\\
$I_{\rm B1} = I_{\rm B2} = I_{\rm E}/(\beta + 1) = 3.557\;\mu$A\\[2ex]

\noindent
\fbox{1c:}\\
offset: ($I_{\rm B1} - I_{\rm B2}$) = 0\\
bias: ($I_{\rm B1} + I_{\rm B2}$)/2 = 3.557 $\mu$A\\[2ex]

\noindent
\fbox{2:}\\
The common-mode gain is defined as
\begin{eqnarray}
\nonumber
A_{\rm cm} = \frac{v_{\rm o1}-v_{\rm o2}}{v_{\rm i}} = 0
\end{eqnarray}
The single-ended common-mode gain can be defined by
\begin{eqnarray}
\nonumber
A_{\rm cm}^{\rm se} = \frac{v_{\rm o1}}{v_{\rm i}} =
\frac{v_{\rm o2}}{v_{\rm i}} = - \frac{\alpha
  R_{\rm C}}{2R_{\rm S}+r_{\rm e}} = - \frac{1\cdot 3900}{2\cdot 10000 +
  36.36} = -0.195
\end{eqnarray}
Note that for unequal resistors $R_{\rm C1} \neq R_{\rm C2}$ the
common-mode gain deviates from zero:
\begin{eqnarray}
\nonumber
A_{\rm cm} = (A_{\rm cm}^{\rm se})_1 - (A_{\rm cm}^{\rm se})_2 = -
\frac{\alpha}{2R_{\rm S}+r_{\rm e}}\Delta R_{\rm C} , \;\;\;\;\;\;\; \Delta
R_{\rm C} = R_{\rm C1} - R_{\rm C2}
\end{eqnarray}

\noindent
\fbox{3:}\\
The differential(-mode) gain is given by
\begin{eqnarray}
\nonumber
A_{\rm d} =  \frac{v_{\rm o1}-v_{\rm o2}}{v_{\rm i1}-v_{\rm i2}} =
- \frac{\alpha R_{\rm C}}{r_{\rm e}} = -\frac{1\cdot 3900}{36.36} = -107.26
\end{eqnarray}
The single-ended differential(-mode) gain can be defined by
\begin{eqnarray}
\nonumber
A_{\rm d}^{\rm se} =  \frac{v_{\rm o1}}{v_{\rm i1}-v_{\rm i2}} =
- \frac{\alpha R_{\rm C}}{2r_{\rm e}} = \frac{g_{\rm m}R_{\rm C}}{2} =
A_{\rm d}/2 = -53.63
\end{eqnarray}

\noindent
\fbox{4:}\\
The common-mode rejection ratio is defined by
\begin{eqnarray}
\nonumber
{\rm CMRR} = \left| \frac{A_{\rm d}}{A_{\rm cm}} \right| = \infty
\end{eqnarray}
The single ended ``common-mode rejection ratio'' can be defined by
\begin{eqnarray}
\nonumber
{\rm CMRR}^{\rm se} = \left| \frac{A_{\rm d}^{\rm se}}{A_{\rm cm}^{\rm
      se}} \right| = \frac{53.63}{0.195} = 275
\end{eqnarray}

\parpic[r]{\scalebox{0.8}{\includegraphics{t1_2.eps}}}
\noindent
\fbox{6:}\\
The output resistance of the current source $r_{\rm o3}$ is
approximately $\beta r_{\rm o}$/4. This can be seen in the
small-signal equivalent circuit of the current source looking into the
collector (see the picture on the right). Solve the current and
voltage equations ($i_{\rm x}$ is the current through $r_{\rm o}$)
will yield a complicated equation which, for nominal values of $R_{\rm
  E}$ and $R_{\rm B}$ will give $r_{\rm o3} = \beta r_{\rm o}/4$.
$r_{\rm o}$ can be calculated if we assume an Early voltage of 120 V:
$r_{\rm o} = V_{\rm A}/I_{\rm C}$ = 120 V/1.43 mA = 84 k$\Omega$.  The
output resistance is then $r_{\rm o3}$ = 4.7 M$\Omega$. This replaces
$R_{\rm S}$ in the above equations. Ergo:
\begin{eqnarray}
\nonumber
A_{\rm cm}^{\rm se} &=& - \frac{\alpha R_{\rm C}}{2r_{\rm o3} + r_{\rm e}}
= -\frac{1\cdot 3.9}{2\cdot 4700 + 0.03636} = - 4.15\cdot 10^{-4}\\
\nonumber
A_{\rm d}^{\rm se} &=& - \frac{\alpha R_{\rm C}}{2r_{\rm e}}
= -\frac{1\cdot 3900}{2\cdot 36.36} = -53.63\\
\nonumber
{\rm CMRR}^{\rm se} &=& 1.29\cdot 10^{5}
\end{eqnarray}\\
 
\noindent
\fbox{8:}\\
In the same way, the output resistance of the current mirror can be
shown to equal $r_{\rm o5}$ = $r_{\rm o}$, see ch.6.4, p.510 of Sedra.
When we assume an Early voltage of 120 V for the PNP transistors too,
this output resistance is 120 V/0.715 mA = 170 k$\Omega$. This
resistance, parallel with the output resistance of Q2 (also 170
k$\Omega$), replace the collector resistances R$_{\rm C}$:
\begin{eqnarray}
\nonumber
A_{\rm cm}^{\rm se} &=& - \frac{\alpha (r_{\rm o5} \| r_{\rm o2} ) }{2r_{\rm o3} + r_{\rm e}}
= \frac{1\cdot (170\| 170) }{2\cdot 4700 + 0.03636} = - 9.0\cdot 10^{-3}\\
\nonumber
A_{\rm d}^{\rm se} &=& - \frac{\alpha (r_{\rm o5} \| r_{\rm 02})}{2r_{\rm e}}
= \frac{1\cdot (170\| 170) }{2\cdot 0.03636} = -1169\\
\nonumber
{\rm CMRR}^{\rm se} &=& 1.29\cdot 10^{-5}
\end{eqnarray}
Note that, although the single-ended common-mode and the differential
mode gains increase, the common-mode rejection-ratio doesn't change.
The benefit in the new circuit is that the (double-ended) CMRR is no
longer sensitive to mismatched resistors at the collectors, since they
are defined equally by means of the current mirror. The (double-ended)
CMRR
will increase dramatically.\\[2ex]
With a 1 k$\Omega$ load resistance, the output resistance of Q2 is
lowered from 280 k$\Omega$ to approximately 1 k$\Omega$ and therefore
the gain drops to \nonumber
\begin{eqnarray}
\nonumber
A_{\rm cm}^{\rm se} &=& - \frac{\alpha
  (1\; {\rm k}\Omega \| r_{\rm o5} \| r_{\rm o2} ) }{2r_{\rm o3} + r_{\rm e}}
= \frac{1\cdot 1}{2\cdot 4700 + 0.03636} = - 1.06\cdot 10^{-4}\\
\nonumber
A_{\rm d}^{\rm se} &=& - \frac{\alpha (1\; {\rm k}\Omega \| r_{\rm o5}
\| r_{\rm 02})}{2r_{\rm e}}
= \frac{1\cdot 1000}{2\cdot 36.36} = -13.75\\
\nonumber {\rm CMRR}^{\rm se} &=& 3850
\end{eqnarray}
 
\vspace{2cm}

\noindent
\emph{\small Further information:\\
  Chapter 12 of ``Electronic Devices and Circuits'', T.\ F.\ Bogart,
  Jr., 4th edition, Prentice Hall, 1997.\\
  Chapter 6 of ``Microelectronic Circuits'', A.\ S.\ Sedra, K.\ C.\ 
  Smith, 4th edition, Oxford University Press, 1982.}


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