By looking through some references I managed to find a really elegant derivation of the gain of a differential amplifier...I never thought the hyperbolic tangent function would show up here! I'll write it out since I need practice with LaTeX and it might help someone who is working through the same material.
I_e = e^\frac{Vbe}{V_t}
So
V_{be} = V_t ln I_e
V_{dif} = V_{be1} - V_{be2} = V_t ln I_{e1} - V_t ln I_{e2} = <br />
Vt ln \frac{I_e1}{I_e2}So
e^\frac{V_{dif}}{V_t} = \frac{I_e1}{I_e2}
\frac{I_{c1}}{I_{c2}} = \frac{\alpha I_{e1}}{\alpha I_{e2}} = \frac{I_{e1}}{I_{e2}}
The differential current ratio is:
\frac{I_{c1} - I_{c2}}{I_{c1} + I_{c2}} = \frac{e^\frac{V_{dif}}{V_t} - 1}{e^\frac{V_{dif}}{V_t} + 1} = tanh(\frac{V_{diff}}{2V_t})
So by multiplying the differential current ratio top and bottom by the collector load resistance we can do this:
\frac{R_c(I_{c1} - I_{c2})}{Rc(I_{c1} + I_{c2})} = \frac{V_{o1} - V_{o2}}{Rc(I_{c1} + I_{c2})} = \frac{V_{od}}{R_c \alpha I_o}
Where Io is the current through both emitters of the differential pair and alpha is the common base current gain.
Setting the two equations equal we finally have:
\frac{V_{od}}{R_c \alpha I_o} = tanh(\frac{V_{dif}}{2V_t})
V_{od} = R_c \alpha I_o tanh(\frac{V_{dif}}{2V_t}).
For small signals, tanh(\frac{V_{dif}}{2V_t}) \approx \frac{V_{dif}}{2V_t} and alpha can be taken to be 1, so we get for small signal gain:
V_{od} = R_c\frac{I_o}{2V_t}.
For large signals, the transfer function of the amplifier behaves just like the hyperbolic tangent function: it's linear in a small region around the quiescent point, but goes asymptotic as the voltage increases or decreases beyond this linear region and the input transistor saturates or goes into cutoff.
