DC Operating Point

Previously we only considered the relationship between the voltage and current at both the input and output ports of a transistor in either CB or CE configuration. Now we need to further find these voltages and currents when the transistor is connected to the rest components of a transistor circuit. Specifically, we treat the transistor as the load of a voltage source and a resistor, and find voltage and current at both the input and output ports.

Example: In the CE circuit shown below, $V_{CC}=12V$, $R_B=6 K\Omega$, $R_C=2 K\Omega$, $\beta=60$. The load line can be determined by two points:

• The open-circuit voltage: $(I_C=0,\;V_{CE}=V_{CC}=12\;V)$
• The short-circuit current: $(V_{CE}=0,\;I_C=V_{CC}/R_C=6\;mA)$

Find output voltage $V_{out}=V_{CE}$ when the input voltage $V_1$ takes each of the following values:

• $V_{BE}=V_{in}=0<0.7V$, $I_B=0$ and $I_C=\beta I_B=0$, $V_C=V_{CC}=12\;V$, the transistor is the cutoff region.

• $V_{in}=1V$. We assume $V_{BE}\approx 0.7\;V$, and get
  

  $$I_B = (V_{in}-V_{BE})/R_B=(1-0.7)/6=0.05\;mA$$

  $$I_C = \beta I_B=60\times 0.05\;mA=3\; mA$$

  $$V_{CE} = V_{CC}-I_C R_C=12\;V-3\;mA \times 2\;K\Omega=6\;V$$ (21)

  
  The transistor is in the linear region.

  

• $V_{in}=2V$. $I_B=(2-0.7)/6=0.22\;mA$, $I_C=\beta I_B=13\;mA$, and $V_{CE}=12\;V-13\;mA\times 2\;K\Omega=-14\;V$.

  This result is unreasonable and incorrect, because $I_B$ is so high that the transistor is no longer in the linear region as in the previous case, but it is in the saturation region, where the linear relationship $I_C=\beta I_B$ is no longer applicable, and the actual voltage $V_{CE}$ can be approximated to be about $V_{CE}=0.2V$, and the actual $I_C$ can be found to be $(V_{CC}-V_{CE})/R_C=(12-0.2)/2=5.9\;mA$.

  In general, when analyzing a transistor circuit we can first find $I_C=\beta\,I_B$, assuming this linear relationship holds. However, this assumption is invalid if $I_C$ exceeds the maximum current (the short-circuit current) $V_{CC}/R_C$, or the corresponding $V_{CE}$ is too low ($<0.2\,V$).

Summarizing the above, we see that the operation of a transistor can be in one of the three possible regions:

• Cutoff region:

  When $V_{BE}<0.7V$, or even negative, $I_B=0$, the output current is $I_C=I_{CE0} \approx 0$, $V_C=V_{CC}$, i.e., the transistor (between collector and emitter) is cut off (immediate above the horizontal axis of the output plot).

• Linear region:

  When $V_{BE}\approx 0.7V$, but $I_B>0$ is small enough so that the transistor is in the linear range where the collector current $I_C=\beta I_B < V_{CC}/R_C$ is proportional to base current $I_B$, and $V_C=V_{CC}-R_CI_C$. The CE transistor circuit in the linear region is widely used for amplification.

• Saturation region:

  When $V_{BE}$ is further increased $I_B$ and $I_B$ is also significantly increased (due to the exponential relationship between $I_B$ and $V_{BE}$), $\beta I_B > V_{CC}/R_C$, the linear relationship $I_C=\beta I_B$ no longer holds as $I_C$ approaches its maximum $V_{CC}/R_C$. The transistor is is saturated and $V_{CE}\approx 0.2V$, independent of $I_B$ (to the immediate right of the vertical axis of the output plot).

A typical CE circuit is shown in the figure below, where $I_E=I_B+I_C$, $V_{in}=V_{BE}=V_B$, and $V_{out}=V_{CE}=V_C$.

The DC steady-state operating condition of the CE transistor circuit, in terms of the currents and voltages $I_B$ and $V_{BE}$ of the input port, and $I_C$ and $V_{CE}$ of the output port, is called the DC operating point (Q-point), which iss determined by

• The external circuit including the voltage source $V_{CC}$ and $R_B$ and $R_C$ (as a non-ideal voltage soource represented by the linear load line);

• The nonlinear input and output characteristics of the transistor.

as the intersect of the two curves, as shown in the figures below.