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

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:

The DC operating point in terms of $I_C$ and $V_{CE}$ can be found in the following steps:

• Find the base current:

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

• Find the collector current:

  If the transistor is in linear region, then

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

• Find the output voltage

  $$V_C=V_{CE}=V_{CC}-I_C R_C$$ (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 $V_{BE}=0.7\;V$ (may not be valid if $R_B$ is too large) and get

$$I_B=\frac{V_{CC}-V_{BE}}{R_B}=\frac{12-0.7}{200}=0.057\,mA,\;\;\;\;\; I_C=\beta I_B=5.7\,mA$$ (32)

For this DC operating point with $I_C=5.7\,mA$ to be in the middle of the load line, we need to have

$$V_{CE}=V_{CC}-I_CR_C=12-5.7 R_C=6\,V,\;\;\;\;\; \;\;\;\;\;\; R_C=\frac{6\,V}{5.7\,mA}\approx 1.05\,k\Omega$$ (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:

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

so that

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

• Open-circuit voltage: $I_C=0,\; V_C=V_{CC}=12\,V$
• Short-circuit current: $V_C=0,\; I_C=V_C/R_C=12\,V/1\,k\Omega=12\,mA$

Same as before, we have

$$I_B=\frac{V_{CC}-V_{BE}}{R_B}=\frac{12-0.7}{200\times 10^3}=0.057 \;mA$$ (36)

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

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

• $\beta=50$, too close to cutoff region.
• $\beta=100$, in the middle of linear region as desired.
• $\beta=200$, too close to the saturation region.

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 $R_E$.

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

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

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

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

• If the resistances of $R_1$ and $R_2$ are small so that the current $I_2$ through $R_2$ is much larger than the base current $I_B$ i.e., $I_2\gg I_B$, then the basis voltage $V_B$ can be approximated to be (voltage divider):

  $$V_B = V_{CC} \;\frac{R_2}{R_1+R_2}$$ (39)

  Applying KVL to the base-emitter loop of the circuit, we get

  $$V_B-V_{BE}-I_ER_E=V_B-V_{BE}-(I_C+I_B)R_E=V_B-V_{BE}-(\beta+1)I_BR_E=0$$ (40)

  Solving for $I_B$, we get

  $$I_B=\frac{V_B-V_{BE}}{(\beta+1)R_E},\;\;\;\;\;\;\;\;\;\;\;\;\; I_... ...beta(V_B-V_{BE})}{(\beta+1)R_E} \approx \frac{V_B-V_{BE}}{R_E} =\frac{V_E}{R_E}$$ (41)

  Note that $I_C$ is independent of $\beta$, as it is completely determined by $R_1$, $R_2$, and $R_E$, as well as $V_{CC}$.

• If the condition $I_B \ll I_2$ is not satisfied, the method above is no longer valid. In this case, we can use Thevenin's theorem to replace the base circuit by an open circuit voltage $V_T=V_B$ (already found above), in series with the internal resistance $R_T$:

  $$R_T=R_B=R_1\vert\vert R_2=\frac{R_1R_2}{R_1+R_2}$$ (42)

  

  Applying KVL to the base loop we get

  $$V_B-I_BR_B-V_{BE}-(I_C+I_B)R_E=V_B-V_{BE}-I_BR_B-(\beta+1)I_BR_E=0$$ (43)

  Solving this equation for $I_B$, we get:

  $$I_B=\frac{V_B-V_{BE}}{(\beta+1) R_E+R_B},\;\;\;\;\; \;\;\;\;\;\; I_C =\beta I_B=\frac{\beta(V_B-V_{BE})}{(\beta+1) R_E+R_B}$$ (44)

  If $R_E$ is not too small (strong negative feedback effect) and $R_B=R_1\vert\vert R_2$ is not too large (approximately a voltage divider), so that $\beta R_E \gg R_B=R_1\vert\vert R_2$ (e.g., $\beta R_E \ge 10 R_B$ even for a small $\beta$), then $I_C$ can be approximated as

  $$I_C \approx \frac{\beta(V_B-V_{BE})}{(\beta+1)R_E} \approx \frac{V_B-V_{BE}}{R_E} =\frac{V_E}{R_E}$$ (45)

  which is the same as what we got previously. In this case, $I_C$, and thereby $V_{CE}$ and the DC operating point, is determined only by the resistors of the circuit, independent of the $\beta$. 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.

  For this approximation above to be valid, we desire to have smaller $R_B=R_1\vert\vert R_2$ so that $V_B$ is less affected by $I_B$, and large $R_E$ for stronger negative feedback. However, as the voltage gain of the circuit will be reduced due to the negative feedback, $R_E$ cannot be too large.

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:

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

determined by these two points at:

• $V_{CE}=0$ and $I_C=V_{CC}/(R_C+R_E)=12/(2+1)=4\,mA$ (short-circuit current)
• $I_C=0\,mA$ and $V_{CE}=12V$ (open-circuit voltage)

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

• Based on the voltage divider approximation, we get the DC operating point independent of $\beta$:
  

  $$R_B = R_1\vert\vert R_2=\frac{R_1R_2}{R_1+R_2}=26.5\,k\Omega$$

  $$V_B \approx V_{CC} \frac{R_2}{R_1+R_2}=3.18\;V$$

  $$V_E = V_B-V_{BE}=3.18-0.7=2.48\;V$$

  $$I_C \approx I_E=\frac{V_E}{R_E}=2.48\;mA$$

  $$V_C = V_{CC}-R_CI_C=12-2\times 2.48 = 7.04\;V$$

  $$V_{CE} = V_C-V_E=7.04-2.48=4.56\;V$$ (47)

  
  

• Based on the Thevenin theorem, we get more accurate results:
  

  $$I_B = \frac{V_B-V_{BE}}{(\beta+1) R_E+R_B}$$

  $$I_C = \beta I_B$$

  $$V_E = I_E R_E=(I_C+I_B)R_E$$

  $$V_B = V_E+V_{BE}$$

  $$V_C = V_{CC}-I_CR_C$$

  $$V_{CE} = V_C-V_E$$ (48)

  
  The DC operating points corresponding to the three $\beta$ values can be found to be:

  $$\begin{array}{c\vert\vert c\vert c\vert c}\hline & \beta=50 &... ...\ \hline V_{CE}\;(V) & 7.17 & 6.15 & 5.46 \\ \hline \end{array}$$ (49)

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

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

Then we get

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

and

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

• The resistor $R_B$ connecting the collector to the base forms a feedback from the output $V_C$ to $V_B$ as well as providing the forward baising needed for the base-emitter PN junction:

  $$I_C \uparrow \Longrightarrow V_C=V_{CC}-R_CI_C \downarrow \Longrightarrow I_B\downarrow \Longrightarrow I_C=\beta I_B \downarrow$$ (53)

• The DC operating point can be found by applying KVL:

  $$V_{CC}=R_C(I_B+I_C)+R_BI_B+V_{BE}=(R_C(\beta+1)+R_B)I_B+0.7$$ (54)

  Solving for $I_B$ we get

  $$I_B=\frac{V_{CC}-0.7}{(\beta+1)R_C+R_B},\;\;\;\;\; I_C=\beta I_B=\frac{\beta(V_{CC}-0.7)}{(\beta+1)R_C+R_B}$$ (55)

  and

  $$V_C=V_{CC}-(I_C+I_B)R_C=V_{CC}-(\beta+1)I_BR_C$$ (56)

  Note that if $R_B \ll (\beta+1)R_C$ (strong negative feedback), then $I_C \approx (V_{CC}-0.7)/R_C$, independent of $\beta$. Although typically $R_B$ is significantly greater than $R_C$, the negative feedback still has the tendency to reduce the affect of varying $\beta$ and thereby stablize the DC operating opint.

• Given $V_{CC}=10V$, $\beta=100$, and a desired $I_C=2\,mA$, find $R_C$ and $R_B$ so that the DC operating point is in the middle of the linear region.

  For the DC operating point with $I_C=2\,mA$ to be in the middle of the linear region, we need $V_C=V_{CC}/2=5V$, and $I_C=2\,mA$ to be half of the short-circuit current $I_{SC}=V_{CC}/R_C=10\,V/R_C=4\,mA$, we need
  

  $$R_C = 10/4=2.5\,K\Omega$$

  $$R_C = (V_{CC}-V_C)/I_C=(10-5)V/2\,mA=2.5\,K\Omega$$ (57)

  
  and

  $$I_B=I_C/\beta=0.02mA, \;\;\;\;\;R_B=(5-0.7)/0.02=4.3/0.02=215K\Omega$$ (58)

  $$I_C=\beta I_B,\;\;\;\;\;\;\;\;V_C=V_{CC}-I_ER_C=V_{CC}-(\beta+1)I_BR_C$$ (59)

• Following the equations given above, we can find the DC operating pointt for each of the three $\beta$ values, which are all approximately in the middle of the linear region:

  $$\begin{array}{c\vert\vert c\vert c\vert c}\hline & \beta=50 &... ... & 2.6 \\ \hline V_C\;(V) & 6.6 & 5 & 3.5 \\ \hline \end{array}$$ (60)

As discussed above, to avoid distortion of the AC signal, we desire that the DC operating point is in the middle of the linear region of the output characterisc, i.e., $V_{CE}=V_{oc}/2=V_{CC}/2$ and $I_C=I_{SC}/2=V_{CC}/2R_C$. In this case, the power consumption is $P=I_C V_{CE}=V^2_{CC}/4R_C$, which is the maximum among all possible DC operating points on the load line:

$$P(I_C) = V_{CE}I_C=(V_{CC}-R_CI_C)I_C=V_{CC}I_C-R_CI_C^2$$

$$\frac{d}{dI_C}P(I_C) = V_{CC}-2R_CI_C=0$$ (61)

Solving the second equation we get $I_C=V_{CC}/2R_C$, and

$$P_{max}=V_{CC}I_C-R_CI_C^2=\frac{V_{CC}^2}{4R_C}$$ (62)

The maximum power consumption is the same as that corresponding to the DC operating point in the middle of the load line.