Biasing

As shown before, the DC operating point of a transistor amplification circuit needs to be set up properly (in the middle of the linear region) to avoid signal distortion. We now consider how the operating point is determined by the biasing circuit, in terms of $R_B$, $R_C$, and $V_{CC}$.

Fixed Biasing

transistorbiasinga.gif

By properly setting the voltage $V_{CC}$ (not too low) and $R_B$ (not too large), the voltage $V_{BE}$ can be approximated as a constant value of $0.7V$, as shown in the input characteristic plot:

fixedbias2.gif

Then the base current can be estimated to be:

$\displaystyle I_B=\frac{V_{CC}-V_{BE}}{R_B} %\approx \frac{V_{CC}}{R_B}
$ (29)

The collector current is $\beta$ times $I_B$ if the transistor is in linear region:

$\displaystyle I_C=\beta I_B=\beta \frac{V_{CC}-V_{BE}}{R_B} %\approx \beta \frac{V_{CC}}{R_B}
$ (30)

The output voltage is

$\displaystyle V_{CE}=V_{CC}-I_C R_C$ (31)

As both $I_C$ and $V_{CE}$ depend on $\beta$, which may differ for different transistors and change depending on the temperatures the operating point may be unstable and inconsistent.

Example 1

In the fixed biasing transistor circuit shown above, $R_B=200\,k\Omega$, $\beta=100$, $V_{CC}=12\,V$, find $R_C$ so that the DC operating point is in the middle of the linear region of the output characteristic plot, i.e., $V_{CE}=V_{CC}/2=6\,V$. We assume $V_{BE}=0.6\;V$ (may not be valid if $R_B$ is too large) and get

$\displaystyle I_B=\frac{V_{CC}-V_{BE}}{R_B}=\frac{12-0.6}{200}=0.057\,mA,\;\;\;\;\;
I_C=\beta I_B=5.7\,mA$ (32)

For this DC operating point with $I_C=5.6\,mA$ to be in the middle of the load line, we need to have

$\displaystyle V_{CE}=V_{CC}-I_CR_C=12-5.7 R_C=6\,V,\;\;\;\;\;$i.e.$\displaystyle \;\;\;\;\;\;
R_C=\frac{6\,V}{5.6\,mA}\approx 1.05\,k\Omega$ (33)

Alternatively, we can also require the short-circuit current $V_{CC}/R_C$ to be twice of $I_C=5.7\,mA$ at the DC operating point:

$\displaystyle \frac{V_{CC}}{R_C}=\frac{12\,V}{R_C}=2\times I_C=11.4\;mA,\;\;\;\;\;$i.e.$\displaystyle \;\;\;\;\;R_C=\frac{12\,V}{11.4\,mA}\approx 1.05\,k\Omega$ (34)

so that

$\displaystyle V_{CE}=V_{CC}-R_CI_C=12\,V-1.05\,k\Omega\times 5.7\,mA \approx 6\,V$ (35)

is indeed in the middle of the load line.

Example 2:

In the same circuit above, $V_{CC}=12V$, $R_B=200\,k\Omega$, $R_C=1\,k\Omega$. Find the operating point $(V_{CE},\;I_C)$ for $\beta=50,\;100,\; 200$.

The load line $V_C=V_{CC}-I_C R_C$ is determined by these two points:

Same as before, we have

$\displaystyle I_B=\frac{V_{CC}-V_{BE}}{R_B}=\frac{12-0.6}{200\times 10^3}=0.057 \;mA$ (36)

We then find the folllowing for each of the three $\beta$ values:

\begin{displaymath}\begin{array}{c\vert\vert c\vert c\vert c}\hline
& \beta=50 &...
...V_C=V_{CC}-R_CI_C\;(V) & 9.15 & 6.3 & 0.6 \\ \hline
\end{array}\end{displaymath} (37)

To minimize distortion, the DC operating point needs to be in the middle of the load line at $(V_C=12/2=6\,V,\;I_C=12/2=6\,mA)$. But in this case, we see that

The DC operating point of this fixed biasing circuit is not completely determined by the parameters of the circuit such as the resistors, as it is also directly affected by factors such as $\beta$ value and temperature. This situation can be improved by introducing negative feedback into the circuit.

Self-Biasing

To correct the problem above, the self-biasing circuit shown below can be used to decrease the effect of changing $\beta$ by negative feed back due to the introduction of $R_E$.

transistorbiasingb.gif

Qualitatively, an increased $I_C$ (caused by reasons such as increased $\beta$ due to temperature change) will cause the following to happen:

$\displaystyle I_C \uparrow \Longrightarrow I_e=I_C+I_B \uparrow
\Longrightarrow...
...wnarrow \Longrightarrow I_B \downarrow \Longrightarrow
I_C=\beta I_B \downarrow$ (38)

This is a negative feedback loop that tends to stabilize the operating point.

Quantitatively, we can further carry out analysis of the circuit:

Example 3:

In the circuit of self-biasing, $V_{CC}=12V$, $R_1=100\,k\Omega$, $R_2=36\,k\Omega$, $R_E=1\,k\Omega$, $R_C=2\,k\Omega$, Assume $V_{BE}=0.7$. The load line is determined by this equation:

$\displaystyle V_{CE}=V_{CC}-I_CR_C-I_ER_E\approx V_{CC}-I_C(R_C+R_E)$ (46)

determined by these two points at:

To minimize distortion, the desired operating point should be in the middle of the load line at $V_{CE}=12\,V/2=6V$ and $I_C=4\,mA/2=2\,mA$.

We see that in all three cases, $I_C\approx 2\;mA$, $V_{CE}\approx 6\,V$, i.e., the DC operating point is always close to the middle of the load line.

Example 4

In a self-biasing transistor circuit, $R_1=R_2=100\,k\Omega$, $R_E=2\,k\Omega$, $\beta=100$, $V_{CC}=12V$, find $R_C$ so that the DC operating point is in the middle of the linear region of the output characteristic plot.

We first convert the base circuit into its Thevenin's equivalent voltage source composed of

$\displaystyle R_{Th}=R_B=R_1\vert\vert R_2=50\;k\Omega,\;\;\;\;\;\;\;
V_{Th}=V_B=V_{CC}\frac{R_2}{R_1+R_2}=6\,V$ (50)

Then we get

$\displaystyle I_C=\beta I_B=\frac{\beta(V_B-V_{BE})}{(\beta+1)R_E+R_B}
=\frac{100\times(6-0.7)}{101\times 2+50}=\frac{530}{250}=2.12$ (51)

and

$\displaystyle V_{CE}=V_{CC}-(I_C+I_B) R_E-I_CR_C\approx V_{CC}-I_C(R_E+R_C)
=12-2.12\times (2+R_C)$ (52)

To set this DC operating point to be in the middle of the load line, we need $V_{CE}=12\,V/2=6\,V$, and solving the equation we get $R_C=0.83\,k\Omega$.

Example 5

The circuit below shows yet another way to introduce feedback to stablize the DC operating point.

Example5.gif