This is another tutorial on BJT DC analysis. This tutorial is important from a design perspective. How to set circuit parameters, to work in an active mode, or in saturation mode or cut-off mode. For example, change in base and collector resistors (RC and RB) will result in changes in the base and collector currents. This change will have a direct impact on the operating mode of the transistor. The design problems are a little bit tricky. So, let's get started.
Example # 1: Calculate RB, RC and RE.
\[I_{CQ}=\frac{1}{2}I_{C(sat)}\]
\[I_{C(sat)}=8mA=4mA\]
\[V_C=18V\]
\[\beta=110\]
Apply KVL at the input loop.
\[-V_B+I_BR_B+V_{BE}+I_ER_E=0 \text { equation 1}\]
\[I_C=\beta I_B\]
\[I_B=36\mu A\]
\[I_E =I_C+I_B\]
\[I_E=4.036mA\]
Calculate RC.
\[ R_C=\frac{V_{CC}-V_C}{I_{CQ}}\]
\[R_C=2.5k\Omega\]
and RE set the Q point value and slope of the load line.
\[I_{C(sat)} = \frac{V_{CC}}{R_C+R_E}\]
\[R_C+R_E = \frac{28}{8m}=3.5k\Omega\]
\[ R_C = 3.5k-2.5k = 1k \Omega\]
Substitute RE in equation 1 and evaluate RB.
\[-28+36\mu R_B+0.7+4.036m*1k=0\]
\[R_B=646k\Omega\]
Example # 2: Calculate VCC, RC, RB
From load line,
\[V_{CC} = 20V\]
\[I_{C(sat)}=8mA\]
\[I_{BQ}=40 \mu A\]
\[I_{C(sat)}=\frac{V_{CC}}{R_C}\]
\[R_C=2.5k \Omega\]
For RB, apply KVL at the input loop.
\[-V_{CC}+I_B R_B+V_{BE}=0\]
\[-20+40 \mu R_B+0.7=0\]
\[R_B=482k \Omega \]
Example # 3: Design a common base NPN transistor with following parameters.
\[I_E=1.5mA\]
\[V_{CB} =7.5V\]
\[V_{CC}=15V\]
\[V_{CE}=-5V\]
Apply KVL at the input loop.
\[-V_{EE}+V_{BE}+I_ER_E=0\]
\[R_E= \frac{V_{EE}-V_{BE}}{I_E}\]
\[R_E=\frac{5-0.7}{1.5m}=4.5k \Omega\]
KVL at the output loop.
Keep in mind that common base configuration has unity current gain and hence IC = IE.
\[-V_{CC}+V_{CB}+I_CR_C=0\]
\[R_C= \frac{V_{CC}-V_{CB}}{I_C}\]
\[R_C=\frac{15-7.5}{1.5m}=5k \Omega\]
Example # 4: Calculate RC & RE.
\[I_E=2mA\]
\[V_{CB} =9V\]
KVL at the input loop.
\[-V_{EE}+V_{BE}+I_ER_E=0\]
\[R_E= \frac{V_{EE}-V_{BE}}{I_E}\]
\[R_E=\frac{10-0.7}{2m}=4.7k \Omega\]
KVL at the output loop.
\[-V_{CC}+V_{CB}+I_CR_C=0\]
\[R_C= \frac{V_{CC}-V_{CB}}{I_C}\]
\[R_C=\frac{20-9}{2m}=5.5k \Omega\]
Example # 5: Calculate RB and RC.
Bias point values.
\[V_{CE}=6V\]
\[I_C=2mA\]
\[I_B=\frac{I_C}{\beta}=20\mu A\]
Find the possible bias point values for β = 50 - 150.
KVL at the input loop, and evaluate RB.
\[-V_{CC}+I_BR_B+0V_{BE}=0\]
\[R_B= \frac{V_{CC}-V_{BE}}{I_B}\]
\[R_B=\frac{12-0.7}{2 \mu}=565k \Omega \text { equation 1 }\]
KVL at the output loop, and evaluate RC.
\[-V_{CC}+I_CR_C+V_{CE}=0\text { equation 2 }\]
\[R_C= \frac{V_{CC}-V_{CE}}{I_C}\]
\[R_C=\frac{12-6}{2m}=3k \Omega\]
For β = 50 and β = 150: Calculate minimum and maximum IC and VCE.
From equation 1:
\[I_B=\frac{12-0.7}{565k}=20 \mu A\]
\[I_{C(min)}=\beta_{min}I_B\]
\[I_{C(min)}=50*20 \mu = 1mA\]
\[I_{C(max)}=\beta_{max}I_B\]
\[I_{C(max)}=150*20 \mu = 3mA\]
From equation 2:
\[V_{CE(min)}=V_{CC}-I_{C(max)}R_C\]
\[V_{CE(min)}=12-3m*3k=3V\]
\[V_{CE(max)}=V_{CC}-I_{C(min)}R_C\]
\[V_{CE(max)}=12-1m*3k=9V\]
IC may vary from 1mA to 3mA.
VCE may vary from 3V to 9V.
Example # 6: Find RB and RC.
\[V_{CE}=5V\]
\[I_C=1mA\]
\[V_{CC}=15V\]
\[\beta = 100\]
\[I_B=10 \mu A\]
Find the possible bias point values for β = 30 - 150.
KVL at the input loop, and evaluate RB.
\[-V_{CC}+I_BR_B+0V_{BE}=0\]
\[R_B= \frac{V_{BB}-V_{BE}}{I_B}\]
\[R_B=\frac{15-0.7}{ 10 \mu}=1.4M \Omega \text { equation 1 }\]
KVL at the output loop, and evaluate RC.
\[-V_{CC}+I_CR_C+V_{CE}=0\text { equation 2 }\]
\[R_C= \frac{V_{CC}-V_{CE}}{I_C}\]
\[R_C=\frac{15-5}{1m}=10k \Omega\]
For β = 30 and β = 150: Calculate minimum and maximum IC and VCE.
From equation 1:
\[I_B=\frac{15-0.7}{1.4M}=10 \mu A\]
\[I_{C(min)}=\beta_{min}I_B\]
\[I_{C(min)}=30*10 \mu = 0.3mA\]
\[I_{C(max)}=\beta_{max}I_B\]
\[I_{C(max)}=150*10 \mu = 1.5mA\]
From equation 2:
\[V_{CE(min)}=V_{CC}-I_{C(max)}R_C\]
\[V_{CE(min)}=15-1.5m*10k=0V\]
\[V_{CE(max)}=V_{CC}-I_{C(min)}R_C\]
\[V_{CE(max)}=15-0.3m*10k=12V\]
IC may vary from 0.3 mA to 1.5 mA.
VCE may vary from 0V to 12V.
Example # 7:
Bias point values VCE and IE are given.
\[V_{CE}=16V\]
\[I_E=4mA\]
\[V_{CC}=24V\]
\[\beta = 100\]
\[I_B=\frac{I_E}{\beta} =0.4 \mu A\]
Apply KVL at the output loop, and evaluate RE.
\[-V_{CC}+V_{CE}+I_ER_E=0\]
\[R_E=\frac{V_{CC}-V_{CE}}{I_E}\]
\[R_E=\frac{24-16}{4m}=2k \Omega\]
Apply KVL at the input loop, and evaluate RB.
\[-V_B+I_BR_B+V_{BE}+I_ER_E=0\]
\[R_B=\frac{V_B-V_{BE}-I_ER_E}{I_B}\]
\[R_B=\frac{24-0.7-4m*2k}{0.4 \mu}=382.5k \Omega\]
Example # 8:
Bias point values VCE and IE are given.
\[V_{CE}=12V\]
\[I_E=10mA\]
\[V_{CC}=30V\]
\[\beta = 100\]
\[I_B=\frac{I_E}{\beta} = 100 \mu A\]
Apply KVL at the output loop, and evaluate RE.
\[-V_{CC}+V_{CE}+I_ER_E=0\]
\[R_E=\frac{V_{CC}-V_{CE}}{I_E}\]
\[R_E=\frac{30-12}{10m}=1.8k \Omega\]
Apply KVL at the input loop, and evaluate RB.
\[-V_B+I_BR_B+V_{BE}+I_ER_E=0\]
\[R_B=\frac{V_B-V_{BE}-I_ER_E}{I_B}\]
\[R_B=\frac{30-0.7-10m*1.8k}{100\mu}=113k \Omega\]
Example # 9:
\[V_C=5V\]
\[I_C=I_E=2mA\]
Apply KVL at the input loop, and evaluate RE.
\[V_{BE}+I_ER_E-V_{EE}=0\]
\[R_E=\frac{V_{EE}-V_{BE}}{I_E}\]
\[R_E=\frac{15-0.7}{2m}=7k \Omega\]
To determine RC, find voltage across RC:
\[I_CR_C=V_{CC}-V_C\]
\[R_C=\frac{V_{CC}-V_C}{I_C}=\frac{15-5}{2m}=5k \Omega\]
Example # 10: Find Out the highest possible voltage at the base for which the transistor is in active mode.
Maximum base voltage means the base voltage at which collector current is maximum. But it should remain in active mode.
Saturation occurs at VCE > 0.2V. Let's calculate maximum collector current just before the saturation mode. At this point, VCE = 0.3V.
Apply KVL at the output loop and figure out the collector current IC.
\[-V_{CC}+I_CR_C+V_{CE}+I_ER_E=0\]
\[I_C=I_E\]
\[-10+I_C(R_C+R_E)+V_{CE}=0\]
\[I_C=\frac{10-0.3}{8k}=1.2 mA\]
Apply KVL at the output loop and figure out the base voltage VB.
\[-V_B+V_{BE}+I_ER_E=0\]
\[V_B=4.7V\]
Example # 11: Find Out RE and RC.
\[V_C=5V\]
\[I_C=I_E=0.5mA\]
\[V_{CB}=2V\]
Reverse biased voltage at base collector junction.
Apply KVL at the output loop and figure out the base voltage RE.
\[-V_B+V_{BE}+I_ER_E=0\]
\[R_E=6.6k \Omega\]
As you know,
\[V_{CB}=V_C-V_B\]
\[V_C=4+2=6V\]
For RC,
\[R_C=\frac{V_{CC}-V_C}{I_C}\]
\[R_C=\frac{10-6}{0.5m}=8k \Omega\]
Example # 12: Find the value of RC to which the transistor remains in active mode.
A PNP transistor is given. Apply KVL at the input loop, and evaluate IE.
\[-10+I_ER_E+V_{EB}=0\]
\[I_E=\frac{10-0.7}{2k}=4.6mA\]
VEC<0.3V, this condition is for saturation mode. The transistor remains in active mode, to fulfill this requirement, suppose VEC=0.3V.
Apply KVL at the output loop and evaluate RC.
\[-10+I_CR_C+V_CE+I_ER_E=0\]
\[I_C=I_E\]
\[R_C=4.65k \Omega\]
Example # 13: Determine RC and RE
\[I_C=I_E=1mA\]
A PNP transistor is given. Apply KVL at the input loop, and evaluate RE.
\[-10+I_ER_E+V_{EB}=0\]
\[R_E=\frac{10-0.7}{1m}=9.3k \Omega\]
In this example, VC (that is collector voltage is given.
\[V_C=V_{CC}+I_CR_C\]
\[-4=-10+1m*R_C\]
\[\text { Solve for }R_C\]
\[R_C=6k \Omega\]
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