Now that we have learnt the impedances seen from a transistor's terminals, we can analyze the most important circuit configurations. They are quite simple, nonetheless they turn out to be very useful, because of their properties. We start with the
common emitter configuration, or as it is usually abbreviated, the
CE configuration.
Of course, we assume that the circuit is already polarized, and the large signal current flowing in the bipolar transistor is known. Furhtermore, we assume that the polarization grants a voltage of about \( \displaystyle 0.7 \text{V}=V_{\gamma} \) between the base collector and ground, so that the BJT is turned on. After that, we will have to check that it is working in the correct region. In fact, it may be whether in saturation (i.e. both BE and CB junctions turned on) or in active region (i.e. BE junction turned on, CB junction turned off).
In analogic applications, we never want the BJT transistor to be saturated, otherwise its parameters are subject to degradation. For example, in saturation it is no more true that \( \displaystyle i_c=\beta i_b \) , in fact the active CB junction injects electrons in the base, and the transistor is forced to satisfy \( \displaystyle i_c < \beta i_b \) .
So, assuming the BJT is in forward active region, we may ask ourselves a few questions:
1) What is the small signal gain of the transistor?
2) What is the input impedance of the circuit?
3) What is the output impedance of the circuit?
Let's answer with order.
1) This is straightforward. Actually, we already analyzed this configuration, studying the BJT impedances. The collector current is, of course,
\( \displaystyle i_c=g_m v_{be}=g_m v_{in} \)
and this is the same current flowing in the load resistor \( \displaystyle R_c \) , so the answer is
\( \displaystyle G_{CE}=\frac{v_{out}}{v_{in}}=-g_m R_c \)
The minus sign is due to the current flowing away from \( \displaystyle R_c \) , which holds for positive \( \displaystyle v_{in} \) . So,
CE is an inverting stage.
2) Again, straightforward. What could it be, other than
\( \displaystyle Z_{in}=\frac{\beta}{g_m}=r_{\pi} \) ?
3) This is not totally straightforward, but it is 98% straightforward. Collector impedance is infinite, but it has \( \displaystyle R_c \) in parallel, so...
\( \displaystyle Z_{out}=R_c \) .
That was easy, wasn't it?