DC 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.

• Fixed Current Biasing This is the simplest biasing.

  

  The base current $I_B$ can be properly set by selecting the voltage $V_{BB}$ (usually $V_{BB}=V_{CC}$) and $R_B$, as the load-line shown in the input characteristic plot:

  

  Or, the voltage $V_{BE}$ is usually estimated to be $0.7V$, much smaller compared to $V_{BB}$ (>10V), the base current can also be estimated to be:
  

  $$I_B=\frac{V_{CC}-V_{BE}}{R_B} \approx \frac{V_{CC}}{R_B}$$
  
  
  The collector current is
  
  $$I_C=\beta I_B+(\beta+1)I_{CB0}=\beta I_B+I_{CE0} \approx \beta \frac{V_{CC}-V_{BE}}{R_B} \approx \beta \frac{V_{CC}}{R_B}$$
  
  
  which is directly proportional to $\beta$. The output voltage is
  
  $$V_{CE}=V_{CC}-I_C R_C$$
  
  

  As $I_C$ and $V_{CE}$ depend on $\beta$, which is different for different transistors and changes as a function of temperature, the operating point is unstable and inconsistent.

  Example 1: In the circuit of fixed current biasing, $V_{CC}=12V$, $R_B=560K\Omega$, $R_C=3K\Omega$.

  The load line is determined by this equation:
  

  $$V_{CE}=V_{CC}-I_C(R_C+R_E)$$
  
  
  corresponding to these two points at $(V_{CE}=0,\;I_C=12/3=4 mA)$ and $(V_{CE}=12V,\;I_C=0 mA)$. For minimum distortion, the desired operating point is in the middle of the load line at $(V_{CE}=6V,\;I_C=2 mA$).

  Assume $V_{BE}=0.7\;V$ Find the operating points $(V_{CE},\;I_C)$ for $\beta=20,\;100,\; 200$.

  We first find $I_B$:
  

  $$I_B=\frac{V_{CC}-V_{BE}}{R_B}=\frac{12-0.7}{560\times 10^3}=0.02 mA$$
  
  
  • When $\beta=20$,
    
    $$\begin{array}{l} I_C=\beta I_B=20\times 0.02=0.4 \;mA \\ V_C=V_{CC}-R_C I_C=12-3\times 0.4=10.8 V\end{array}$$
    
    
    The operating point is too close to cut-off region.

  • When $\beta=100$,
    
    $$\begin{array}{l} I_C=\beta I_B=100\times 0.02=2 \;mA \\ V_C=V_{CC}-R_C I_C=12-3\times 2=6 V \end{array}$$
    
    
    The operating point is in the middle of linear region, as desired.

  • When $\beta=200$,
    
    $$\begin{array}{l} I_C=\beta I_B=200\times 0.02=4 \;mA \\ V_C=V_{CC}-R_C I_C=12-3\times 4=0 V \end{array}$$
    
    
    In this case $I_C$ can no longer be determined by $I_C=\beta I_B$, as the operating point is in the saturation region with $V_{CE}=0.2$ and the maximum $I_C$ is
    
    $$I_C=\frac{(V_{CC}-V_{CE})}{R_C}=\frac{12-0.2}{3}=3.7\;mA$$
    
    

• Self-Biasing

  

  To correct the problem above, self-biasing circuit is used to decrease the effect of changing $\beta$ by negative feed back due to the introduction of $R_E$. Qualitatively, if $I_C$ is increased due to increased $\beta$ or temperature, the following happens:
  

  $$I_C \uparrow \Longrightarrow V_E \uparrow \Longrightarrow V_{... ...rrow I_B \downarrow \Longrightarrow I_C=\beta I_B \downarrow$$
  
  
  This is a negative feedback loop which tends to stabilize the operating point.

  To analyze this circuit quantitatively, we first find the base voltage $V_B$ and base current $I_B$. Note that only when the base current is much smaller than the current through $R_2$ ($I_B \ll I_2$), can we approximate $V_B$ by voltage divider as:
  

  $$V_B=V_{CC} \frac{R_2}{R_1+R_2}$$
  
  
  Applying KVL to the base-emitter loop of the circuit, we get:
  
  $$V_B-V_{BE}-(\beta+1)R_E=0$$
  
  
  and
  
  $$I_B=\frac{V_B-V_{BE}}{(\beta+1)R_E},\;\;\;\;\;\;\;\;\;\;\;\;\; I_C=\beta I_B=\frac{\beta(V_B-V_{BE})}{(\beta+1)R_E}$$
  
  
  If the condition $I_B \ll I_2$ is not satisfied, we have to use Thevenin's theorem to replace the base circuit by a voltage source
  
  $$V_{BB}=\frac{R_2}{R_1+R_2} V_{CC}$$
  
  
  in series with a resistance
  
  $$R_B=\frac{R_1R_2}{R_1+R_2}$$
  
  

  

  Next we use KVL to the base loop to get
  

  $$V_{BB}-I_BR_B-V_{BE}-(I_C+I_B)R_E=V_{BB}-V_{BE}-I_BR_B-(\beta+1)I_BR_E$$
  
  
  where we have used $I_C=\beta I_B$. Solving this, we get $I_B$ and also $I_C=\beta I_B$:
  
  $$I_B=\frac{V_{BB}-V_{BE}}{(\beta+1) R_E+R_B},\;\;\;\;\; I_C =\beta I_B=\frac{\beta(V_{BB}-V_{BE})}{(\beta+1) R_E+R_B}$$
  
  
  If we could assume that even for the minimum possible $\beta$, it is still true that $R_B \ll \beta_{min} R_E$, e.g., $R_B =0.1\times \beta_{min} R_E$, then $I_C$ can be approximated as
  
  $$I_C \approx \frac{V_{BB}-V_{BE}}{R_E}$$
  
  
  i.e., $I_C$, and thereby $V_{CE}$ and the DC operating point is determined only by the resistors of the circuit, independent of the $\beta$ value which may change for different transistors or at different temperatures. Comparing this with fixed biasing with $I_C \approx \beta (V_{CC}-V_{BE})/R_B$ directly proportional to $\beta$, the self-biasing circuit has a much more stable operating point.

  Example 2: In the circuit of self-biasing, $V_{CC}=28V$, $R_1=90 K\Omega$, $R_2=10 K\Omega$, $R_E=2 K\Omega$, $R_C=14 K\Omega$, Assume $V_{BE}=0.7$.

  The load line is determined by this equation:
  

  $$V_{CE}=V_{CC}-I_C(R_C+R_E)$$
  
  
  corresponding to these two points at $(V_{CE}=0,\;I_C=28/16=1.75 mA)$ and $(V_{CE}=28V,\;I_C=0 mA)$. For minimum distortion, the desired operating point is in the middle of the load line at $(V_{CE}=14V,\;I_C=1.9 mA$).

  Find the operating points for $\beta=20, 100, 200$.
  

  $$\begin{array}{l} R_B=R_1R_2/(R_1+R_2)=10\times 90/(10+90)=9 K... ..._C=2\times 1.05=2.1\;V \\ V_{CE}=V_C-V_E=11.2\;V \end{array}$$
  
  
  Alternatively, these currents and voltages can be found more accurately for each $\beta$ value:
  • When $\beta=20$,
    
    $$I_C=\beta I_B=\beta \frac{V_{BB}-V_{BE}}{(\beta+1) R_E+R_B} =\frac{20\times (2.8-0.7)}{21\times 2+9} \approx 0.82 mA$$
    
    
    
    
    $$\begin{array}{l} V_C=V_{CC}-I_CR_C=28-0.82 \times 14=16.5 V... ...2K\Omega=1.64V \\ V_{CE}=V_C-V_E=16.5-1.64=14.8V \end{array}$$
    
    

  • When $\beta=100$,
    
    $$I_C=\beta I_B=\beta \frac{V_{BB}-V_{BE}}{(\beta+1) R_E+R_B} =\frac{100\times (2.8-0.7)}{101\times 2+9} \approx 1 mA$$
    
    
    
    
    $$\begin{array}{l} V_C=V_{CC}-I_CR_C=28-1 \times 14=14 V \\ ... ...prox 1mA\;2K\Omega=2V \\ V_{CE}=V_C-V_E=14-2=12V \end{array}$$
    
    

  • When $\beta=200$,
    
    $$I_C=\beta I_B=\beta\frac{V_{BB}-V_{BE}}{(\beta+1) R_E+R_B} =\frac{200\times (2.8-0.7)}{201\times 2+9} \approx 1.02 mA$$
    
    
    
    
    $$\begin{array}{l} V_C=V_{CC}-I_CR_C=28-1.02 \times 14=13.7 V... ...02mA\;2K\Omega=2V \\ V_{CE}=V_C-V_E=13.7-2=11.7V \end{array}$$